PHY112 M

What is a dielectric? What is the purpose of a dielectric?

-A dielectric is the insulator placed between the two conducting plates of a capacitor -Dielectrics will increase the overall capacitance of the capacitor -allow the capacitor components to be rolled up, and put into smaller containers rather than relying on parallel plate designs

What is the significance of the spacing between two equipotential lines?

-Voltage changes equally from one equipotential line to the next (just like elevation changes equally from one line on a topographical map to the next) -Closely spaced lines show that voltage changes very quickly relative to position. Widely spaced lines show that voltage changes slowly relative to position.

what is the difference between electric fields and electric force?

-electric fields are a way of visualizing the space near a single charge -electric force is a push or pull between to charges

what is the difference between potential and potential energy?

-electric potential (voltage) provides info about various positions near a point source charge (hypothetical) -potential energy is the specific amount of electric energy a particular charged particle has stored due to work done as the particle moves through an electric field

Explain why large plate areas and small plate separation distances result in larger capacitances

-large plate areas allow charge to spread out, which reduces repelling static electric forces and allows for more charge to be added -small distances allow for greater static electric forces between the two charged plates. The attracting force between the oppositely charged plates allows more charge to be added since the attracting force can be used to counter some of the repelling force experienced between the charges on the same plate

why are gravity forces only attractive, but electrical forces are attractive and repulsive?

-mass can only interact in one way (attractive) -there are 2 types of charge and the different combos of charge type lead to both attractive and repulsive forces

what is the difference between uniform and non-uniform fields?

-non-uniform fields are created by point sources --the direction and the size of the field depend on how far away from the source you measure -uniform fields are created by 2 oppositely charged objects -- have the same strength everywhere

Where does positive and negative voltages occur?

-positive voltage near positive point sources -negative voltage near negative point sources

equation to use for a uniform field vs nonuniform field

-uniform: E = F / q -non-uniform: E = kq/r^2

Two charged particles, each with a charge of +8.0 μC are at some position near each other. What is the distance between the particles if the static electric force between them is 5.0 N?

0.34 m

Three charged particles are held in place. Particle A has a net charge of +2 μC and is at (0, 0) m. Particle B has a net charge of -4 μC and is placed a (0.5, 0) m. Particle C has a net charge of -4 μC and is at (0, 0.5) m. What is the magnitude of the net force on particle A?

0.41 N

What is the strength of an electric field measured 0.5 m from a point source with a net charge of 5.0 μC?

1.8 x 10^5 N/C use E=kq/r^2

Charged particles A and B experience a static electric force of F. If the distance between the charge particles triples, then the static electric force becomes

1/9 F because, (⅓)^2= 1/9 F

Three capacitors, each with a capacitance of 40 μF, are wired together in series. What is the overall capacitance?

1/C = 1/C + 1/C + 1/C 1/C = 1/(40 μF)x3 C = 13.3 μF

Three charged particles are held in place. Particle A has a net charge of +2 μC and is at (0, 0) m. Particle B has a net charge of +4 μC and is placed a (0.5, 0) m. Particle C has a net charge of -4 μC and is placed at (0, 0.5) m. What is the angle of direction of the net force acting on particle A?

135°

Consider a uniform electric field of 50 N/C directed towards the east. If the potential measured at some point in the field is 80 V, what is the potential at a position 1.0 m directly east of that point?

30 V

Two positively charged particles experience a static electric force of F. One of the particles is replaced with a new charged particle that has three times more charge than the original. The new static electric force will be

3F

A positive particle begins at a position near a negative plate and ends at a position near a positively charged plate. Which statement is most accurate? a. Work is done on the particle so that the particle will end with more energy. b. Work is done by the particle so that the particle will end with more energy. c. Work is done on the particle so that the particle will end with less energy. d. Work is done by the particle so that the particle will end with less energy.

A

Explain how a negative charge can have negative energy but a positive potential

A negative charge would lose energy moving from a position infinitely far away to a position near a positive source charge. As a result, the negative charge would have negative energy. However, since potential is a function of the source charge and not the test charge, the test charge near a positive source charge would be at a positive potential

Explain why combining capacitors together in series results in an overall smaller capacitance

As the distance between plates increases, the attraction between plates decreases, which means the repelling forces between the excess charge particles will become more significant and the same applied voltage will have a harder time pushing charge onto the plate and capacitance will decrease

Which of the following statements regarding energy and potential is accurate? a. Energy is a vector measurement, but potential is scalar. b. Energy is a scalar measurement, but potential is vector. c. Both energy and potential are vector measurements. d. Both energy and potential are scalar measurements.

C

A positive particle is moved towards a positive source. A negative particle is also moved towards a positive source. In each case, a)the positive particle moved to a higher potential while the negative moved to a lower potential. b)the positive particle moved to a lower potential while the negative moved to a higher potential. c)both particles moved to a higher potential. d)both particles moved to a lower potential.

C The potential has nothing to do with the test charge. Potential depends only on the distance from the source charge. Positions closer to the positive source have a greater potential than positions far away. Since both particles moved towards the source, both particles moved to positions with higher potential

Three capacitors, each with a capacitance of 40 μF, are wired together in parallel. What is the overall capacitance?

C = C + C + C C = (40 μF)x3 C = 120 μF

Two identical parallel plate capacitors are wired together in series. The overall capacitance is 10 μF. How much charge can be stored in the capacitors if they are connected to a 9.0 V battery? What is the individual capacitance of one of the capacitors?

C = Q / V (10 μF) = Q / (9.0 V) Q = 90 μC 1/C = 1/C + 1/C 1/(10 μF) = 2 (1/C) C = 20 μF

A capacitor with a capacitance of 25 μF is connected to a 9V battery. How much charge will be stored in the capacitor?

C = Q/V (25x10^-6 F) = Q / (9 V) Q = 2.25x10^-4 C

How much excess charge could be stored in a capacitor if it has a plate area of 0.05 m2, a plate separation distance of 3.0 mm, has a dielectric with a dielectric constant of 5, and is connected to a 12 V battery?

C = Q/V = kεoA/d Q / (12 V) = 5 (8.85x10^-12 F/m)(0.05 m^2) / (3x10^-3 m) Q = 8.85 x 10^-9 C

An electronics designer needs a parallel plate air capacitor that can store 0.25 mC of charge using a voltage of 25 V. If the distance between the plates is 12 cm, what must the area of the plates be? How reasonable is this design?

C = Q/V = εoA/d (0.25x10^-3 C) / (25 V) = (8.85x10^-12 F/m)(A) / (0.12) A = 1.36x10^5 m2 This is an extremely large area so this is not very reasonable.

A parallel plate air capacitor has a plate area of 0.10 m2 and a plate separation distance of 0.02 m. What is the capacitance of the capacitor? How much charge can be stored in the capacitor if it is connected to a 12 V battery?

C = εA/d C = (8.85E-12 F/m)(0.1 m^2) / (0.02 m) C = 4.4E-11 F This is a small capacitance. C = Q / V (4.4E-11 F) = Q / (12 V) Q = 5.3E-10 C This is a small amount of charge, which is not surprising given the small capacitance.

A parallel plate air capacitor has a plate separation distance of d (2 mm), and the plate area measures L (0.4 m) by W (0.5 m). What is the capacitance of the capacitor How much charge can this capacitor hold if connected to a 12V battery?

C = εA/d C= (8.85 x 10^-12)(.2)/(2 x 10^-3) C= 8.85 x 10^-10 F C=Q/V (8.85 x 10^-10)= Q/12 Q= 1.06 x 10^-8 C

How is a capacitor made? What is a capacitor used for?

Capacitors are made by alternating layers of conductors and insulators. They are used to store charge and energy.

A metal sphere with a net charge of +5 μC is brought into contact with a second, identical metal sphere that has a net charge of -3 μC. What is the final charge on both spheres after they are brought into contact?

Charge will conduct between the two metal spheres until they have the same total charge. Negative charge will continue to move off the negative sphere until they both reach a final charge of +1 μC

A proton is placed into a uniform electric field with a field strength of 100 N/C. What is the electric force on the proton?

E = F / q (100 N/C) = F / (1.6x10^-19 C) F = 1.6x10^-17 N

A particle with a net charge of -6 nC is placed in a uniform electric field with a field strength of 500 N/C. What is the magnitude of the force on the particle?

E = F / q (500 N/C) = F / (6 x 10^-9 C) F = 3 x 10^-6 N

Particle A has a net charge of +15 nC, and particle B has a net charge of -10 nC. The two particles are placed 1.0 m apart with particle A on the left and B on the right. What is the net field strength measured 0.25 m to the left of particle A?

E = kq / r2 EA = (9x10^9 N-m2/C2)(15x10^-9 C) / (0.25 m)^2 EA = 2160 N/C EB = (9x10^9 N-m2/C2)(10x10^-9 C) / (1.25 m)^2 EB = 57.6 N/C There are no y components so the net field strength will be the sum of the x components. Note the field created by particle A points to the left so it should have a negative value. ΣE = ΣEx = EA + EB ΣE = (-2160 N/C) + (57.6 N/C) ΣE = -2102 N/C

What is the electric field strength of a -8 nC charged particle at a position measured 0.5 meters away?

E = kq/r2 E = (9x10^9 N-m2/C2)(8x10^-9 C) / (0.5 m)^2 E = 288 N/C

The electric field strength measured a distance r from a point source is determined to be some value E. By what factor will the field strength change if the field is measured at a point 2r?

E is inversely proportional to r^2 E will decrease by a factor of 4 The field at the new position will be ¼ E

If the amount of charge on a particle doubles, what will happen to the field strength measured at some distance r from the particle?

E is proportional to q If q doubles, E will also double at all locations

A particle with a net charge of +1 μC is placed in a uniform electric field that has a field strength of 1000 N/C. What is the force on the particle?

E= F/q 1000 N/C= F/ 1 x 10^-6 F= 1 x 10^-3 N

What is the electric field strength at a position measured at 2 m from a 4.0 mC point source charge? If a particle with a charge of 4E -3 C is placed at the position indicated in the previous question, what is the magnitude of the force on the particle?

E=kg/r^2 E= (9 x 10^9)(4 x 10^-3)/2^2 E= 9 x 10^6 N E= f/q 9 x 10^6= F/ 4 x 10^-3 F= 3.6 x 10^4

A parallel plate air capacitor has a capacitance of C (20 μF). The charge on each plate is Q (40 μF). There are no batteries connected to the capacitor. How much work is required to double the separation distance between the plates?

E_cap=Q^2/2C E_cap= (40 x 10^-6)^2/2(20 x 10^-6) E_cap= 4 x 10^-5 J

what is the difference between electric field lines and equipotential lines?

Electric field lines are drawn in the direction of the force the source charge would put on a positive test charge (i.e., field lines are drawn away from positive sources and towards negative sources). Equipotential lines are drawn perpendicular to the electric field lines

An electron is released from rest from a position 2 cm from a -5 nC source charge. What is the speed of the electron when it has reached a position 3 cm from the source charge?

Energy start = energy end Uinitial = Ufinal + K 1/2mv2 = (kqq/r)initial - (kqq/r)final ½(9.11x10^-31 kg)(v2) = (9x10^9 N-m2/C2)(-1.6x10^-19 C)(-5x10^-91.62 C)(1/0.02 m - 1/0.03 m) v = 1.62x107 m/s

Two charged particles each with a net charge of -q are a distance r apart. The electric force F is measured. The distance between the particles then doubles to 2r, and the amount of charge on BOTH charges doubles to -2q. Relative to the original force, what is the new electric force?

F the double charge and double distance cancel each other out

Particle A has a net charge of -8.2 μC, and is placed 0.5 m to the left of Particle B, which has a net charge of +25 nC. What is the static electric force between the two? Which applies the largest force, Particle A on B or Particle B on A?

F = kqq/r^2 F = (9 x 10^9 N-m^2/C^2)(8.2 x 10^-6 C)(25 x 10^-9 C) / (0.5 m)^2 F = 7.38 x 10^-3 N attractive equal force b/c Newton's 3rd law

T or F protons have a charge of +1 C

F: 1.6 x 10^-19 C

T or F if an object has a neutral charge, it contains no positive or negative charge

F: it contains equal amounts of positive and negative

A Styrofoam ball suspended from a string is placed into an electric field with a field strength of 2x10^3N/C. The ball has a mass of 5.0 grams and a net charge of +6 μC. What is the tension in the string?

Fg = mg Fg = (0.005 kg)(9.8 N/kg) Fg = 0.049 N E = Fe / q (2x10^3 N/C) = F / (6x10^-6 C) Fe = 0.012 N FT2 = Fg2 + Fe2 FT2 = (0.049 N)^2 + (0.012 N)^2 FT = 0.0504 N tanθ = (0.049 N) / (0.012 N) θ = tan-1 (4.08) θ = 76.2° above the -x axis

A particular resistor has 15 C of charge pass through it every 20 seconds. If the potential difference across the resistor is 12 V, what is the resistance of the resistor?

I = Q / t I = (15 C) / (20 seconds) I = 0.75 A V = IR (12 V) = (0.75 A)(R) R = 16 Ohms

A particular wire has 0.5 A of current. How many electrons will pass through the wire in one minute?

I = Q/t (0.5 A) = Q / (60 sec) Q = 30 C Number electrons = total charge / charge on 1 electron (30 C) / (1.6x10-19 C) = 1.88x1020 electrons

Consider a uniform electric field of 50 N/C directed towards the right. If the potential measured at some point in the field is 80 V, what is the potential at a position 1.0 m directly to the right of that point?

If the field is directed to the east, then the positive plate is on the left and the negative plate is on the right. Since you're looking at a point to the right of the given point, you should expect the voltage to be smaller (the position is closer to zero). Solve for the position of the 80 V potential reading. V = Ue / q = Er (80 V) = (50 N/C)r r = 1.6 m Solve for the voltage with an additional 1.0 m subtracted from that position. V = Er V = (50 N/C)(1.6 - 1.0) V = 30 V

A proton is accelerated from rest a distance of 0.5 m through a voltage difference of 1000 V. What is the final speed of the proton when it leaves the electric field?

K = ΔUe 1/2mv2 = qV ½ (9.11 x 10^-31 kg)(v^2) = (1.6-19 C)(1000 V) v = 1.87 x 10^7 m/s

Explain why combining capacitors together in parallel results in an overall larger capacitance

Linking capacitors in parallel is similar to increasing the total area of the top and bottom capacitor plates -makes it possible to collect more excess charge on a plate since they have more room to spread out

The units used to measure the strength of an electric field are

Newtons per Coulomb (N/C)

How much work is done on a charged particle if it is moved along an equipotential line? Does it matter if the particle has a net positive or negative charge? Explain

No work is done moving a charged particle along an equipotential line. There is no change in voltage, which means the energy per charge ratio (definition of voltage) remains constant. Voltage is a determined by the source charge and not the charge of the particle in the electric field so it does not matter if the test charge is positive or negative.

In equipotential drawings, how do positive and negative point sources act?

Positive point sources act like electrical hills, and negative point sources act like electrical valleys

What is the difference in overall capacitance if three 5 μF capacitors are wired together in series versus parallel?

Series: 1/C = 1/C1 + 1/C2 + 1/C3 1/C = 1/(5 μF) + 1/(5 μF) + 1/(5 μF) 1/C = 0.6 C = 1.67 μF Parallel: C = C1 + C2 + C3 C = (5 μF) + (5 μF) + (5 μF) C = 15 μF

T or F all matter carries both type of charge

T

T or F charge is the fundamental property of matter that is responsible for electrical effects

T

T or F the total amount of charge present must remain constant for any process

T

A positively charged plastic balloon filled with air will stick to a neutrally charged nonmetal ceiling. Which of the following is the most likely explanation for this behavior?

The balloon caused the ceiling to polarize, which created a net attracting force between the polarized ceiling and the balloon

Particle A is placed at position (-2, 2) m, particle B is placed at (2, 2) m, particle C is placed at (-2, -2) m, and particle D is placed at (2, -2) m. Solve for the net electric field strength at position (0, 0) if particles A and C have a charge of -5 nC, and particles B and D have a charge of +10 nC

The distance between any of the particles and the origin is: r^2 = 2^2 + 2^2 r = 2.83 m The electric fields for Ed and Eb will have the same strength. The electric fields for Ec and Ea will have the same strength. The field strength for Ea is half the field strength of Ed. E = kq / r^2 E = (9 x 10^9 N-m^2/C^2)(10 x 10^-9 C)/ (2.83 m)^2 ED = EB = 11.25 N/C EA = EC = ½ ED EA = EC = ½ (11.25 N/C) EA = EC = 5.63 N/C All of the y components cancel out due to symmetry so the total field strength will be equal to the sum all four x components. Note that all four components are in the negative direction. ΣE = ΣEx = 2 (-5.63 N/C)(cos 45) + 2 (-11.25 N/C)(cos 45) ΣE = -23.9 N/C at 180°

Particle A has a net charge of +5.5 μC, and is located at (0,0) m. Particle B has a net charge of -2 μC, and is located at (2, 2) m. Particle C has a charge of -2 μC, and is located at (-2, 2) m. What is the net charge on particle A?

The distance between particles C and A is: r2 = 2^2 + 2^2 r = 2.83 m Due to symmetry, the force of C on A is the same force as the force of B on A. F = kqq/r^2 F = (9 x 10^9 N-m^2/C^2)(2 x 10^-6 C)(5.5 x 10^-6 C) / (2.83 m)^2 F = 0.0124 N The x components of the forces cancel out so the net force will be equal to the sum of the two y-component forces. ΣF = ΣFy = (0.0124 N)(sin 45) ΣF = 8.75 x 10^-3 N at 90°

Two charged particles have opposite net charges. They are placed near each other, and released from rest. The particles will...

The particles will accelerate towards each other with an increasing acceleration acceleration increases b/c as the particles move closer together, the force increases which also increases the acceleration F=ma

Two charged particles both have positive net charges. They are near each other and released from rest. Which of the following statements best describes the interaction of the particles?

The particles will repel. As they get further apart, the repelling force will decrease

Particles A and B both have a net charge of +q, and are separated by distance d. Explain how static electric force would change if the charge on particle B is increased to +3q and the distance is decreased to 1/3 d

The x3 charge will cause the force to triple. The 1/3 distance will cause the force to change by a factor of 1/(1/32) = 9. Combined the effects will be 3/9 = 1/3 F. The force will decrease by a factor of 3.

Particle A has a net charge of -5 μC. Particle B has a net charge of +10 μC. Which of the following is the largest force?

They apply the same force on each other

A capacitor with a capacitance of 15 mF is connected to a 12 V battery. How much energy will be stored in the capacitor?

U = ½ qV U = ½ (CV)V U = ½ (15E-3 F)(12 V)^2 U = 1.08 J

A capacitor with a 25 μF capacitor is connected to a 12 V battery. How much energy is stored in the capacitor?

Uc = ½ qV = ½ (CV)V Uc = ½ (25x10^-6 F)(12 V)2 Uc = 1.8x10^-3 J

An electron accelerates from rest through a uniform electric field with a voltage difference of 12 V. What is the speed of the electron as it leaves the electric field?

Ue = K Vq = 1/2mv^2 (12 V) (1.6x10^-19 C) = ½ (9.11x10^-31 kg)v2 v =2.05x10^6 m/s

A proton starts from rest, and is accelerated through a uniform electric field. If the proton reached a velocity of 4x105 m/s after traveling 1 cm in the field, what is the field strength?

Ue = K qEr = 1/2mv2 (1.6x10^-19 C)(E)(0.01 m) = ½(1.67x10^-27 kg)(4x10^5 m/s) E = 0.21 N/C

What is the electric potential energy of a particle with a net charge of +1.5 μC located at a position 20 cm from an object with a net charge of +3 mC?

Ue = kqq/r Ue = (9x10^9 N-m2/C2)(1.5x10^-6 C)(3x10^-3 C) / (0.2 m) Ue = 202.5 J

How much electric potential energy does a -7.3 nC charged particle have at a position 2.0 m from a +0.05 C source?

Ue = kqq/r Ue = (9x10^9 N-m2/C2)(7.3x10^-9 C)(0.05 C) / (2.0 m) Ue = 1.64 J However, when a negative charge moves towards a positive source, the charge will lose electric potential energy. To move the negatively charged particle from infinitely far away to the given position, the particle would have to lose 1.64 J of energy. The final answer is then: Ue = -1.64 J

How much energy does it take to move a particle with a charge of +2 nC, from an initial position infinitely far from a second particle with a charge of +3 mC, to a final position 0.20 m from the second particle?

Ue=kqq/r 0.27 J

A parallel plate air capacitor has an initial voltage difference of 12 V and a plate separation distance of 15 cm. A dielectric is inserted between the two plates. If the dielectric constant is 3.4, what is the new electric field strength between the plates? There are no voltage sources connected to the plates

V = Er (12 V) = E(0.15 m) E = 80 N/C Adding a dielectric with no external constant voltage source will cause the field strength to decrease by a factor equal to the dielectric constant. Can you explain why? E = (80 N/C) / 3.4 E = 23.5 N/C

Two parallel plates separated by a distance of 20 cm are connected to a 9 V battery. What is the electric field strength between the plates? What is the potential at a point halfway between the two plates?

V = Er (9 V) = E (0.2 m) E = 45 N/C V = Er V = (45 N/C)(0.1 cm) V = 4.5 V

Two charged parallel plates have a uniform field strength of 1200 N/C between them. If the separation distance between the plates is 0.15 m, what is the potential difference between the plates?

V=Er 180 V

If the rms voltage of a standard U.S. wall outlet is 120 V, what is the peak voltage?

Vrms = Vo /√2 (120 V) = Vo / (√2) Vo = 170 V

How much work is required to move a 1.0 g particle with a net charge of +3 μC to a point halfway between two oppositely charged parallel plates separated by a distance of 0.5 m? The electric field between the plates is a uniform 900 N/C.

W = Ue W = qEr W = (3x10^-6)(900 N/C)(0.25 m) W = 6.75x10^-4 J

induction is...

a charge transfer process that involves polarizing an object and then allowing charge to transfer to another object through conduction. While still under the influence of the dipole, the charge transfer mechanism is removed (the two objects are moved apart). The two objects are now charged

charge is...

a property of matter that causes electrical effects

polarization is...

a redistribution of charge rather than a transfer of charge; the total amount of charge on the polarized object does not change

Which of the following statements regarding charge is most accurate? -Charge is a fundamental property of matter. -Protons and electrons carry equal amounts of charge. -Two types of charge are possible. -All of the above are accurate.

all the above

conductors...

allow electrons to move easily through them

A particular object has a net 6.5 μC of the positive type of charge. Were electrons transferred onto or off the object? How many?

because the object has a positive charge, electrons were transferred off the material 6.5 μC= 6.5 x 10^-6 C q_e= 1.6 x 10^-19 6.5 x 10^-6/1.6 x 10^-19= 1.06 x 10^13 electrons removed

how is charge transferred?

by the moving of protons and electrons

what makes a good capacitor?

can hold a lot of excess charge with very little electrical push

electric fields are created by

charge

What must happen for a material to gain a net charge?

charge must be transferred from one object to another

the total charge present in the system and surroundings is...

constant in any process

what causes static electric forces?

created by the interaction of charge

what is a capacitor?

device used to store electrical charge

insulators...

do not allow electrons to move freely

When drawing the equipotential 3D maps, positive point sources will form

hills

Which of the following processes will NOT result in a change in net charge on an object? -Conduction -Induction -Polarization -all the above

polarization

Equipotentials are best defined as

positions in an electric field where multiple positions have the same voltage

with a positive test charge, what is the direction of electrical fields on a positive and negative source

positive source charges will always create fields directed away while negative source charges will always create fields directed towards the charge

in regards to static electric charge, force is...

proportional to the inverse square distance between the charged particles

Particle A is placed at position (3, 3) m, particle B is placed at (-3, 3) m, particle C is placed at (-3, -3) m, and particle D is placed at (3, -3) m. Particles A and B have a charge of -5 μC and particles C and D have a charge of 9 μC Solve for the magnitude and direction of the net electric field strength at position (0, 0) m

r^2= 3^2+3^2 r^2= 18 EA=EB= kq/r^2 =(9 x 10^9)(5 x 10^-6)/18 =2.5 x 10^3 N/C EC=ED= (9 x 10^9)(9 x 10^-6)/18 =4.5 x 10^3 N/C Enet= (EA+EB+EC+ED) Sin(theta) =2 (2.5+4.5) x 10^3 Sin(45) =9.899 x 10^3 N/C in the north direction

Electric potential is

the ratio of energy per amount of charge on a particle in an electric field

conduction is...

the transfer of charge between objects