Chapter 16 P202

electroscope

An instrument used to detect electric charge

make the force on Q3 zero. In where could place a fourth charge, Q4= -50 microCoulomb, so that the net force on Q3 wold be zero?

By the principle of superposition, we need a force in exactly the opposite direction to the resultant F due to Q2 and Q1 that we calculated. Our force must have a magnitude of 290 N, and must point down and to the left of Q3 opposite F.

k is a constant. the value is

8.99 x 10^9 N*m^2/C^2

the force between 2 charges separated by a distance d is F. If the charges are pulled apart to a distance of 3d, what is the force on each charge? original is F

F/9

example: find ratio of electric force compared to gravitation force between two equal particles

F1=F2= Gm1m2/r^2 m1=m2= 9.11 x 10^-31 kg -attractive Fe1= Fe2= Fe= kq1q1/r^2 Fe/Fg = kq1q2/ Gm1m2 q1=q2= -1.6 x 10^-19 ratio= 4.16 x 10^42 -repels -electric is much larger than gravitational

three charged particles are arranged in a line. calculate the net electrostatic force on particle 3 due to the other 2 charges (Q1)----(Q2)---(Q3) Q1: -8 microcoulombs .3 m between Q2: 3 microcoulombs .2 m between Q3: -4 microcoulombs

F= F31 + F32 F32. F31. <--(-Q3)--> F31= 1.2 N F32= 2.7 N -no y components -add x components -2.7 + 1.2 = -1.5 N (to the left)

coulomb's law: force is proportional to the charge on the charged objects and inversely proportional to the square of the distance what is the equation?

F= [k]*[ (Q1*Q2)/r^2] -magnitude only

electric force on electron by proton: determine magnitude and direction of the electric force on the electron of H exerted by a proton (q= +e). r= 0.53 x 10^-10 m

F=8.2 x 10^-8 N

practice in notes: rank the Gaussian surfaces in order of increasing electric flux, starting with the most negative:

Phi= Q/E_0 A- all encompassing Phi= 2q+-q+q/E_0 B Phi 2q+-q=q/E_0 C Phi= -q + q= 0/E_0 D Phi= -q/E_0 D<C<B<A

Gauss's Law

Phi= Q_enclosed/ E_0 Phi (flux) Q (enclosed charge) E_0 (permittivity of free space)

electric charge: symbol

Q or q

calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q= -3 x 10^-6.

The magnitude of the electric field due to a single point charge is E=k(Q/r^2) = 3x10^5 N/C direction is toward the charge Q, to the left, since we defined the fiection as that of the force on a positive test charge, to the left as shown on page 454. If Q had been positive, the electric field would have pointed away

which exerts a greater force on the other? Q1= 50 microcoulombs Q2= 1 microcoulomb

They're equal in magnitude! -Coloumb's law and Newton's third

law of conservation of electricity: the object that loses e- has an excess of + charge while the object that gains e- has

an excess of - charge -charges can be transferred but not changed total charge

coulomb's law and gravitation seem similar but differ in

attractiveness -gravitation is always attractive (toward each other)

electric field is a vector. if source charge is positive, electric field points

away from the source charge

electrons in insulators are

bound tightly to nuclei (few free e-) -think of diagram

a charged rod *hits* a ball, if the ball is a good conductor, the

charge is spread out to all over the surface of the ball

coloumb is

charge unit

the electrostatic force between A and T, and between C and G exists because these molecules have

charged parts due to some electrons in each of these molecules spending more time orbiting one atom than another. -when H+ is involved the weak bond it can make with a nearby negative charge, such as O-, is relatively strong (partly bc H+ is so small) and is referred to as a hydrogen bond

2 ways to charge:

conduction and induction

calculate the total electric field at point A due to Q1 and Q2 A |. \ (30 cm) |. \ (60 cm) Q2-----Q1. (30 degrees) (50 mC). (-50 mC).

dealing with fields instead of force Ea1= 1.25x10^6 N/C Ea2= 5x10^6 N/C Eax= 1.1x10^6 N/C Eay= Ea2-Ea1sin30= 4.4x10^6 N/C Ea= 4.5x 10^6 N/C direction: 76 degrees

By doing what will the electric field in a parallel-plate capacitor increase?

decreasing the area of each plate or by increasing the charge of each plate

if you enclose a positive charge

electric field lines point away from the charge

when you enclose a negative charge the

electric field lines point towards the charge

electric flux is a measure of the

electric field perpendicular to a surface

the conductor shields any charge within it from

electric fields created outside the conductor

Gauss's law states that the

electric flux through a closed surface is proportional to the charge enclosed by the surface -shape of surface doesn't matter

charging by rubbing: transferring --- from one object to another

electrons ---ex. plastic takes electrons from towel to become charged

Conductors and Insulators

electrons move easily in good conductors and poorly in good insulators

1 coulomb is rarely

encountered

insulators are a lattice of

lattice of + nuclei w small negative clouds -tight e- ex. rubber, wood

charging by rubbing: charge only hold for a limited time, excess charge ----------. ----------, and then are transferred back to the positively charged object

leaks onto polar water molecules; e- leak onto the positive dipole of H2O in the air

dont forget to add from missing part

lecture 2 and book problems

what is the electric *field* at the center of the square when there are only two charges (-2C) top left and (-2C) bottom left? up from center, down, right, left, or zero

left! -towards the negative charges E produced by charge one (top left) has a vector from the center pointing top left E produced by charge two (bottom left) has a vector from the center pointing bottom left sum: middle to the left; vertical cancels

electrons in metals (conductors) are

loosely bound cannot "leave" per se -"free electrons" -think of diagram

Calculate the net electrostatic force on charge Q3 due to charges Q1 and Q2. (pg 452 in book) Q3 (+65 mC) | \ | 30 degrees Q2---------------Q1 (+50 mC) (-86 mC)

magnitudes: F31= 140 N F32= 325 N F31x= 140cos30= 120N F31y= -70N F32 y= 325 N Fx= 120 N Fy= 255 N F= 280 theta= 2.13/ 65 degrees

F=ma; force causes ---, moon orbits Earth (centripetal a), if v is changing-> a force is acting

motion

when a rod is brought close to a positively charged scope, the leaves get closer. what charge is the rod?

negative

a metal ball hangs from the ceiling by an insulating thread. the ball is attracted to a positive-charged rod held near the ball. the charge of the ball must be:

negative or neutral -separation of charge allows for attraction (neutral)

the electric *field* is always ---- to the *surface* of a conductor- if it weren't, the charges would move along the surface

perpendicular -if not, charges would be moving up and down the conductor and would contract that at equilibrium they're at rest -inside still zero

example: glass rods when rubbed becomes () and a plastic ruler when rubbed becomes ()

positive (glass), negative (plastic)

2 types of charge:

positive and negative

grounding: electroscope is wired to earth, electrons flow from the electroscope to the earth; positive charge ---

positive charge goes to metal far from the wall; negative charge goes to side with wire and flow to earth

conductors are a lattice of

positive ions in a big negative cloud made of many electrons -loose e- ex. human body, earth, metals

disconnecting the ground wire leaves the electroscope

positively charged (uniform spread)

electric charge is

quantized (discrete)

Between the red (+2) and blue charge (+1) which experiences a greater electric FORCE due to the green charge (+1)

red I think Force = q* E-field

if electric charge is put on one point of a conductor, it will

repel and place itself as far on the surface so that its distributed -charges at rest on equilibrium

charges that are in equilibrium are at

rest

objects can become charged from

rubbing ("electric charge"... static/shocks)

each type of charge repels the --- and attracts a ---

same type; different type (unlike attract)

if there's a positively charged rod touched to electroscope (conduction),

scope takes up same charge as rod (all over) -some e- transfer to metal rod -both end up positive

if there's a positively charged rod put near neutral electroscope (induction),

separation of charge occurs -negative charge in metal ball -positive charge on leaves and bottom of middle rod

electric forces are vectors: have direction that depends on ----

signs of charges -like repel; different attract -action-reaction forces -opposite directions + same magnitude

semi-conductors have

some free e-

place a small test charge at area of interest and determine the force that acts on it by the

source charge

amber effect

static electricity, the ancient Greeks called it this

can electric flux be negative

yes, depending on the angle chosen -so two thetas on each side of the normal

Conductors and shielding: at equilibrium, the electric field is ---- at any point *within* a conducting material

zero

a dipole composed of two charges of +Q and -Q separated by a 2a and enclosed within a Gaussian surface that is a cylinder of a diameter 2r and length L as shown in the figure below. The *total* electric flux through the surface is

zero -video on phone -one side has a +Q charge, so the flux points out -one side has -Q charge so flux points in -same magnitude and opposite directions so 0 -shape doesn't matter

Law of conservation of electric charge 1. Net amount of electric charge produced in any process is -- 2. no net electric charge can be ----

zero, be created nor destroyed

a proton and electron are released at a distance of 1m. What happens to the force between them as they approach each other?

it gets bigger

case 1: source charge is a point charge: can use coulomb's law to determine the force then use

E=force/q0 or E=k|q1|/r^2 to determine the field created by the charge

electric field also equals

E=k|q1|/r^2 -q1 is the source charge

extreme case where theta = 90

EAcos(90)=0 -lines are parallel to surface

another way to use an electroscope: 1. give scope a positive charge ---leaves are repelling 2. bring unknown rod close to ball (not touching/induction) -if positive, -if negative,

-if positive, leaves go further (pushes more charge towards leaves) -if negative, leaves get closer (negative charge moves up toward metal ball and less negative in leaves)

to be safe during a lighting storm it is best to be -in a grassy meadow -tall tree -metal car -wood building

-metal car

when a separation of charge occurs, can treat each side like a different "q" (positive and negative). these qs have different distances from the rod, so can use coulomb's law to find the net force

-net force depends on the different distances *look in notebook at figure*

example with a wire cage, electroscope and acrylic rod

-rubbing the rod and placing it near the scope pushes the leaves apart -putting a cage around the scope then allows the leaves to come together showing the inside is zero -shielding

smallest charge is of one electron; its called the elementary charge. what is the value?

1.6022 x 10^-19 Coulomb's notice e is positive so the charge of the electron is -1.0622 x 10^-19

you are sitting a distance from a pt charge, and you measure an electric field of E_0. If the charge is doubled and the distance is doubled, what is the magnitude of the field now? 4E_0, 2E_0, E_0, 1/2E_0, 1/4E_0

1/2 E_0 E_0= k|q|/r^2 E'= k|2q|/(2r^2)

E_0 (permittivity)

1/4*pi*k =8.85 x 10^-12 C^2/N*m^2 (k= 8.99x10^9 Nm^2/C^2)

in terms of permittivity of free space (curly E0), E=

1/4piE_0 (Q/r^2)

atoms are the smallest constituent of regular matter which is

10^-10 m

atoms have -- types of charge

2 (makes neutral)

electric field

E=Force/q0 (q0 is a test charge, it serves to allows the electric force to be measured, but it is not large enough to create a significant force on any other charges)

phi_e=

E*A*cos(theta) (N*m^2/C) E (mag of field) A (area of surface) Theta (angle between field lines and where the lines would be perpendicular to the surface) -same thing as E_perpendicular * A or E*A_perpendicular

extreme case where theta = 0

E*A*cos(theta) -it's perpendicular -maximum number of possible lines cross the area

Based on Gauss's law, we can find the the electric field in the parallel-plate capacitor with equal and opposite charges:

E=1/E_0 * Q/A = sigma/ E_0 Q-> total charge on each plate A-> area of each plate sigma= Q/A -> surface charge density

Fg=

Gm1m2/r^2 G= 6.673 x 10^-11 N*m^2/kg^2 -due to mass

E at a point between two charges. Two point charges are separated by a distance of 10 cm. One has a charge of - 25 microcoulombs and the other +50 microcoulombs. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2 cm from the negative charge. (b) if an electron mass = 9.11 x 10^-31 kg is placed at rest at P and then released, what will be its initial acceleration and magnitude. (pg 455)

The electric field at P will be the vector sum of the fields created separately by Q1 and Q2. The field due to the negative charge Q1 points toward Q1, and the field due to the positive charge Q2 points away from Q2. Thus both point to the left. E= k(q1/r^2) + k(q2/r^2) = 6.3 x 10^8 F=ma: a=F/m= qE/m= 1.1 x 10^20 m/s^2

a thin spherical shell of radius r0 possess a total net charge Q that is uniformly distributed on it. determine the electric field at points (a) outside the shell, and (b) inside the shell

field must be symmetric bc charge is distributed symmetrically. the field outside the shell must be directed radially outward (inward if Q<0) and must depend only on r. choose an imaginary gaussian surface as a sphere of radius r (r> r0) concentric with the shell. then, by symmetry the electric field will have the same magnitude at all points on the gaussian surface. bc E is perpendicular to this surface, Gauss's law gives (Qencl=Q) sigmaEperp delta A= E sigma delta A= E(4pir^2) = Q/E_0 where 4pir^2 is the surface area of our sphere gaussian surface of radius r this E=1/4piE_0 (Q/r^2) [r>r0] we see that the field outside a uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge inside the shell the electric field must also be symmtric. so e must again have the same valu eat all points. thus E can be factored out of the sum and with Qencl-0 bc the charge inside surface A2 is zero we have sigmaEperp*delta A= E(4pir^2)= Qencl/E0= 0 E=0

explain what "F12" means

force on q1 by q2

the direction is always along the line joining the 2 charges, but the way it faces depends on whether the

forces repel or attract

effects of charge: generate --- and cause ---

forces, motion

the surface we use to relate the electric field to the electric flux is the

gaussian surface

electroscope: greater the charge,

greater the separation of leaves

a neurtal hollow box is placed between two parallel charged plates. what is the field *inside* the box?

if our metal box has been solid, not hollow, free electrons in the box would have redistributed themselves along the surface until all their individual fields would have canceled each other inside the box. the net field inside the box would have been zero. for a hollow box, the external field is not changed since the electrons in the metal can move just as freely as before to the surface. Hence, the field inside the hollow metal box is also zero. a conducting box is an effective device for shielding delicate instruments and electronic circuits from unwanted external electric fields.

a photocopy machine works by arranging positive charges (in the pattern to be copied) on the surface of a drum, then gently sprinkling negatively charged dry toner (ink) particles onto the drum. The toner particles temporarily stick to the pattern on the drum and are later transferred to paper and "melted" to produce the copy. Suppose each toner particle has a mass of 9 x 10^-16 kg and carries an average of 20 extra electrons to provide an electric charge. Assuming that the electric force on a toner particle must exceed twice its weight in order to ensure sufficient attraction, compute the required electric field strength near the surface of the drum.

the electric force on a toner particle of charge q= 20e is F=qE, where E is the needed electric field. This force needs to be at least as great as twice the weight (mg) of the particle. qE= 2 mg E= 2mg/q = 2(9x10^-16)(9.8)/20(1.6x10^-19)= 5.5 x 10^3 N/C ?question about signs here

consider the 2 charges shown: +q1= + 3.2 microC and -q2= -1.6 microC -correctly describe the magnitude of the electric force?

the force on q1 has the same magnitude as the force on q2 -action-reaction forces

positively charged metal object is put near neutral object: what happens

the neutral rod stays neutral overall but there's a separation of charge so negative charges over by the positive rod *charge by induction*

positively charged metal object touches neutral object: what happens

the objects end up to the same charge -there's a passage of some e- from the neutral object to the charged *charge by conduction*

Between the red (+2) and blue charge (+1) which experiences a greater electric FIELD due to the green charge (+1)

the same for both -field only depends on source charge (green +1 for both) -however, if distances were different, it would change - E-field= k*(source charge)/ d^2

Conductors and shielding: at equilibrium under electrostatic conditions, any excess charge resides on

the surface of a conductor

if several charges are present, the net force on any of them will be

the vector sum of the force due to each of the others -principle of superposition

if you rub a rod with a towel then place it over pith balls, the balls stick to the ruler. this means

there must be a force acting on the balls (electric)

two charged balls repel each other; what can you say about their charges?

they're either both + or both -

electric flux through an area is essentially the

total number of field lines crossing the area

if the source charge is -, field points

toward the source charge