Physics II exam 1

*experimentally, R =*

*ρ (L/A)* - L is length of wire - A is cross-section area - ρ is called is resistivity (material) unit Ω • m can calculate it, but usually given -longer = more resistance -bigger area = bigger resistance

*1 electron (q = - e = - 1.6 x 10^-19 C) moving through a potential difference of 1 V gains/loses*

1.6 x 10^-19 J of potential energy (from/into kinetic energy)

*U =*

1/2 QV = 1/2 (EAε0) (Ed) = 1/2 ε0 E^2 Ad

*I =*

1/R V *IR = V*

*F = |q|vBsinθ ->*

F = It (L/t) Bsinθ --> (because q = It and v = L/t) *F = I L Bsinθ* -depends on current in wire -depends on length of wire exposed to magnetic field -proportional to the magnetic field -proportional to the sine of angle between I and B

*Given E, we can calculate the force on a charge q0*

F = q0E (if q0 in negative, F and E have opposite directions)

Responsible for weight

F/m = G(Mm/r^2) (1/m) = G(M/r^2)

*E =*

F/q q positive = same direction q negative = opposite direction -for electric field, force was in direction of the field or opposite

B perpendicular to current 2

F21 = I2 L2 B1

Force on wire 2 from wire 1

F21 = I2 L2 B1 sinθ

In order to remain in circular motion:

FB=Fcp ∣q∣v B sin90 = m (v^2/r)

Charged Conductors

Electrons move freely in conductors -Repulsive force between like charges = charges on the surface = Electric field on surface = No electric field inside (no electron movement, E = 0)

Resistors come with color code

Example: R = 47 x 1000 Ω +/- 5% = 47 kΩ +/- 5% -this color chart on slide 8 will be helpful for lab (not needed for class exams)

*Electromotive Force*

No force! Electric potential difference of battery under ideal conditions -Flow of current reduces electric potential difference (we use it anyway)

small sphere: V =

kQ/(R/2) = kσ'(4πR^2/4) / (R/2) = 2πkσ'R electric field (σ) off the small sphere is twice as strong because more charge density

large sphere: V =

kQ/R = kσ(4πR^2) / R = 4πkσR

A wire with a current of 2.9 A is at an angle of 37.0º relative to a magnetic field of 0.77 T Find the force exerted on a 2.35 −m length of the wire.

F = I L Bsinθ F = (2.9) (2.35) (0.77) sin37 *F = 3.2 N*

Large current means

large magnetic field

*Negative charges accelerate towards*

larger V

*Voltmeters measure*

voltage -to measure V in circuit you have to plug in the voltmeter in parallel with what you want to measure

*When 2 opposite charges come together*

cancels out wherever the lines cross (vector sum) -no electric field because -Electric field tangent to lines -add up lines (more lines = stronger field/# of lines proportional to magnitude of charge)

The current depends on

voltage and resistance (body)

σ =

Q/surface area = 4πr^2

*Magnetic Field*

"B" Extends out of magnet (like E extends out of charges) -Lines go around / inside magnet: north → south poles (*always a closed loop*)

Neutral atoms:

# protons = # electrons (same charge)

*Magnetic field right-hand rule: grasp the wire with your right hand so that your thumb points in the direction of the conventional current; then your fingers will encircle the wire in the direction of the magnetic field. How to calculate B?*

-turn on current = magnetic fields lines in a circle close to current -right hand pretends to grab axis pole and curve hands in the direction of the field -strongest current = stronger magnetic field generated

A 15-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.7 μF , 11 μF , and 30 μF . Find the voltage across the 30 μF capacitor.

1 / (1/4.7) + (1/11) + (1/30) = C C = 2.97 Q = CV -> Q = 44.55 V = Q/C V = 44.55 / 30 *V = 1.49*

10^9 eV =

1 GeV

10^6 eV =

1 MeV

1.6 x 10-19 J ≡

1 eV (electron volt)

10^3 eV =

1 keV

1 electron moving through a potential difference of 1000 V gains/loses 1000 eV =

1 keV (kiloelectron volt) of potential energy -Used for dealing with molecules and particles but not SI

When we are wet

R < 103 Ω

The circuit shown below(Figure 1)consists of four different resistors and a battery. You don't know the strength of the battery or the value any of the four resistances. 1.) Select the expressions that will be equal to the voltage of the battery in the circuit, where VA, for example, is the potential drop across resistor A.

VA+VB VA+VC VD

*Electric Potential due to Point Charges*

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Chapter 19:

Electric Charge, Insulators and Conductors

Chapter 21

Electric Current, Resistance, Ohm's Law and Electric Circuits

chapter 20

Electric Potential Energy, Electric Potential and Energy Conservation

Electric Hazards

Electric current stimulates nerves and muscles (shock) and heats and burns human body

*Electric potential unit*

V (volt) = 1 J/C

*E = 0:*

V = constant

*Electric Current*

amount of charge passing by cross section of wire in interval of time

*Electric potential difference =*

voltage

*For example, if the electric potential difference between two plates is 10 V, a +1 C charge moving from a to b gains:*

(1 C)(- 10 V) = - 10 J (lost electric potential energy)

Like B from magnet

(magnet made from orbiting electrons)

*Magnetic Force Formula*

*F = ∣q∣ v Bsinθ* If θ = 0 (v∥B): no force If θ = 90º (v ⊥ B): Fmax = qvB -parallel = 0º because sin0 = 0 -sin90 = 1

Net torque τ =

*I ABsinθ* with area = height * width

*E stronger where equipotential lines are*

*closer* E = -ΔV/d

*Superposition of Forces*

-Coulomb's law only describes the interaction of 2 charges -Electric force is a vector (magnitude + direction) -Net force of more than 2 charges: vector sum of forces between each pair F1 = F12 + F13 + F14 .....

The Heart

-Cells have net + charge on the exterior and net - charge on the interior -In one cell about 10^-8 C in each surface -Positive ions from exterior pass inside the wall and neutralize negative ions and reverse polarization -Depolarization moves in one direction on muscle -The muscle repolarizes again (all in less than 1 s) -The potential increases and decreases (1 cell) ions go in and out in order to reverse it

*Electric Charge*

-Convention -Charged is quantized

*ε0 =*

1/4πk = 8.85 x 10^-12 C2/(N m^2)

τ =

2 (IhBsinθ)(w/2) -lever arm = w/2 -2x toque because both sides -no force on the bottom and top (only the sides)

*k =*

9x10^9 Nm^2 /C^2 is huge!

*From calculus: ΔU*

= UA − UB = k q0Q/rA - k q0Q/rB (to bring q0 from B→ A)

*material (insulator) with K*

> 1 will help you

*Superposition of Electric Potential*

Algebraic sum ☺ with signs

Example 1.) 1 cm2 of membrane has a capacitance of 7 x 10^-7 F. If the potential difference across the membrane is 0.1 V, find the electrical energy stored in 1 cm^2 of membrane.

A = (10^-2 m)^2 C = 7 x 10^-7 F V = 0.1 V U = ? ---> U = 1/2 CV^2 U = 1/2 (7 x 10^-7) (0.1)^2 *U = 3.5 x 10^-9 J*

Even in relatively simple setups as in the figure shown, electric field lines are quite helpful for understanding the field qualitatively (understanding the general direction in which a certain charge will move from a specific position, identifying locations where the field is roughly zero or where the field points a specific direction, etc.).

A good figure with electric field lines can help you to organize your thoughts as well as check your calculations to see whether they make sense.

*Magnetic Force*

All moving charges experience force from magnetic field -Perpendicular to v and B (magnetic force is perpendicular)

*action and reaction pair*

According to Newton's third law, for every action force there is an equal (in size) and opposite (in direction) reaction force. Forces always come in pairs - known as "action-reaction force pairs." -ex: one negative charge (1) and one positive charge (2) repeal each other by induction. there is the same force on object 1 by object 2 that there is on object 2 by object 1

*Using V = RI*

P = I^2 R *P = V^2/R*

Replacing B1

B1 = (μ0 / 2π) (I1/d) *F21 = μ0 I1 I2 L2 / 2π d* pointing left - F12 points right (action-reaction pair): attractive ! -If the currents are anti-parallel: wires repel each other -both up and down = attractive -if they were opposite = repel/push -magnetic field from 1(B) causes a force on are 2 -magnetic field points into the board at 2 so force pointing left = attractive force

It is desired that a solenoid 39 cm long and with 440 turns produce a magnetic field within it equal to the Earth's magnetic field (5.0×10^−5T). What current is required?

BN = μ0 n I (0.39) (5.0×10^−5) = (4πx10^−7) (440) I *I = 0.035 A*

*Electric Potential*

For test charge q0 with electric potential energy U (in relation to some point) V = U/q0 -Electric potential does not depend on test charge!

Example: Three capacitors are connected in a circuit. a) Find the equivalent capacitance b) How do the voltages across the various capacitors compare? c) How does the stored energy in each capacitor compare?

C1 and C3 are not in series, C1 and C2 are in parallel = Cp, C3 and Cp are then in series a.) Cp = C1 + C2 -> Cp = 0.3 MF Ceq = 1/(1/C3 + 1/Cp) -> Ceq = *0.2 MF* b.) Qtotal = Ceq Vbattery => Q = 2.4 MF V3 = Q3/C3 -> V3 = 4V Vp = 2.4/0.3 -> Vp = 8V (V3 and Vp add up to 12V because in series) *Vp > V3* c.) U = 1/2CV^2 U1 = 3.2 MJ U2 = 6.4 MJ U3 = 4.8 MJ U1 > U2 > U3

*Q =*

CV

Conductors

Charges move freely on surface (W = 0) equipotential surface (ΔV = 0) -Charges homogeneously distributed on sphere -Charges concentrated on tips -picture: circles with positive charges on the edges and field lines pointing outwards -can conduct electrons = conductor -positive = surface electric field going out -equipotential surface (means same potential/charges can move around freely) on all surfaces of electric field

How do magnets pick up small pieces of unmagnetized material?

Domain aligned in same direction temporarily (paramagnetism)

*Molecules of insulators rearrange in presence of*

E

*defining charge per unit of area σ=Q/A*

E = (1/2) (σ/ε0) -normal surface will point in same direction as field -> Φ = Ea -> Ea/2A -2A is the area through which electric field is passing

*Electric field inside / outside spherical shell with charge q inside*

E = (k q/r^2) except inside conductor (E=0) -point charge

*If the electric field is due to more than 1 charge, the net electric field is the vector sum*

E = E1 + E2 + E3 + . . . Problem solving: same as before but for electric field

(Magnitude of E (does not depend on q0)*

E = F/q0 = k (q0Q/r^2) (1/q0) = k(Q/r^2)

*V = 0:*

E ≠ 0

*Electric field outside a thin infinite sheet of charge (Gaussian surface is a cylinder with area 2A)*

E=Φ/2A=(1/2A)(Q/ε0)

*Electric Flux*

Electric flux ∝ area Φ = E A→E Acosθ Units N•m^2/C -measure of the distribution of the electric field through a given surface. although an electric field in itself cannot flow. It is one way of describing the electric field strength at any distance from the charge causing the field -ex: when you open your window in the morning the amount of light coming through depends on the size (area) of the window and the amount of light there is outside

*Force on test charge due to q:*

F = k q0q/r^2 (repulsive)

Learning Goal: To learn to calculate the equivalent resistance of the circuits combining series and parallel connections. Resistors are often connected to each other in electric circuits. Finding the equivalent resistance of combinations of resistors is a common and important task. Equivalent resistance is defined as the single resistance that can replace the given combination of resistors in such a manner that the currents in the rest of the circuit do not change.

Finding the equivalent resistance is relatively straight forward if the circuit contains only series and parallel connections of resistors.

*Maximum force for θ = 90º (I⊥B): *

Fmax = I LB

What about 2 wires?

Force Between Parallel Wires

For Earth's surface, g =

G (MsubE/rsubE^2)

*Resistance*

How much I increases with V depends on resistance of wire -Interaction of electrons with atoms from wire (pipe filled with rocks) -Resistance allows less current for same V

Example 2.) Calculate all the different currents in the circuit. (slide 16. class 8)

I3 = I1 + I2 + ε1 − (R1 + R2) I3 − R3 I1 = 0 + ε1 − (R1 + R2) I3 + ε2 − (R5 + R4) I2 = 0 have everything except currents

*Unit for Electric Power*

J / s = W

*C =*

K ε0 (A/D) where K ε0 = ε (permittivity of the material)

*Considering kinetic energy K = 1/2 mv^2 and electric* *potential energy U = qV*

KA + UA = KB + UB -only use kinetic and electric potential

*Glow from collisions*

KE → thermal energy electrons moving inside PE -> KE (- to +) collide and create energy

*when the main charge is positive*

Lines point radially outward from charge

B =

N μ0 I / 2R in the center of a loop with N turns and radius R -B of N loops adds up

unit of torque

N•m

*Unit Ohm*

Omega Ω = V/A

Calculate the speed of a proton after it accelerates from rest through a potential difference of 260 V .

PE = KE qV = 1/2mv^2 v = √2qV/m v = √2(1.6 x 10^-19) (260) / (1.67 x 10^-27) v = 2.23 x 10^5 m/s

Calculate the speed of an electron after it accelerates from rest through a potential difference of 260 V .

PE = KE qV = 1/2mv^2 v = √2qV/m v = √2(1.6 x 10^-19) (260) / (9.11 x 10^-31) v = 9.56 x 10^6 m/s

Example: How many electrons make up a charge of - 30 µC?

Q = -30 x 10^-6 C (6.25 x 10^18 e- / 1 C) --> *1.9 x 10^14 e-* µ means 10^-6 Q is charge -just a change of units

Example 1.) A particle with charge of -5x10^-4 C and a mass of 2x10^-9 kg moves at 1000 m/s in the + x-direction. It enters a uniform magnetic field of 0.2 T that points in the + y-direction. a) Which way will the particle deflect as it enters the field? b) What is the magnitude of the force on the particle when it is in the field? c) What is the radius of the circular arc that the particle will travel while in the field?

Q = 5x10^-4 C m = 2x10^-9 v = 1000 m/s B = 0.2 T enters perpendicular (90º) a.) *-z* b.) F = qvBsinθ -> *F = 0.1 N* c.) r = mv/qB -> *r = 0.02 m*

*so equivalent single resistance that would draw same current I is*

Req = R1 + R2 + R3 valid for any number of resistors

*Resistivity*

Resistance depends on setup and on material

Current carrying wires can be made of superconducting materials below critical temperature

Resistivity goes to zero

*Static Electricity*

Rubbing pieces of rubber, glass, plastic, amber (elektron in Greek)... with cloth Produces static electricity (separated electric charges) -not produces, but transfers the charges around -depends on materials

*Table: Resistivity (slide9, class 7)*

Silver is too expensive! Copper and aluminum (light) are used in wires -table lists resistance from lowest to highest

*Kirchhoff's Rules*

So far: calculated Req (resistors in parallel and series) and used ε to find Itotal ε = RI -Then calculated other currents and potentials across different resistors -Kirchhoff (1845) derived 2 rules: conservation of charge and conservation of energy

*Net charge on edges of dielectric*

Some electric field lines end (created on dielectric) Electric field inside the dielectrics is less than in air

*What about force on a current carrying wire? *

Sum of forces on positive charges !

*Vector Review*

Tail-to-tip method and parallelogram method

*Unit of B is*

Tesla (non SI gauss 1G = 10^−4T) Largest magnetic field produced in lab ~ 10T Magnetars (stars with very high B) ~ 10^18 G = 10^14 T Magnetic field of earth ~ 0.5 G = 0.5 x 10^-4 T

*total voltage across each resistor in the series*

Total voltage V is ε = V1 + V2 + V3 = R1 I + R2 I + R3 I *voltage different across each, but same current*

*U = 1/2 QVf , so*

U = 1/2 CV^2 and U = 1/2 (Q^2/C)

*Defining the volume A d*

U = 1/2 ε0 E^2 volume *UE = U/volume = 1/2 ε0 E^2* energy density = E2 in parallel plate capacitor -Valid for ANY electric field !

*V =*

U/Q --> U = VQ

*Unit of B is Tesla = *

[1T = 1N ]

*Direction and magnitude obtained using*

a test charge (does not depend on q0) -Using large number of test charges: field lines -electric field can have the same direction as the force when it has the same sign, but a different direction with opposite signs

Ratio between electric and gravitational forces between electron and proton:

about 10^39

*unalike charges*

attract

*Ohm's law:*

constant resistance

*Current is not a vector:*

conventionally from + to - on wire -but can be + or - -negative just means going the opposite direction

*Ammeters measure*

current -to measure A in circuit you have to plug in the ammeter in series with what you want to measure

Example 2.) The separation between the plates of a parallel plate ca- pacitor is doubled while the charge on them is held con- stant. What happens to the a.) electric field between the plates? b.) potential difference? c.) capacitance? d.) stored energy?

d' = 2d, Q' = Q, A' = A a.) E' = ? -> E = (1/A) (Q/ε0) -> E' = (1/A') (Q'/ε0) -> *E' = E* b.) V' = ? -> V = Ed -> V' = E' d -> E2d = 2v -> *V' = 2V* c.) C' = ? -> C = Q/V -> C' = Q'/V' -> Q/2V = 1/2 (Q/V) = 1/C -> *C' = 1/2 C* d.) U' = ? -> U = 1/2 C V^2 -> U' = 1/2 C'V'^2 = 1/2 (1/2C)(2V)^2 = 1/2 x 2 x C x V^2 = 2U -> *U' = 2U* ε0 is a constant

Vc and q (and I)

decrease with time

*Electron has very small charge - e*

e = 1.6 x 10^-19 C C is Coulomb so... 1 C has 6.25 x 10^18 electrons

*Electric field lines closer on*

edges -Excess of charges on edges

*If the path is closed :*

electric circuit

*whenever you have charges, you have an*

electric field (E) + -> -

*Magnetic Field Analogous *

electric field F = ∣q∣v Bsinθ But force on particle is perpendicular to field (and also velocity) instead of parallel -Use right-hand rule to find direction of F -moving charge in magnetic field = force

Vc, q, I change in time

exponentially

*release charge =*

falls = KE

*release on the right =*

falls to the left (reduces PE)

*release on the left =*

falls to the right (loses PE)

*Number of lines ∝*

field strength -stronger force the closer to the magnet

*Formula above for "point charges"*

for spheres use r as distance between sphere's centers -if you have 2 point charges/masses, the distance between their centers is "r" -if you have more than 2 objects: in the same direction you add them up, but if in different directions you use vector ratios

*Amount of energy transformed ∝ to*

how much charge flows

*Problem Solving: recipe for adding vectors in different directions*

i. Draw the free-body diagram for each object with coordinate system and direction of each force (attraction or re-pulsion) ii. Use Coulomb's law to calculate the magnitude of the force for desired pairs (ignore the sign) iii. Add all forces acting one the body (bodies) taking signs into account and using symmetry (if possible)

*Problem Solving: how to go about circuit*

i. Labels currents with #s and directions (different #s in different branches) (negative sign found means opposite direction) ii. Write down variables you have and identify which you don't (at least the same # of equations as the # of un- known variables) iii. Apply 1st Kirchhoff's rule at each junction iv. Apply 2nd Kirchhoff's rule at each loop (positive sign passing battery - → + ) (decrease V at resistor if you're following the current) v. Solve system of equations (check your answers)

*pushing against will =*

increase PE (energy comes from food you eat so you can transfer energy to your muscles and move your arm) -if you push a negatively charged object towards the negative side = against will

Capacitors in series have

less equivalent capacitance -Opposite than resistors!

*Thick wires have*

less resistance

*Req is*

less than each resistance

Batteries, generators, ... produce electrical energy to be transformed into

light, thermal energy, ...

Increase torque:

more current loops

*resistance symbol*

more resistance = less current through

If the circuit is grounded

most current passes through it (low resistance) towards Earth -grounded = dump excessive charges onto earth

*mechanical:*

motors - to come!

Free electrons

move within the object and sometimes outside

radius of orbit (r) =

mv/∣q∣B

Most objects (atoms, everyday objects, stars, ...) are charged

neutral

*Minimum force for θ = 0 (I ∥ B): *

no force

Minimum torque for θ = 0 (I ∥ B):

no force

*South pole of needle attracted to*

north pole of magnet and north pole of needle attracted to south pole of magnet

Ex: If V = 0 at a point in space, must E = 0 there? If E = 0 at some point, must V = 0 there?

picture for problem on slide 6, top right, class 5 a = ΔV/ΔT

*Newton's third law*

push the table with your hand and the table pushes back with the same amount of force

Example 2.) How large a magnetic field is needed to cause an O2+ ion to move in a circular orbit of radius 2 m at 10^6 m/s? (the mass of the O2+ ion is approximately 32 u, where 1 u = 1.66 x 10^-27 kg)

q = +e = +1.6 x 10^-19 C r = 2m v = 10^6 m/s m = 32 x 10^-22 kg B = ? B = mv/rq -> (32 x 10^-22 kg) (10^6) / (2) (1.6 x 10^-19) -> *B = 1.66 x 10^-4 r*

*like charges*

repel

*2 positive charges*

slide 4 at the saddle point the gradient is zero -picture: 2 hills doesn't contain V=0

*Capacitance: Capacitors*

store electric charge (to be released later) -camera flash -energy back up for computers for power blackouts -circuit protection for surges of power -RAM memory in computer to store "zeroes" and "ones" ... -need to store because need to use a lot of charge at once

*Direction of magnetic field*

tangent to lines (see iron fillings)

For the situation shown in the figure(Figure 3), indicate whether there will be a tendency for the square current loop to rotate clockwise, counterclockwise, or not at all, when viewed from above the loop along the indicated axis.

the loop will not rotate at all

Many applications (time delay):

traffic lights, car wind- shield wipers, camera flashes, pacemakers, ...

*work means*

transform into PE

*picture (b) slide 9*

vertical (ex: mid-day) means cos90 = 0 = flux is zero

If there is a break or loose wire inside a metal appliance

we can become part of the circuit by touching it (when current is open) I = V/R = 120 V / 100000 Ω = 1.2 mA I = V/R = 120 V / 1000 Ω = 120 mA Rubber shoes and gloves, no rings, no water around, no 2 hands may also help

*What about many charges?*

when not a sphere = parallel plates

*power =*

work / time Watts (J/s)

*Energy stored =*

work done to charge capacitor

*slope equation*

y = mx + b slope in resistance = 1/R

*Electric potential:*

ΔV = VA − VB = 1/q0 (UA - UB) = k Q/rA - k Q/rB

when an atom loses an electron =

atom has an extra proton = positively charge

*Ions are*

atoms with gained or lost electrons

Same direction I:

attraction -> -> means solenoids stay together

*Charges don't move immediately but*

"push" others (like a water pipe) -electrons are already in the wire, just pushed when switch is flipped -takes a long time for new electrons to move to bulb/device

*Electric field between metal plates*

(Gaussian surface is a cylinder with area A because E and Q are zero otherwise) E = Φ/A = (1/A) (Q/ε0) = σ/ε0 -don't have 1/2 in this formula because before the electric field was passing through 2 sides

*so, W =*

(Vf / 2) (Q − 0) *W = 1/2 QVf* -work is related to charge and voltage

*When connected to a battery, capacitor is charged*

(electrons move - → +) Q = CV -> ΔV -Q is magnitude of charge in each plate -C is capacitance of the capacitor (depends on size, shape and material) -V is potential difference

*Power transformed by any device with resistance*

(energy transfer to atoms)

*Electric dipole*

(equal charges with opposite sign) -not a dipole when one charge is stronger

RC is called τ (tau)

(has units of time) and relates to how fast the capacitor is charged Vc (0) = 0 and q (0) = 0 Vc (∞) = ε and q(∞) = Q Vc (RC)= 0.63ε and q(RC) = 0.63 Q

*Net field lines never cross*

(no 2 directions of field at the same place) -cancels out edges (no arrows pointing outside plates) and the center charges add up (move towards the negative plate)

Turns off circuit

(open) bending or melting metal strip

Find the net charge of a system consisting of 212 electrons and 158 protons.

(protons-electrons) x (1.6 x 10^-19) (-54) x (1.6 x 10^-19) 8.64 x 10^-18

Find the net charge of a system consisting of 6.08×10^6 electrons and 7.75×10^6 protons.

(protons-electrons) x (1.6 x 10^-19) (1.67 x 10^6) x (1.6 x 10^-19) 2.672 x 10^-13 C

*Determine which groups of resistors are in series and which are in parallel Reduce all groups to equivalent resistances until you get to a single loop with one resistance (slide 9, class 8)*

(slide 10) -to be in series, you cannot have anything in between them, like the junction in this circuit -only parallel are 3 and 4 3 and 4 -> Ra Ra and R5 are in series -> Rb R2 and Rb are in series now -> Rc Rc and R1 in series (curves don't matter) -> Req

Vectors: Adding components

(slide 6) adding vectors F1 and F2 F1x = F1 cosθ1 F1y = F1sinθ1 F2x = F2 cosθ2 F2y = −F2sinθ2 Fx = F1x + F2x Fy = F1y + F2y F = square root Fx^2 + Fy^2 θ = arc tan Fy / Fx

B =

(μ0 / 2π) (1/r) where μ0 = 4πx10^−7 T•m/A

*Work done when there is displacement in direction of force*

*W = F d cosθ* =Fd, ifF∥d *W = q0 E d* ifF∥d independent of path for conservative forces (associated with PE)

Capacitors in Series

*picture on slide 3, class 9* Positive charge (fictitious) flows from battery to C1 and from C3 to battery -Nothing happens at first between points A and B -Inside plates of C1 and C3 become charged -Move positive charges from and to C2 -C2 is also charged! -*Same charge in all capacitors*

*Example*: Three particles travel through a region where the magnetic field points as shown in the figure. Find the charge of each particle.

*slide 9, class 10* = picture magnetic field is going into the board (know because "x"s) 1.) enters the field moving right (thumb up = deflected up) = *positive* 2.) enters field going down (arm down, fingers toward the board, thumb is down) = *negative* 3.) no charge because no effect = *neutral* 4.) enters going up = *negative*

*Change in potential energy equation*

*ΔU = −W = −q0Ed* , ifF∥d opposite for: opposite d direction or opposite charge sign

*Convention*

+ (lost) and - (gained) -postive charge with lost electrons and negative charge with gained electrons -ex: start with a charge of zero and gain an electron means it has an -e charge; if you gain another electron the charge is -2e; and so on -cannot have 1.5 electron charge in this course = can only have an integer amount of electrons/whole number or zero

Gravitational Field

-Analogous to electric field -Gravitational field generated by any mass -Always point radially inward towards the object -Responsible for weight

*Magnetic Force on Wire*

-Assume uniform magnetic field -Assume straight wire in between poles of a magnet -Right-hand rule

*Free electrons flow from*

- to + terminal of battery

*Resistor in Series*

-2 or more resistors connected end to end on single path in a wire -Can be simple resistors, light bulbs, electric heaters,... -We ignore the resistance of wire -Same current passes through all resistors (otherwise charge would appear/disappear) top pic *current is the same throughout the circuit at any part*

Current Loops: What about a wire that is not straight?

-A circular wire generates a magnetic field -Also not uniform (stronger inside the loop) -picture slide 2, class 12: current out = thumb toward you = current moves counterclockwise -larger loop = less current flow and weaker magnetic field lines

*magnets*

-Attract (some) metal pieces -Have 2 ends called poles (stronger effect) -When suspended one side points north (compass = magnetized needle because earth has a magnetic field = needle charged south points north) -Side that points close to geographic north is called north (other is called south)

Polarization (distortion) of atoms

-Attraction of small/light objects (weak)

*Storage of Electric Energy*

-Capacitors store energy (charge separation) -Battery moves charges from one plate to another

Atom

-Circulating charges (I) produce magnetic field -Normally cancel each other -In some cases (iron, nickel, cobalt, ...) there is net magnetic field -More details in the future (quantum mechanics) -All magnetic fields generated classically by currents Reason why magnetic field lines are closed! -magnetic field lines directed by current = close lines around current

*Coulomb's Law*

-Coulomb studied magnitude of electric forces (~1780) using a torsion balance -Force proportional to each of the charges (same distance) -Distance proportional to 1/ distance squared (1/r^2) -Proportionality constant (formula in picture) -force that charges exert on other charges = Coulomb's force -double one charge = force repelled away is stronger from the other charge (can double any *one* charge), or triple, etc. -force is proportional to any of the charges (F∝Q1Q2 x 1/r12^2 --> F = k (Q1Q2 x 1/r12^2))

Ampère's Law

-Current generates magnetic field lines around wire -B decreases with distance (long straight wire approximation) -B increases with the current

Torque on a Current Loop

-Electric current in presence of uniform magnetic field -Magnetic force on closed loop current -Opposite directions on each side by right-hand rule -No effect on parallel sides -Net torque rotates the coil -apply force on rotation axis = rotation = torque (depends on force and lever arm(distance from the axis)) -rotation -picture on the right, slide 5 class 11: current is flowing up and the force is into the screen, the current pointing down on the other side has the force out of the screen (rotation = push in one direction around axis)

Circular Motion

-Electric field does work on particle (unless F ⊥ v): changes speed - like gravity -Magnetic field does NOT do work on particle (F ⊥ v): constant speed -If v⊥B: circular path -doesn't change speed -force perpendicular to motion -magnetic field holds the charge in

Water Analogy

-Electric potential (gravitational) difference creates current (flow rate) -Battery (pump) creates current (flow rate) in closed circuit (pipe) -Current (flow rate) continuous throughout circuit (pipe) -Reversing the battery (pump) reverses the current (flow rate) (electrons move in opposite direction)

EX: Heart Defibrillator

-Energy stored in capacitors can electrocute you (even when power is off !) -Energy stored in a capacitor can also save lives -A heart attack (fast irregular beating) prevents heart to pump blood properly and can cause death -Defibrillators have capacitors charged to thousands of volts -Fast discharge through 2 paddles that spread current over chest -Discharge causes heart to stop followed by resumption of normal heart beating -stops your heart by releasing voltage to it -pile up charges to increase V

Torque on Loops

-Formula τ=I ABsinθ applies to any planar loop Corresponding A

charging Permanently by Induction

-Free electrons move within the neutral object Charge separated in neutral object -Neutral object connected to the ground (reservoir of charge) -Electrons move in wire -Wire cut in the presence of charged object -Net charge in object (no contact)! -grounding (object put into earth) = electrons go away (since electrons go away, then the resulting object is positively charged) -remove rod = go back

Electric Fields

-Gravitational and electrostatic forces different from contact forces -Idea of forces acting at long distances is counter intuitive (even for Newton) -Faraday (1791-1867) introduced the idea of field -Electric field extends outward from every charge into space -Other charges feel the force exerted by the electric field -field comes off of all the mass or charge (gravitational field of earth = has a charge around it) -good to know for non-contact forces

Electromagnets

-Have a piece of iron inside solenoid Iron piece becomes magnet -Resultant magnetic field (solenoid + iron piece) much higher -Many applications like switches (I produces B that moves metal) -Used in doorbells (iron rod strikes bell), car starters (starter starts sniping the engine), magnetic circuit breakers (iron plate breaks contact points) -picture: magnetic field moves around through the metal (out one end and in the other), while current moves with the wire

Aurora Borealis

-Ions from Sun (solar wind) have v component along B -Follow helical motion along B to north pole -B increases on poles (closer field lines): radius decreases -High concen- tration of ions ionizes air -Light emitted: aurora borealis or northers lights -charged particles from the sun -> hit atmosphere and ionizes particles -> emits light -can only see these very close to the poles

*Earth's Magnetic Field*

-Lines as if there was a bar inside the planet -North pole of needle points near geographic north (defined as magnetic south) -South pole of needle points near geographic south (defined as magnetic north) -Geographic poles (related to rotation) apart from magnetic ones by ~ 900 km (560 miles) -When using a compass to geographic north /south you have to correct -Earth's field changes over time (motion of molten iron in outer core)

Solenoids

-Long coil of wire consisting of many loops -Generates very large constant magnetic field inside (sum of field due to each loop) -Almost no magnetic field outside solenoid -Force on loops is attractive -configuration with many very close loops -magnetic field on the inside

*Uniform Magnetic Field*

-Magnetic field between 2 flat parallel pole pieces (area of pole pieces much larger than separation) -Mostly uniform -Fringes on edges -Just like electric field between 2 parallel plates

Force Between Parallel Wires

-Magnetic field exerts force on current -Current generates magnetic field -Does it exert force on current? Yes, but cancels out (points radially in)! -What about 2nd wire parallel to 1st? Yes, resultant force! -net force on 2nd wire, not one

Ferromagnetism

-Materials like iron have small regions called domains -Mini magnets (N and S) normally randomly aligned -Permanently aligned in presence of large magnetic field: magnets -Heating material first helps: increases of random thermal motion -Followed by cooling -Can be reversed by heating, dropping, hitting it, ... -How geomagnetic field is recorded in rocks in the bottom of the ocean -can't generate a net magnetic field, can just align -could be magnetized = could be aligned -domains can align with magnetic field -magnetic field out = misalign

*Dielectrics*

-Name of insulating sheet between plates of capacitor -Better than air (sometimes can conduct ): high V -Can be very thin: small d -Increases capacitance by factor K (dielectric constant) -larger area and smaller distance between = better capacitor -use a thin and solid insulator -note: this formula uses capital "K"

*Electric Circuits: EMF*

-Need source (battery or generator) to transform energy -Mechanical, chemical, light into electric energy -Potential difference generated by source : emf (electromotive force) of source ε

*Charge Conservation*

-No net charge can be created or destroyed: net amount of electric charge produced in any process is zero -Static electricity charges "go away" quickly into air -The air is full of water molecules

not on exams, but helpful: Superconductivity

-Normally resistivity ρ increases with temperature (atoms move more) → resistivity increases with time -For certain metals like mercury ρ → 0 (~ 10^-25 Ω . m) at very low temperatures! - Tc (superconductivity critical temperature) much below 0 ºC close to 0 K -Nowadays Tc up to 160 K (-113ºC) -Such compounds are difficult to bend (wires expansive) -lower temp = less resistance

The Atom

-Positively charged nucleus (protons + neutrons) -Negatively charged electrons

Example: Pacemakers

-RC circuits can charge and discharge capacitor continuously (regular voltages pulses) -Send voltage pulse (60 to 80 per minute) to induce each beat -Electrodes implanted close to heart -Contain capacitor and resistor -Capacitor charges and discharges -R and C based on required pulse rate

*Batteries in Series*

-Their potentials sum up or subtract (when connected oppositely) (wasteful) -But that's how batteries (like in a car) are charged! (not all batteries can be charged) -Positive particles (fictitious) forced back into positive terminal of battery -if batteries are pointing in the same direction (+ with -) you add them up -if batteries are pointing in the opposite direction (like signs together) you subtract them

Galvanometer

-Type of ammeter -Coil connected to electric circuit -Depending on current, coil rotates and moves needle (torque rotates the system) τ=NI ABsinθ -Many loops allow large τ for small I (very sensitive)

Christmas Lights

-Usually in series -Old Christmas lights used to have simply bulbs connected in series -Guess what happened when 1 bulb burned out ? -Modern Christmas lights have shunts in parallel with each bulb (new low resistance path for current) -Other lights more bright when one burns out -have a secondary/alternate path

*Same as opening 2 instead as 1 pipe of a dam (slide 7, class 8)*

-V → gravitational potential same across 2 pipes -I → water flow twice as much (than 1 pipe) -resistance is 1⁄2 (than 1 pipe) 2 x more area

Vc = ε (e^−t/RC) and q(t) = Qe^−t/RC

-Vc (0) = ε and q(0) = Q -Vc (RC )= 0.37 ε and q (RC) = 0.37 Q (fall 63%)

Helical Path

-What about angle between v and B θ≠90º? -Helical path: non-zero component of v in B direction (movement in B direction) -Still constant speed -Still F perpendicular to v and B -usually the particle enters at an angle, then goes round to perpendicular

Domain aligned in opposite direction temporarily (diamagnetism)

-all materials - not only for metals -very weak -levitation (because H2O aligning in opposite direction of the magnetic field)

*Parallel circuits are used in houses and buildings*

-can disconnect one resistance without interrupting the whole circuit -same voltage across all devices

(=*Magnetic Force Depends on*

-charge of particle -speed of particle (force of magnetic field on moving charge, particle needs to be moving) -magnitude of magnetic field (unit Tesla - huge) -angle between v and B

Example 1.) According to the Bohr model of the hydrogen atom, the electron in orbit around the proton can exist only in certain sized circular orbits. The smallest orbit has a radius of 0.0529 nm, and the next largest has a radius of 0.212 nm, which orbit is at a higher electric potential (with respect to infinity)?

-closer to point charge in the center =larger potential V1 = kQ/r1 -> (9 x 10^9) (+e) / (0.212 x 10^-9) -> *V1 = 27.2 V* V2 = kQ/r2 -> (9 x 10^9) (+e) / (0.0529 x 10^-9) -> *V2 = 6.8 V*

*For charges distributed along plates:*

-electric field is constant (parallel and equally spaced lines) -edge effect on borders (ignored for small separation between plates) -positive charges are repelled by left plate and attracted to right plate

Coulomb's law is very similar to gravitational force law, but

-gravity is always attractive -there is only one "sign" of mass - G = 6.67 x 10^-11 Nm^2/kg^2 -gravitational pull -gravity is always attractive, but the farther the distance, the lesser the force of attraction -gravity is weak

*Electrical case: the greater the distance (from plate with opposite charge sign) and the charge magnitude, the*

-more electric U the charge has -more K it will have at the other side -more work it can do -same electric potential at same distance -drop on the positive side = drops down to the negative side

*Gravity analogy: the greater the height and the mass, the*

-more gravitational U the rock has at the top -more K it will have at the bottom -more work it can do -same gravitational potential at same height -more potential energy at the top and more KE at the bottom

*Positive and negative charges*

V = k Q/r when Q>0 (graph of negative slope) V = k Q/r when Q<0 (graph of positive slope starting in negative y axis)

*Electric potential difference definition*

-positive plate is at higher electric potential than negative plate -positive charges move from high electric potential to low potential -negative charges move from low electric potential to high potential -when a negative charge moves towards a positive charge it moves from a high U to a low U, and when a positive charge moves towards a negative charge it moves from a high U to a low U (picture of plates on slide 4) -positive charges have a high V and negative have a low V -object falls towards the lowest/lower PE (PE decreases when falling) -drop charges = moves to reduce PE

*Gauss Law*

-sphere with positive charge -points out radially

The bent wire circuit shown in the figure (Figure 1) is in a region of space with a uniform magnetic field in the +z direction. Current flows through the circuit in the direction indicated. Note that segments 2 and 5 are oriented parallel to the z axis; the other pieces are parallel to either the x or y axis. 1.) Determine the direction of the magnetic force along segment 1, which carries current in the -x direction 2.) Determine the direction of the magnetic force along segment 2, which carries current in the -z direction 3.) Determine the direction of the magnetic force along segment 3, which carries current in the +y direction. 4.) Determine the direction of the magnetic force along segment 4, which carries current in the +x direction. 5.) Determine the direction of the magnetic force along segment 5, which carries current in the +z direction. 6.) Determine the direction of the magnetic force along segment 6, which carries current in the +x direction. 7.) Determine the direction of the magnetic force along segment 7, which carries current in the -y direction.

1.) +y 2.) 0 3.) +x 4.) -y 5.) 0 6.) -y 7.) -x

(Figure 1)A neutral conducting sphere contains a spherical cavity. A point charge q is placed at the center of the cavity. 1.) What is the total surface charge qint on the interior surface of the conductor? 2.) What is the total surface charge qext on the exterior surface of the conductor? 3.) What is the magnitude Eint of the electric field inside the cavity as a function of the distance r from the point charge? Let k, as usual, denote 1/4πϵ0. 4.) What is the electric field Eext outside the conductor?

1.) -q (because the negative charges will be on the inside closer to the positive point charge) 2.) q 3.) kq/r^2 4.) The same as the field produced by a point charge q located at the center of the sphere

The placement of resistors in a circuit is one factor that can determine the current passing through the resistor. You will be given three circuits, and for each circuit you will be asked to compare the current through the various resistors. In each of the circuits, all resistors are identical. 1.) Rank below the three identical resistors (A, B, and C) in (Figure 1) on the basis of the current that flows through them. (largest to smallest) 2.) Rank below the three identical resistors (A, B, and C) in (Figure 2) on the basis of the current that flows through them. (largest to smallest) 3.) Rank below the four identical resistors (A, B, C, and D) in (Figure 3) on the basis of the current that flows through them. (largest to smallest)

1.) A > B=C 2.) C > A=B 3.) D > A > B = C

A parallel-plate capacitor has plates of area 3.62 × 10^−4 m^2 . 1.) What plate separation is required if the capacitance is to be 1430 pF ? Assume that the space between the plates is filled with air? (Dielectric constant for air is 1.00059) 2.) What plate separation is required if the capacitance is to be 1430 pF ? Assume that the space between the plates is filled with paper. (Dielectric constant for paper is 3.7)

1.) C = εrε0A/d -> 1430 = (1.00059) ε0 (3.62 × 10^−4) / d -> d = (1.00059) (8.85×10^−12)(3.62 × 10^−4) / (1430 x 10^-12) -> *d = 2.24 μm* 2.) C = εrε0A/d -> 1430 = (8.85×10^−12) (3.62 × 10^−4 m^2)(3.7) / d -> d = (8.85×10^−12) (3.62 × 10^−4 m^2) (3.7) / (1.430 x 10^-9) -> d = 0.000008289 m -> *d = 8.3 μm*

(Figure 1) shows two different ways to visualize an electric field. On the left, vectors are drawn at various points to show the direction and magnitude of the electric field. On the right, electric field lines depict the same situation. Notice that, as stated above, the electric field lines are drawn such that their tangents point in the same direction as the electric field vectors on the left. Because of the nature of electric fields, field lines never cross. Also, the vectors shrink as you move away from the charge, and the electric field lines spread out as you move away from the charge. The spacing between electric field lines indicates the strength of the electric field, just as the length of vectors indicates the strength of the electric field. The greater the spacing between field lines, the weaker the electric field. Although the advantage of field lines over field vectors may not be apparent in the case of a single charge, electric field lines present a much less cluttered and more intuitive picture of more complicated charge arrangements. 1.) Which of the following panels (labelled A, B, C, and D) in (Figure 2) correctly depicts the field lines from an infinite uniformly negatively charged sheet? Note that the sheet is being viewed edge-on in all pictures 2.) In (Figure 2), what is wrong with panel B? (Pick only those statements that apply to panel B.) 3.) Which of the following panels (labelled A, B, C, and D) in (Figure 3) shows the correct electric field lines for an electric dipole? 4.) In (Figure 3), what is wrong with panel D? (Pick only those statements that apply to panel D.) 5.) In (Figure 4), the electric field lines are shown for a system of two point charges, QA and QB. Which of the following could represent the magnitudes and signs of QA and QB?

1.) D (the lines on both sides going straight into the single plate) 2.) field lines cannot cross each other and the field lines should be parallel because of the sheet's symmetry. 3.) B (the one with the arrows curved, moving from the positive charge to the negative charge and out of the positive charge on the other side) 4.) the field lines should be smooth curves and the field lines should always end on negative charges or at infinity. 5.) QA=+7q, QB=−3q (Very far from the two charges, the system looks like a single charge with value QA+QB=+4q. At large enough distances, the field lines will be indistinguishable from the field lines due to a single point charge +4q.)

*Potential difference across each resistor is*

V1 = R1 I V2 = R2 I V3 = R3 I

A +5.0−μC charge experiences a 0.47-N force in the positive y direction. 1.) If this charge is replaced with a −2.7μC charge, what is the magnitude of the force will it experience? 2.) Find the direction of the force in this case.

1.) E = F/q = 0.44 / 5.0 x 10^-6 = 9.4 x 10^4 N/C (in the positive y direction) F = Eq = (9.4 x 10^4) x (-2.7 x 10^-6) = *-0.25 N* 2.) negative y direction

Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2A. 1.) If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2? 2.) If the spacing between the plates in capacitor 1 is d, what should the spacing between the plates in capacitor 2 be to make the capacitance of the two capacitors equal?

1.) E = Q / Aε0 and ∣V∣ = Ed , so V = (Q / Aε0)(d) V1 = (Q / Aε0)(d) then V2 = (Q / 2Aε0)(d) then V2 = *V/2* (Even though the spacing between the plates is the same in both capacitors, the capacitor with the larger plates has a lower voltage between its plates. In fact, because the capacitors are equally charged, capacitor 2 has a smaller surface charge density, and therefore a weaker electric field between its plates. Since the voltage between two parallel plates is proportional to the electric field, capacitor 2 also has a lower voltage.) 2.) *2d* (Since the capacitance of a capacitor depends only on the geometry of the capacitor, capacitors 1 and 2 have the same capacitance only when the spacing between the plates in capacitor 2 is twice that in capacitor 1. This balances the effect of the different areas of the two capacitors.)

For each of the situations below, a charged particle enters a region of uniform magnetic field. Determine the direction of the force on each charge due to the magnetic field. 1.) Determine the direction of the force on the charge due to the magnetic field. (Figure 1) 2.)Determine the direction of the force on the charge due to the magnetic field. (Figure 2) 3.) Determine the direction of the force on the charge due to the magnetic field. Note that the charge is negative. (Figure 3)

1.) F⃗ = 0 2.) F⃗ points into the page. 3.) F⃗ points into the page.

A flashlight bulb carries a current of 0.33 A for 90 s 1.) How much charge flows through the bulb in this time? 2.) How many electrons?

1.) I = ΔQ/Δt -> 0.33 = ΔQ/90 -> *ΔQ = 29.7 * 2.) Qx = 29.7 -> (1.602 x 10^-19)x = 29.7 -> *x = 1.9 x 10^20 electrons*

from previous info... 1.) For the combination of resistors shown, find the equivalent resistance between points A and B. (series) 2.) For the set-up shown, find the equivalent resistance between points A and B. (parallel) 3.) For the combination of resistors shown, find the equivalent resistance between points A and B. (Figure 5) 4.) For the combination of resistors shown, find the equivalent resistance between points A and B.

1.) Req = 2 + 3 + 4 -> Req = *9Ω* 2.) 1/Req = 1/6 + 1/3 -> Req = *2Ω* 3.) 1/4 + 1/12 = 1/Req --> Req = 3 + 2 = *5Ω* 4.) 1/4 + 1/6 + 1/12 = 1/Req = 1/5 + 1/20 = 1/Req = those 2 answers + 1 + 3 + 2 = *3Ω*

In the diagram below,(Figure 1)the two resistors, R1 and R2, are identical and the capacitor is initially uncharged with the switch open. 1.) How does the current through R1 compare with the current through R2 immediately after the switch is first closed? 2.) How does the current through R1 compare with the current through R2 a very long time after the switch has been closed? 3.) How does the current through R1 compare with the current through R2 immediately after the switch is opened (after being closed a very long time)?

1.) The current through R1 is greater than the current through R2 2.) The current through R1 is equal to the current through R2 3.) The current through R1 is less than the current through R2

To study the behavior of a circuit containing a resistor and a charged capacitor when the capacitor begins to discharge. A capacitor with capacitance C is initially charged with charge q. At time t=0, a switch is thrown to close the circuit connecting the capacitor in series with a resistor of resistance R. (Figure 1) 1.) What happens to the charge on the capacitor immediately after the switch is thrown? 2.) What is the current I0 that flows through the resistor immediately after the switch is thrown?

1.) The electrons on the negative plate eventually pass through the resistor and neutralize the charge on the positive plate. 2.) V at t=0 is V = q/C *I0 = q/CR* Note that since current is charge per time, the preceeding formula shows that the units of RC must be time. The combination of variables τ=RC is called the time constant. It will occur frequently in problems involving a resistor and a capacitor.

A car battery does 280 J of work on the charge passing through it as it starts an engine. 1.) If the emf of the battery is 12 V, how much charge passes through the battery during the start? 2.) If the emf is doubled to 24 V, does the amount of charge passing through the battery increase or decrease? By what factor?

1.) V = E/Q -> 12 = 280/Q -> *Q = 23.33 C* 2.) 24 = 280/Q -> Q = 11.67 C *decrease* *by factor of 2*

A particle with a charge of q = -5.10 nC is moving in a uniform magnetic field of B⃗ =( -1.20 T ) z^. The magnetic force on the particle is measured to be F⃗ =( −7.60×10^−7 N )y^. 1.) Can vy, the y component of velocity be determined? 2.) Calculate vx, the x component of the velocity of the particle. 3.) Can vz, the z component of velocity be determined?

1.) YES The magnitude of the force F on a charge q=−6.00nC, moving with velocity of magnitude v in a magnetic field of magitude B=−1.30T, is F=qvBsinθ=qvperpB, where θ is the angle between v and B=−1.30T, and vperp refers to the component of the velocity that is perpendicular to the magnetic field. If you know F, you can determine vprep. In this particular case Fx=0, which implies vy=0. 2.) F = qvBsinθ -> 7.60×10^−7 = (5.10x10^-9) (v) (1.2) sin90 -> *vx = 124.18 m/s* 3.) NO The magnitude of the force F on a charge q=−6.00nC, moving with velocity of magnitude v in a magnetic field of magitude B=−1.30T, is F=qvBsinθ=qvperpB, where θ is the angle between v and B=−1.30T, and vperp refers to the component of the velocity that is perpendicular to the magnetic field. If you know F, you can determine vperp. However, you still don't know the component of the velocity that is parallel to the magnetic field. This component of velocity does not affect the force on the charge, since it is parallel to the magnetic field.

Consider the combination of capacitors shown in the diagram, where C1 = 3.00 μF , C2 = 11.0 μF , C3 = 3.00 μF , and C4 = 5.00 μF . (Figure 1) 1.) Find the equivalent capacitance CA of the network of capacitors. 2.) Two capacitors of capacitance C5 = 6.00 μF and C6 = 3.00 μF are added to the network, as shown in the diagram.(Figure 2) Find the equivalent capacitance CB of the new network of capacitors.

1.) combine ones in parallel C4 + C3 + C2 = 19μ = Cp in series = Cp + C1 = 19 x 3 / 19 + 3 = *CA = 2.59 μF* 2.) C4 + C3 + C5 = 14 μF C2 and C6 = 11 x 3 / 11 + 3 = 2.36 μF 14 + 2.36 = 16.36 μF CB = 16.36 x 3 / 16.36 + 3 *CB = 2.54 μF*

A conducting wire is quadrupled in length and tripled in diameter. 1.) Does its resistance increase, decrease, or stay the same? 2.) By what factor does its resistance change?

1.) decrease 2.) R = ρL / πr^2 as diameter is tripled (and thus radius is tripled), resistance decreases by a factor of 9, while quadrupling the length increases the resistance by a factor of 4 *resistance decreases by a factor of 4/9*

A dozen identical lightbulbs are connected to a given emf. 1.) Will the lights be brighter if they are connected in series or in parallel? 2.) Choose the best explanation from among the following:

1.) in parallel 2.) When connected in parallel each bulb experiences the maximum emf and dissipates the maximum power.

Consider the following configuration of fixed, uniformly charged spheres in (Figure 1): -a blue sphere fixed at the origin with positive charge q, -a red sphere fixed at the point (d1,0) with unknown charge qred, and -a yellow sphere fixed at the point (d2cos(θ),−d2sin(θ)) with unknown charge qyellow. The net electric force on the blue sphere has a magnitude F and is directed in the − y direction. 1.) What is the sign of the charge on the yellow sphere? 2.) What is the sign of the charge on the red sphere? 3.) Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. Calculate the charge qred on the red sphere (Express your answer in terms of q, d1, d2, and θ)

1.) negative 2.) positive 3.) From the problem statement, you know that the x component of the net force acting on the blue sphere is zero.The red sphere and the yellow sphere each exert a force on the blue sphere. You know the charge of the yellowsphere. This allows you to calculate the x component of the force that the yellow sphere exerts on the bluesphere. You need to find the appropriate charge for the red sphere such that the x components of the twoforces sum to zero Find , the Fxyellow component of the force that the yellow sphere exerts on the blue sphere Use Coulomb's law to find the force due to the yellow charge on the blue charge. Then find the xcomponent of the force Fxyellow = k2q^2cosθ / (d2)^2 find the force due to the red sphere (find the Fxred, the x component of the force that the red sphere exerts on the blue sphere) Fxred = -kqredq / (d1)^2 *qred = q•2(d1/d2)^2 cosθ*

Find the direction of the magnetic field at each of the indicated points. 1.) What is the direction of the magnetic field B⃗ A at Point A? 2.) What is the direction of the magnetic field B⃗ B at Point B? 3.) What is the direction of the magnetic field B⃗ C at Point C? 4.) What is the direction of the magnetic field B⃗ D at Point D? 5.) What is the direction of the magnetic field B⃗ E at Point E?

1.) out of the page 2.) into the page 3.) out of the page 4.) out of the page 5.) into the page

Figure 1)Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces shown in the figure when brought very near to the individual balls. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D. Assume throughout this problem that the balls are brought very close together. 1.) What is the nature of the force between balls A and B? 2.) What is the nature of the force between balls A and C? 3.) What is the nature of the force between balls A and D? 4.) What is the nature of the force between balls D and C?

1.) strongly attractive 2.) weakly attractive (Recall that ball C is composed of insulating material, which can be polarized in the presence of an external charged object such as ball A. Once polarized, there will be a weak attraction between balls A and C, because the positive and negative charges in ball C are at slightly different average distances from ball A. If ball C had a very small negative charge the test charge would have the same response (weakly attractive) but it would have a weak repulsive interaction with ball A. However, a smaller negative charge is not one of the options) 3.) attractive 4.) neither attractive nor repulsive (Because the test charge T is neither strongly attracted to nor repelled from ball C, ball C must have zero net charge. Since ball D also has zero net charge, there will not be any force between the two balls.)

*The capacitance can be calculated from*

1.) the electric field between 2 plates E = Φ/A = (1/A)(Q/ε0) = σ/ε0 2.) the magnitude of the potential ΔV = Ed C = Q/V = EAε0/Ed *C = Aε0/d* (arrangement) -Φ/A for 2 plates -capacitor = area (A) of one plate times ε0 / the distance between the 2 plates

1.) Gaussian surface 1 has twice the area of Gaussian surface 2. Both surfaces enclose the same charge Q. Is the electric flux through surface 1 greater than, less than, or the same as the electric flux through surface 2? 2.) Choose the best explanation from among the following:

1.) the same as 2.) The two surfaces enclose the same charge, and hence they have the same electric flux.

This problem explores the behavior of charge on realistic (i.e. non-ideal) insulators. We take as an example a long insulating rod suspended by insulating wires. Assume that the rod is initially electrically neutral. For convenience, we will refer to the left end of the rod as end A, and the right end of the rod as end B (Figure 1). In the answer options for this problem, "weakly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two balls, one of which is charged, and the other acquires a small induced charge". An attractive/repulsive force greater than this should be classified as "strongly attracted/repelled". 1.) A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time? 2.) Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into contact with end A of the rod. After several contacts with the charged ball, how is the charge on the rod arranged? 3.) How does end A of the rod react when the ball approaches it after it has already made several contacts with the rod, such that a fairly large charge has been deposited at end A?

1.) weakly attracted (Currently, you can think of this in the following way: When the sphere is brought near the rod, a positive charge is induced at end A (and correspondingly, end B acquires a negative induced charge). This means that some charge must have flowed from A to B. Since charge flow is inhibited in an insulator, the induced charges are typically small. Later you will learn how to model insulators more accurately and formulate a slightly more accurate argument.) 2.) negative charge on end A with end B remaining almost neutral (When the sphere is touched to end A, some of its negative charge will be deposited there. However, since charge cannot flow easily through an insulator, most of this charge will just sit at end A and will not distribute itself over the rod, as it would if the rod was a conductor.) 3.) strongly repelled (You may have learnt that any material is made of atoms, which in turn consist of a nucleus and electrons. In the atoms of some materials, some of the electrons are "bound" to the nucleus very weakly, which leaves them free to move around the volume of the material. Such electrons are called "free" electrons, and such materials are called conductors, because the charge (i.e. electrons) can move around easily. In insulators, all the electrons in the atom are bound quite tightly to the nucleus, i.e. there are no free electrons available to move through the insulator.)

The number of turns in a solenoid is doubled, and at the same time its length is doubled. 1.) Does the magnetic field within the solenoid increase, decrease, or stay the same? 2.) Choose the best explanation from among the following:

1.) μ0 N/L I = B *stays the same* 2.) The magnetic field remains the same because the number of turns per length is unchanged.

*1 kilowatt-hour =*

1000 J/s . 3600 s = 3.6 x 10^6 J 1 kilowatt-hour not an international unit

Example 2.) At $0.095 per kWh, how much does it cost to leave a 25 W porch light on every night (~12 h) for an year?

1000 Wh -> 0.095/kWh P = 25 W Δt = 12h x 365 ΔU = ? P = ΔU/Δt ΔU = (25)(12 x 365) x (0.095/1000) *ΔU = $10.40* make sure units cancel out!

*electric field (Φ) =*

Ea cosθ a is area

*Conservative forces have*

associated potential energy

*how capacitors work*

2 closed spaced conductors that can be rolled up -2 plates with charge (one positive plate and one negative) -plates can't touch or else they would become neutral = insulator between -usually in a syndical form rolled up with the insulator between the plates

*Resistor in Parallel*

2 or more resistors connected to each other at both ends -Current from source splits into separate paths -Their sum has to be equal to I (conservation of current) I = I1 + I2 + I3

2 beats: Electrocardiogram

Electric potential difference (mV) detected by electrodes on skin Electrocardiogram can identify heart problems -reflection of the depolarization wave means dead regions of the muscle caused by heart attacks measures potential in heart at a function of time

Example 2.) cont...

45 V - 41Ω I3 - 30Ω I1 = 0 --> I1 = 45V - 41Ω I3 / 30Ω 45 V - 41Ω I3 + 80V - 21Ω I2 --> I2 = 125V - 41Ω I3 / 21Ω I3 = I1 + I2 --> (45V - 41Ω I3 / 30Ω) + (125V - 41Ω I3 / 21Ω) --> *I3 = 1.73 V* *plug I3 into equations above to find I1 and I2*

What is the magnitude of the electric field produced by a charge of magnitude 8.40 μC at a distance of (a) 1.00 m and (b) 3.00 m?

E = kQ / r^2 k = 9 x 10^9 Nm^2/C^2 Q = SOURCE charge = 8.40 x 10^-6 C r = 1.00 m or 3.00 m Plug the values into the equation for E and get the field in N/C. a.) E = (9 x 10^9 Nm^2/C^2) (8.40 x 10^-6 C) / (1)^2 *Ea = 7.6 x 10^4 N/C* b.) E = (9 x 10^9 Nm^2/C^2) (8.40 x 10^-6 C) / (3)^2 *Ea = 8.4 x 10^3 N/C*

Different electric field:

E = kQ/R^2 = kσ(4πR^2) / R^2 = 4πkσ E' = kQ/(R^2/4) = kσ'(4πR^2/4) / (R^2/4) = 4πkσ' tips concentrate charges

*It can also be thought that energy is stored in*

E between 2 capacitor plates E = Q / Aε0 ∣V∣ = Ed area of one plate x distance between 2 = volume --> energy / volume = energy density

Electric field lines are a tool used to visualize electric fields.

A field line is drawn beginning at a positive charge and ending at a negative charge. Field lines may also appear from the edge of a picture or disappear at the edge of the picture. Such lines are said to begin or end at infinity. The field lines are directed so that the electric field at any point is tangent to the field line at that point.

How far must the point charges q1 = 6.60 μC and q2 = -25.8 μC be separated for the electric potential energy of the system to be -126 J ?

E = k.q1.q2 / r -126 = (9.0 x 10^9) (6.60 x 10^-6) (-25.8 x 10^-6) ./ r r = 1.2 x 10^-2m *r = 1.21 cm*

*Charges move at the rate*

C/s = Ampere (A) *I = ΔQ/Δt* current = change in charge/change in time

*Electric Currents and Magnetic Fields*

Connection between electricity and magnetism -Compass needles change orientation around electric wire when battery is connected -So electric current produces magnetic field -Field forms circles around the wire (not uniform) -Needles point along magnetic field -force of magnetic fields on charges -current generates magnetic field

*Energy Conservation*

Conservative forces -electricity also changes form

Once again, you are given an unknown material that initially generates no magnetic field. When this material is placed in a magnetic field, it produces a strong internal magnetic field, parallel to the external magnetic field. This field is found to remain even after the external magnetic field is removed. Your material is which of the following?

Ferromagnetic Materials that exhibit a magnetic field even after an external magnetic field is removed are called ferromagnetic materials. Iron and nickel are the most common ferromagnetic elements, but the strongest permanent magnets are made from alloys that contain rare earth elements as well.

RC circuits

Has resistors and capacitors -Turn on circuit by closing switch (closed switch = charged capacitor, open = uncharged)

In the figure(Figure 1)there are two point charges, +q and −q. There are also six positions, labeled A through F, at various distances from the two point charges. You will be asked about the electric potential at the different points (A through F). Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric potentials as higher than negative electric potentials.

Highest B A C = D) F E Lowest -closer to the positive charge = higher potential

Current decreases

I(t )= (ε/R) e^−tC with I (0) = ε/R, I(∞) = 0 X^0 = 1 if time is zero after you close the switch = 1-1 = 0

A rectangular loop of 270 turns is 35 cm wide and 15 cm high. What is the current in this loop if the maximum torque in a field of 0.48 T is 24 N⋅m ?

N = 270, A = (0.35 x 0.15), m, torque = 24, B = 0.48, θ = 90 (because max torque), I = ? τ = NI ABsinθ 24 = (270) I (0.35 x 0.15) (0.48) sin90 I = 3.53 A

for N loops, τ =

N I A Bsinθ -Magnetic moment of the loop N I A (related to amount of torque a set of loops can exert) -Used for voltmeters, ammeters and motors (to come !) -N = number of loops/rotations

An example of a series connection is shown in the diagram: (Figure 1) For such a connection, the current is the same for all individual resistors and the total voltage is the sum of the voltages across the individual resistors. Using Ohm's law (R=VI), one can show that, for a series connection, the equivalent resistance is the sum of the individual resistances. Mathematically, these relationships can be written as: I=I1=I2=I3=... V=V1+V2+V3+... Req−series=R1+R2+R3+... An example of a parallel connection is shown in the diagram: (Figure 2) For resistors connected in parallel the voltage is the same for all individual resistors because they are all connected to the same two points (A and B on the diagram). The total current is the sum of the currents through the individual resistors. This should makes sense as the total current "splits" at points A and B. Using Ohm's law, one can show that, for a parallel connection, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances. Mathematically, these relationships can be written as: V=V1=V2=V3=... I=I1+I2+I3+... 1Req−parallel=1R1+1R2+1R3+... NOTE: If you have already studied capacitors and the rules for finding the equivalent capacitance, you should notice that the rules for the capacitors are similar - but not quite the same as the ones discussed here.

In this problem, you will use the the equivalent resistance formulas to determine Req for various combinations of resistors.

*Equipotential Lines*

Lines where all points have same electric potential (zero potential difference) -Proximity of lines indicates gradient -Never cross -Zero work to move charges along equipotential lines -Same for gravitational equipotential lines

chapter 22:

Magnetic Fields, Magnetic force and Motion of Particles

non conductors / insulators (like wood)

Material with electrons bound tightly no nuclei -don't conduct electricity well

conductors (like metals)

Material with electrons loosely bound to nuclei

Mass Spectrometer

Measures mass of atoms (ions) -device physicists use to measure the mass of magnetic field particles

*Takes about 1 hour for 1 electron to move 1 m in a wire*

Move in a random manner = move slow (about 1 m per hour) -Many electrons passing by cross section in interval of time (current)

ex: Calculate the electric field at the center of a square with sides of 52.5 cm if one corner is occupied by a +45 µC charge and the other 3 occupied by -27 µC charges.

Picture in notes: positive charge in top left corner, arrow to each negative charge, and an extra arrow going to the negative charge across from the positive charge. the whole box is the length Ecenter = ?, l = 0.525 m 2 negative arrows across from each other cancel out 2 arrows towards the one negative charge across from the positive charge add up E = (k (45 x 10^-6) / √2l/2) + (k (27 x 10^-6) / √2l/2) *E = 4.7 x 10^6 N/C* theta = -45

An equivalent capacitor has to hold the same total charge:

Q = Ceq ε which means *Ceq = C1 + C2 + C3 ....* capacitors in parallel sum up their capacitances

An equivalent capacitor has to hold the same charge and have the sameεacross:

Q = Ceqε and ε = V1 + V2 + V3 Using Q = C1V1, Q = C2V2, Q = C3V3 --> Q/Ceq = Q/C1 + Q/C2 + Q/C3 --> *1/Ceq = 1/C1 + 1/C2 + 1/C3*

Total charge of capacitors in Parallel is

Q = Q1 + Q2 + Q3 = C1ε + C2ε + C3ε = (C1 + C2 + C3)ε

*conservation of charge*

at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction I = I1 + I2 + I3 I3 = I1 + I2 -splits in 3 -equation is the same for both junctions

For the situation shown in the figure(Figure 2), indicate whether there will be a tendency for the square current loop to rotate clockwise, counterclockwise, or not at all, when viewed from above the loop along the indicated axis.

The loop tends to rotate clockwise, as viewed from above.

For the situation shown in the figure(Figure 1), indicate whether there will be a tendency for the square current loop to rotate clockwise, counterclockwise, or not at all, when viewed from above the loop along the indicated axis.

The loop tends to rotate counterclockwise, as viewed from above.

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the plates is 305 V , and the plate separation is 0.279 mm

U = ?, C = 225 x 10^-6 F, V = 305 V, d = 0.000279 m U = 1/2 ε0 E^2 E = V/d E = (305)/(0.000279) E = 1093189.964 U = 1/2 (8.85×10^−12) (1093189.964)^2 *U = 5.29 J/m^3* Here we learn how to determine the electric energy density of a capacitor and that it is independent of the capacitance of the device.

*Electric potential energy (point charges and with respect to infinity)*

UA = q0 VA = k (q0Q/rA) U is also smaller for larger r (like V)

*Electric potential difference is a measure of how much electric potential energy an electric charge can acquire or loose*

Ub − Ua = q0 (Vb−Va)

Shielding

Uncharged conductor in electric field -Induced charge separation -Field generated cancels out external field -No electric field inside conductors ! -Conductor almost not noticeable -Many applications ... -experiment with charged ball in class -electrons will try and go against the field because (picture in notes: charge between plates is running from positive -> negative, but the electron wants to go towards the positive = against the field) -nothing happens the center, charges attract to side of opposite charge

Example 2.) Three point charges are arranged at the corners of a square of side L. What is the potential at the fourth corner point A (with respect to infinity)?

VA = k (Q/r) +Q to A distance = √2L +3Q to A = L -2A to A = L VA = ((1/√2) + 3 -2) KQ/L *VA = 4.7 KQ/L*

*Setting rB to infinity (VB to zero):*

VA = k Q/rA

How Vc and q increase

Vc(t) = ε(1 − e^−t/RC) q(t) = Q (1 − e^−t/RC) where Q is the maximum charge

*V average =*

Vf + Vi / 2 = *Vf / 2* because Vi = zero -V average = voltage of battery / 2

Calculate the work done by a 2.5-V battery as it charges a 6.4-μF capacitor in the flash unit of a camera.

W = ?, V = 2.5 V, C = 6.4 x 10^-6 F W = 1/2 CV^2 W = 1/2 (6.4 x 10^-6)(2.5)^2 *W = 20 μJ*

*Work done to move charge against electric potential V*

W = V Δq (constant V average) -changing charge of the capacitor

*From the definition of work (if F is parallel to d)*

W = q0 Ed

*amount of work battery does to move charges around a circuit*

W = ΔQε

*Work done by the electric field to move a charge*

W = −ΔU = −q0 ΔV

*C unit [C/V] =*

[F(Farad)]

*voltage and current are combined in*

a multimeter (analog or digital) -Have specified sensitivity -Ammeters have to be placed in circuit in series (more R) -Voltmeters have to be placed in circuit in parallel (less I)

To understand the electric force between charged and uncharged conductors and insulators. When a test charge is brought near a charged object, we know from Coulomb's law that it will experience

a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.

*electric field is*

a vector ⃗E = F⃗ / q0

Example 1.) What is the equivalent resistance of 3 resistors with resistances of 1Ω, 2Ω and 3Ω a) connected in series? b) connected in parallel? c) How much total current will be delivered by a 12 V battery in each of these arrangements? d) What is the voltage drop across each resistor in each of these arrangements? e) How much current will be in each resistor in each of these arrangements?

a.) 1 + 2 + 3 = *6 Ω* b.) 1/1 + 1/2 + 1/3 = *0.55 Ω* c.) Parallel: I = V/R -> I = 12/6 -> I = *6A* Series: I = V/Req -> I = 12/0.55 -> I = *2 A* d.) Parallel: V1 = V2 = V3 = *12V* Series: V1 = R1I = (1)(2) -> *V1 = 2V* ; *V2 = 4V* ; *V3 = 6V* e.) Parallel: I1 = I2 = I3 = *2A* Series: I1 = V/R1 -> 12/1 -> *I1 = 12A* ; *I2 = 6A* ; *I3 = 4 A*

Example 1.) A 12 V battery causes a current of 0.6 A through a resistor. a) What is the resistance? b) How many joules of energy does the battery lose in 1 minute?

a.) V = 12V, I = 0.6A, R = ? -> R = V/I -> *V = 20 Ω* b.) ΔV = ?, t = 60 sec -> P = ΔU/ΔT = VI -> ΔU = (12)(0.6)(60) -> *ΔU = 432 J*

Example 1.) Because a current-carrying wire is acted on by a magnetic field, it would seem possible to suspend such a wire at rest above the ground using the Earth's magnetic field. a) Assuming this could be done, consider long, straight wire located at equator. What would the current direc- tion have to be to perform such a feat? Up, down, east or west? b) Calculate the current required to suspend the wire, assuming that the Earth's magnetic field is 0.4 G at the equator and the wire is 1 m long with a mass of 30g.

a.) into the board / up b.) B = 0.4 x 10^-4 T, L = 1 m, m = 30 x 10^-3 ILB = mg -> *I = 7.35 x 10^3 A*

Ex: Imagine moving a proton from the negative plate to the positive plate of two parallel plates. The plates are 1.5 cm apart, and the field is uniform with a magnitude of 1500 N/C. a) What is the electric potential difference? b) What is the change in the potential energy? c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? -first 2 have to do with repulsion, while (C) is release/falling

a.) ΔV = ?, E = 1500 N/C, d = 0.015 m ΔV = -Ed --> ΔV = -(1500)(0.015) --> *ΔV = 22.5 V* b.) ΔU = ΔVq = (22.5)(1.6 x 10^-19) = *3.6 x 10^-18 J* c.) (bottom pic on slide 11, falls to negative) ΔU = 3.6 x 10^-18 J -> ΔKfinal = 3.6 x 10^-18 J K = 1/2 mv^2 v = √2 (3.6 x 10^-18)/(1.67 x 10^-27) -> *v = 6.57 x 10^4 m/s*

Everyday charge (rubbing objects)

about µC µ means 10^-6

*potential difference (ΔV, Δ, or E)*

aka voltage or electromotive force measure in volts (V)

Capacitors in Parallel

all capacitors have the same potential across -Their respective charges are Q1 = C1ε, Q2 = C2ε, Q3 = C3ε -voltage is the same, but each has a different charge because each has a different capacitance

*Direction of force*

along line that connects the objects -fore is a vector (has direction and magnitude) -force goes along a line

Lightning rods

are just pointing things that attract lightning (ex: tops of tall buildings) positive charges go to the edge and arrows point out

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be distributed uniformly throughout the cloud. The charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air "breaks down." In general, the term "breakdown" refers to the situation when a dielectric (insulator) such as air becomes a conductor. In this case, it means that, because of a very strong electric field, the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air then emits light as the electrons and ionized atoms recombine to form excited molecules that radiate light. The resulting large current heats up the air, causing its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder. The point of this problem is to estimate the maximum amount of charge that a cloud can contain before breakdown occurs. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 km. Take the breakdown electric field of air to be Eb = 3.00 × 10^6 N/C. 1.) Estimate the total charge q on the cloud when the breakdown of the surrounding air begins. using ϵ0 = 8.85 × 10^−12C^2/(N⋅m^2). 2.) Assuming that the cloud is negatively charged, how many excess electrons are on this cloud?

area = 4 pi r^2 1.) q = EA ϵ0 --> q = (3.00 × 10^6) (3141592.654) (8.85 × 10^−12) *q = 83.4 C* 2.) Each electron contributes 1.60 x 10^-19 C N x 1.60 x 10^-19 C = 83.4 C *N = 5.21 x 10^20*

*Electric potential depends on*

battery -current depends on system's resistance (and voltage)

*Equipotential Lines same charges*

bottom picture

*no spheres means*

can't use F = kQ/r^2 --> Gauss's law

Current charges capacitor, no more current means

capacitor isn't charged until current returns -takes time to charge (resistance) -electrons moving clockwise (from negative side of battery) first, but, removing the battery makes the electrons move back counterclockwise uncharging the capacitor

*Computer keyboard uses*

capacitors Pressing key decreases d → increases capacitance → sends electric signal to be interpreted as the letter

Dry days means

charge accumulation

example of conduction

charged object on the left and neutral object on the right (charged object means it lost electrons, 6 in this case) ---> a metal screw connects the objects , so since metal is a good conductor, 3 electrons will move from the right atom (neutral one) to the left atom (the positively charged one). So, now both of the atoms have a positive charge of 3 -electrons move only, no protons

Smaller RC means capacitor

charges faster

*3D: B vector as arrow*

circle with a dot inside = out of the board circle with an x inside = already into the board

No battery:

close switch

*example of static electricity*

clothes rubbing against one another = one becomes more negative and one more positive = attraction = stuck together -exchange of charges -other examples include brushing hair, drying clothes in the dryer, etc.

No electrical power needed to maintain

current -Huge saving of energy and no heating dissipation -Used in particle accelerators: accelerate charged particles -Used in MRI machines: image atomic nuclei inside body (without metal) -Used in research

Ohm experiment:

current (I) is proportional to V 2XV→2xI -stronger battery = move electrons faster and move more electrons (better current)

*Conservation of electric charge:*

current is the same throughout the circuit -charge is not created in the battery -charge does not disappear in a light bulb -battery just pushes electrons around -electrons just pass through device -picture on slide 3, class 7

*Depends on:*

current passing through it and electric po- tential across it

*If the circuit is open:*

current stops -path can be any shape, but it must be closed

Fuse breaker (if threshold is not too high) should

cut the current

*E points toward*

decreasing V (ΔV < 0)

What type of magnetism is characteristic of most materials?

diamagnetism Almost all materials exhibit diamagnetism to some degree, even materials that also exhibit paramagnetism or ferromagnetism. This is because a magnetic moment can be induced in most common atoms when the atom is placed in a magnetic field. This induced magnetic moment is in a direction opposite to the external magnetic field. The addition of all of these weak magnetic moments gives the material a very weak magnetic field overall. This field disappears when the external magnetic field is removed. The effect of diamagnetism is often masked in paramagnetic or ferromagnetic materials, whose constituent atoms or molecules have permanent magnetic moments and a strong tendency to align in the same direction as the external magnetic field.

*thermal:*

electric heaters, ovens, stoves, toaster ..

*Energy unit J (joule) is too large to deal with*

electrons

Only parts of an atom that move between atoms

electrons -only electrons can transfer charges between atoms

*picture on slide 3 in class 6: circuit with 2 plates*

electrons on the picture move counterclockwise from battery to the negative plate (pile up), so the positive charges have to go through the negative plate to the other plate, then back to the battery = same charge of both plates -depends on total voltage of the battery and characteristics of the capacitors

*Electric Power (P) =*

energy transferred / time = ΔQV/Δt = IV -the energy formula from last class had 1⁄2 due to non- constant charge during charging of the capacitor

*electric field between plates (picture in notes)*

ex: cylinder between arrows between the 2 plates = no net flux

*electric field in plate (picture in notes)*

ex: cylinder set on positive plate = has a net flux

*dielectric constants table*

ex: storms = air becomes conductor

*When magnets are brought together they*

exert force -Repulsive (same poles) -Attractive (opposite poles) But isolated poles (monopoles) haven't been observed (unlike electric charges) -when you split a magnet in half it just creates new poles -Cutting a magnet in half produces 2 new magnets

Torque =

force * lever arm

*If charges have opposite sign:*

force is negative (attraction)

*If charges have same sign:*

force is positive (repulsion)

*Definition of electric field:*

force per charge at a given location -Electric field is a vector (same direction as force for positive q0) -Units N/C (Newton per Coulomb) -main charge pushing away test charge -stronger force = stronger field = stronger push the charge will feel when close to it

*If current is reversed:*

force reversed

Charging by induction (no contact):

free electrons move within the neutral object -Charge is separated in neutral object -ex: a neutral metal rod has a positively charged rod brought up to it (but not touching) = separation of charges (negative charges in the neutral rod move towards the positive charge)

Charging by conduction (contact):

free electrons move from one to another object -Both end up with same charge density -ex: 2 balls touch and transfer charges to create 2 similarly charged objects

*Lines are closer together where field is*

greater -field is stronger the closer you get to the charge = stronger force

Is the electric potential at point 1 in the figure (Figure 1) greater than, less than, or equal to the electric potential at point 3? Choose the best explanation

greater than The electric potential decreases as we move in the direction of the electric field, as shown in the figure. Therefore, the electric potential is greater at point 1 than at point 3.

3rd hole on outlet

ground

*Force exerted on a magnet (by other) described by*

interaction with field

*Materials that produce strong magnetic effects are ferromagnetic:*

iron, cobalt, nickel,

*E=*

k (Q / r^2) bottom picture slide 3: Q-->2Q means 2 times more lines

Suppose the three particles in the figure(Figure 1) have the same mass and speed. 1.) Rank the particles in order of decreasing magnitude of their charge. (largest to smallest)

largest A B C smallest A and B are positive but A has a sharper curve = more charge C is negative

Hollow metal boxes

like a car -keeps you safe from electricity like lightning

*Negative flux for*

lines entering enclosed surface

*Positive flux for*

lines leaving enclosed surface

Living tissue has

low resistance (many ions) but dry skin has high resistance (104 - 106 Ω)

*All charges accelerate towards*

lower U (U = qV)

*electric field points in the direction of*

lower electric potential

Damage from electric current depends on current

magnitude, time it lasts and part of body it reaches (like heart or brain) -1 mA is a normal shock (some pain) -above 10 mA is a more serious shock (severe muscle con- traction) that can prevent the person from letting the object go -above 80 mA across torso is very serious (heart muscles con- tract irregularly and pump blood improperly) can cause death -around 1 A is unlikely to cause heart failure (stops whole heart and starts over - like defibrillator) but causes severe burns

*Electric energy can be transformed into other forms of energy:*

mechanical and thermal

*Switch Symbol*

open = stops current

*when the main charge is negative*

or radially inward towards charge

*For direction use the right-hand rule:*

orient your right hand until your outstretched fingers can point in the direction of the particle's velocity, and when you bend your fingers they point in the direction of the magnetic field lines. Then your outstretched thumb will point in the direction of the force on the particle (down in picture) -For negative particles the force points in the opposite direction (up in picture) (opposite of what thumb is telling you) *slide 8, class 10*

water molecules are

overall neutral in charge, but are polar (one side is more positive and the other is more negative) -ex: a negatively charged object will push the negative charges of the water molecule to the opposite side and attract the positive charges (because unalike charges attract)

*C of one capacitor is about*

pF to μF 10^-12 to 10^-6

You are given a material which produces no initial magnetic field when in free space. When it is placed in a region of uniform magnetic field, the material produces an additional internal magnetic field parallel to the original field. However, this induced magnetic field disappears when the external field is removed. What type of magnetism does this material exhibit?

paramagentism When a paramagnetic material is placed in a magnetic field, the field helps align the magnetic moments of the atoms. This produces a magnetic field in the material that is parallel to the applied field.

*Electric potential drops when*

passing through a resistor -But current stays the same! -*important picture in the middle of slide 8, class 7*

*picture (a) slide 9*

perpendicular (ex: morning and night) means cos0 = 1, so flux = EA

*When the battery is on: force on the wire is*

perpendicular to B and I

*Electric field is*

perpendicular to surface

*Equipotential lines are always*

perpendicular to the electric field

*picture on slide 3, class 11*

picture on the left: current moving into the screen = hand towards the screen, curve to the right = thumb is down = force is down

Example 2.) Two long parallel wires carry current in opposite direc- tions. Wire 1 carries a current of 5 A up and the current in wire 2 is 10 A down. Both have a length of 50 cm, and they are separated by 3 mm. Determine the mag- netic force that each wire exerts on the other.

pictures in notes (9/19) -force on 1 by 2 is to the left -force on 2 by 1 is to the right -this means repulsion F12 = μ0 I1 I2 L2 / 2π d F12 = (4π x 10^-7) (5) (10) (50 x 10^-2) / (2π) (3 x 10^-3) *F12 = 1.67 x 10^-3 N* F12 is the force for both of them (left = left and right = right)

*Absolute potential calculates difference between*

point and infinity (where V = 0) -between 2 objects (object and earth)

reminder: force in electric field points in the

positive direction of field

*Historically, + charges move (*conventional current* is the one we use) from*

positive terminal to negative terminal -Both are equivalent ! (always draw) -conventional current is just the opposite of electrical current (moves + to -) -*when nothing is specified, go with conventional current*

*Batteries produce*

potential difference

*There is E (parallel to wire) due to*

potential difference (V) -there is always an electric field (E) (+ -> -) -will move with conventional current

*Electric Power*

power = how fast you transform electrical energy

A particle with charge 8.00×10^−19 C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction. 1.) The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 6.40 × 10^−19 J . In what direction and through what potential difference Vb−Va does the particle move? 2.) If the particle moves from point b to point c in the y direction, what is the change in its potential energy, Uc−Ub?

q(ΔV) = -ΔK q(Vb - Va) = -ΔK q(Va - Vb) = -ΔK (8.00×10^−19)(Va - Vb) = -(6.40 × 10^−19) (Va - Vb) = -0.8 *The particle moves to the left through a potential difference of Vb−Va= -0.800 V* (In general, if no forces other than the electric force act on a positively charged particle, the particle always moves toward a point at lower potential.) 2.) *0* (Every time a charged particle moves along a line of constant potential, its potential energy remains constant and the electric field does no work on the particle.)

*Gauss's law*

q/Eo = E*A (flux); where E is electric field, E0 is permitivity constant, and A is area

J. J. Thomson is best known for his discoveries about the nature of cathode rays. His other important contribution was the invention, together with one of his students, of the mass spectrometer, a device that measures the ratio of mass m to (positive) charge q of an ion. The spectrometer consists of two regions as shown in the figure.(Figure 1) In the first region an electric field accelerates the ion and in the second the ion follows a circular arc in a magnetic field. The radius of curvature of the arc can be measured and then the m/q ratio can be found. After being accelerated to a speed of 1.41×10^5 m/s , the particle enters a uniform magnetic field of strength 0.800 T and travels in a circle of radius 30.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q. Find the ratio m/q for this particle.

r = mv/qB 0.3 = m (1.41×10^5) / q (0.8) 0.24 q = m (1.41×10^5) 0.24q/(1.41×10^5) = m *m/q = 1.702 x 10^-6 kg/C*

*when 2 same charges come together*

repel (negative charges would look the same except the arrows would be pointing in)

Opposite direction I:

repulsion -> <-

*All electric devices offer resistance (to current flow) but this is not necessarily bad !*

resistors -control current in circuits -light bulb filament -electric heaters -ovens -stoves -toaster -hair dryer

Example: Consider the 3 point charges q 1= 2.5 nC, q2 = 2.5 nC and q3 = 3 nC in the diagram below. What is the magnitude of the electrostatic force on on q3?

slide 11 (use recipe on slide 7) -2 arrows have the same force, so they push the object straight across to the right square root (0.3^2 + 0.4^2) = 0.5 m cosθ = adjacent/hypotenuse = 0.4/0.5 F3y = 0, F3x = 2kF31, F3x = 2kF32 F31 = F31cosθ --> k (Q3Q1/r31^2) --> 9 x 10^9 Nm^2/C^2 (2.5 x 10^-9 C) (3 x 10^-9 C) / 0.5 m^2 --> F31x = 2.16 x 10^-7 N F3x = 2kF31 --> *4.32 x 10^-7 N right*

*1 positive and 1 negative charge*

slide 4 the potential has a large gradient (steep) between the charges -picture: one big bump and one ditch contains V=0 -to get to positive from negative (or vise versa) must pass through almost zero (can only reach zero if far from the charges)

*Positive charges accelerate towards*

smaller V

stronger B =

smaller r

Consider the three electric charges, A, B, and C, shown in the figure.(Figure 1) Rank the charges in order of decreasing magnitude of the net force they experience. --- -qA ------------- qB ------------- qC ---

smallest C A B largest

*picture (c) slide 9*

some angle (ex: other times of the day), so use flux = EA cosθ

*ε is potential difference from*

source and across the 3 resistors

People build devices to maintain potential difference:

source of electrical energy

It costs 2.6 cents to charge a car battery at a voltage of 12 V and a current of 15 A for 120 minutes. What is the cost of electrical energy per kilowatt-hour at this location?

t = 2 h for 2.6 cents V = 12 V I = 15 A power is 12 x 15 = 180 W 180 x 2h = 360 W/h = 0.36 kWh 2.6 cents / 0.36 kWh = 7.22 cents/kWh = *$0.0722 /kwh*

*Test charge acquires potential energy because of*

the electric potential

*The more resistance added in series*

the smaller the current

*conservation of energy*

the sum of changes in potential around any closed path of circuit must be zero ε = V1 + V2 + V3 ε1 - R1I3 - R2I3 -R3I1 = 0 ε1 - R1I3 - R2I3 + ε2 - R5I2 - R4I2 = 0 -get one equation for each loop -from + side = add up voltage -subtract current - voltage when passing a resistor

*Energy (like the one we pay for) is*

the sum of power of devices x time they were on

*Electric field investigation:*

use positive small test charge q0 (does not exert significant force on charge Q) -Move q0 around and measure Fq0Q (along q0Q direction) -Fq0Q proportional to inverse square of distance rq0Q

*work is only done*

vertically, no work is done in the x direction -parallel or anti-parallel to force

*Connecting wires have*

very low resistance (ignored)

mastering physics homework 1

vvvvvvv

Force between 2 objects with charge 1 C separated by 1 m:

~10^10 N (~ weight of 1 million tons!) -electromagnetic force is strong -force is measured in Newtons (N)

A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates are 1.05 cm apart, and have a potential difference of 2.70 × 10^4 V , find the change in electric potential energy for an electron that moves from one accelerating plate to the other.

ΔU = q0 ΔV = (−e)(2.7 x 10^4 V)= −27000 eV (−27000 eV) × (1.60 × 10^−19 J/eV) = *-4.32 x 10^-5 J*

*For example, the energy transformed when 1 C charge provided by 12 V car battery to light headlight is*

ΔU = qΔV = (1C)(12V) = 12J

The Electric Potential of the Earth: The Earth has a vertical electric field with a magnitude of approximately 100 V/m near its surface. What is the magnitude of the potential difference between a point on the ground and a point on the same level as the top of the Washington Monument (555 ft high)?

ΔV = Ed --> ΔV = (100) (169.164) --> ΔV = 16916.4 V --> *ΔV = 2 x 10 ^4 V*

*We only measure differences of potential*

ΔV = Vb − Va = Ub/q0 - Ua/q0 = Ub-Ua/q0 = ΔU/q0 = -W/q0 -only measure difference = PE -> KE ΔV = -work/q1

The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments. (Figure 1) 1.) What is the work done by the electric force to move a 1 C charge from A to B? 2.) What is the work done by the electric force to move a 1 C charge from A to D? 3.) The magnitude of the electric field at point C is

ΔV = ΔU/q W = -ΔU W = -q(V1 - V2) 1.) W = -(1) (1-1) -> *W = 0 J* 2.) W = -(1) (0-1) -> *W = 1 J* 3.) greater than the magnitude of the electric field at point B (The equipotential lines are closer at point C than the equipotential lines at point B.)

*Therefore for a constant electric field and F ∥ d*

ΔV = −Ed E = −ΔV/d [V/m] or [N/C] (combining formulas)

E goes with

ΔV and not V

*Electric flux on surface of sphere with charge q inside*

Φ=EA=(k q/r^2)(4πr^2)=4πkq -theta always zero, so Φ = EA = 4πkq flux depends on charge for enclosed surface

*Resistor in Parallel: Potential difference ε across all resistors is the same*

ε = R1 I1 = R2 I2 = R3 I3

*but, voltage is also*

ε = Req I

*Rewriting the total current I = I1 + I2 + I3 as*

ε/Req = ε/R1 + ε/R2 + ε/R3 1/Req = 1/R1 + 1/R2 + 1/R3 where Req is the equivalent resistor that would draw the same total current from the battery (slide 6, class 8)

*with permittivity of free space*

ε0=1/(4πk)=8.85x10^−12 C^2/(N m^2) *Φ=4πkq=q/ε0* (general formula) -Valid for any surface enclosing charge q !!! -Helps calculating electric field

B of solenoids =

μ0 N/L I = μ0 n I N is number of turns L is length of solenoid n is number of loops per length

Equipotential surface (same V):

σ ' = 2 σ

ex: A thin flat membrane separates a layer of positive ions outside a cell from a layer of negative ions inside. If the electric field due to these charges is 10^7 N/C, find the charge per unit of area σ in the layers on either side of the membrane.

σ = E ε0 (10^7)(8.85 x 10^-12) *88.5 x 10^-6 C/m^2* answer is inside and outside the circle, so one is positive and one is negative

*Q =*

σA

Maximum torque for θ = 90º (I ⊥ B):

τ = I A B

*Dielectric strength is the maximum electric field allowed before the dielectric starts conducting (breakdown)*

∣V∣ = Ed C = Q/V -strength = how much change you can build up before conducting (want strength number high)