Chapter 23

Electric PotentialLecture 5

Dr. Armen Kocharian

Electrical Potential EnergyWhen a test charge is placed in an electric field, it experiences a force

F = qoEThe force is conservativeds is an infinitesimal displacement vector that is oriented tangent to a path through space

Electric Potential Energy, contThe work done by the electric field is F.ds = qoE.dsAs this work is done by the field, the potential energy of the charge-field system is changed by ΔU = -qoE.dsFor a finite displacement of the charge from A to B,

B

B A o AU U U q dΔ = − = − ⋅∫ E s

Electric Potential Energy, finalBecause qoE is conservative, the line integral does not depend on the path taken by the chargeThis is the change in potential energy of the system

Electric PotentialThe potential energy per unit charge, U/qo, is the electric potential

The potential is independent of the value of qo

The potential has a value at every point in an electric field

The electric potential is o

UVq

=

Electric Potential, cont.The potential is a scalar quantity

Since energy is a scalarAs a charged particle moves in an electric field, it will experience a change in potential

B

Ao

UV dqΔ

Δ = = − ⋅∫ E s

Electric Potential, finalThe difference in potential is the meaningful quantityWe often take the value of the potential to be zero at some convenient point in the fieldElectric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field

Work and Electric PotentialAssume a charge moves in an electric field without any change in its kinetic energyThe work performed on the charge isW = ΔV = q ΔV

Units1 V = 1 J/C

V is a voltIt takes one joule of work to move a 1-coulomb charge through a potential difference of 1 volt

In addition, 1 N/C = 1 V/mThis indicates we can interpret the electric field as a measure of the rate of change with position of the electric potential

Electron-VoltsAnother unit of energy that is commonly used in atomic and nuclear physics is the electron-voltOne electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt

1 eV = 1.60 x 10-19 J

Potential Difference in a Uniform Field

The equations for electric potential can be simplified if the electric field is uniform:

The negative sign indicates that the electric potential at point B is lower than at point A

B B

B A A AV V V d E d Ed− = Δ = − ⋅ = − = −∫ ∫E s s

Energy and the Direction of Electric Field

When the electric field is directed downward, point B is at a lower potential than point AWhen a positive test charge moves from Ato B, the charge-field system loses potential energy

More About DirectionsA system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field

An electric field does work on a positive charge when the charge moves in the direction of the electric field

The charged particle gains kinetic energy equal to the potential energy lost by the charge-field system

Another example of Conservation of Energy

Directions, cont.If qo is negative, then ΔU is positiveA system consisting of a negative charge and an electric field gainspotential energy when the charge moves in the direction of the field

In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge

EquipotentialsPoint B is at a lower potential than point APoints A and C are at the same potentialThe name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential

Charged Particle in a Uniform Field, Example

A positive charge is released from rest and moves in the direction of the electric fieldThe change in potential is negativeThe change in potential energy is negativeThe force and acceleration are in the direction of the field

Potential and Point ChargesA positive point charge produces a field directed radially outwardThe potential difference between points A and B will be

1 1B A e

B A

V V k qr r⎡ ⎤

− = −⎢ ⎥⎣ ⎦

Potential and Point Charges, cont.

The electric potential is independent of the path between points A and BIt is customary to choose a reference potential of V = 0 at rA = ∞Then the potential at some point r is

eqV kr

=

Problem 30.

Plot a graph for potential along x direction

= eqV kr

( ) ( ) ( )( )

1 2

2 2 221 2

e ee e k Q k Qk Q k QV x

r r x a x a

+ += + = +

+ + −

( )( )

( )( ) ( )

2 2 2

2

2 2

1

2

1

e e

e

k Q k QV x

ax a x a

V x

k Q a x a

⎛ ⎞⎜ ⎟= =⎜ ⎟+ +⎝ ⎠

=+

Problem 30, cont.Now let charge at –a to be negative (-Q) and plot along y direction

= eqV kr

( ) ( ) ( )1 2

1 2

e ee e k Q k Qk Q k QV y

r r y a y a

+ −= + = +

− +

( )

( )( )

1 1

1 1

1 1

1 1

e

e

k QV y

a y a y a

V y

y a y ak Q a

⎛ ⎞= −⎜ ⎟− +⎝ ⎠

⎛ ⎞= −⎜ ⎟− +⎝ ⎠

Electric Potential of a Point Charge

The electric potential in the plane around a single point charge is shownThe red line shows the 1/r nature of the potential

Electric Potential with Multiple Charges

The electric potential due to several point charges is the sum of the potentials due to each individual charge

This is another example of the superposition principleThe sum is the algebraic sum

V = 0 at r = ∞

ie

i i

qV kr

= ∑

Electric Potential of a DipoleThe graph shows the potential (y-axis) of an electric dipoleThe steep slope between the charges represents the strong electric field in this region

Today: Electric Potential Energy

You should be familiar with the concept of gravitational potential energy from Physics 1Let's reviewIf a force acts on a particle as the particle moves from a b, then

is the work done by the force( is the infinitesimal displacement along the path)

a

b

Careful: the force does not necessarily line up with the displacement For example, a block sliding down an inclinedplane under the influence of gravity:

Conservative forceA force is conservative if the work done by the force is independent of path

Only depends on the initial and final points

Then the work done can be written as function of the difference between some properties of the begin and final point

a

b

U is the potential energyW = -ΔUWork energy theorem: work = change in kinetic energyWa→b = K(b) – K(a)K(a) + U(a) = K(b) + U(b)

Potential energy defined up to additive constant

Today: Electric Potential Energy

You should be familiar with the concept of gravitational potential energy from Physics 1Let's reviewIf a force acts on a particle as the particle moves from a b, then

is the work done by the force( is the infinitesimal displacement along the path)

U is the potential energyW = -ΔUWork energy theorem: work = change in kinetic energyWa→b = K(b) – K(a)K(a) + U(a) = K(b) + U(b)

Potential energy defined up to additive constant

Remember gravitational field, force, potential energy

Near the surface of the earth, constant force (F=const)Think of it as mass times constant gravitational fieldThen gravitational potential energyU=mgh (uniform field)Then gravitational potential energy for two masses is

= 1 2

12G

m mU Gr

Potential Energy of Multiple Charges

Consider two charged particlesThe potential energy of the system is

1 2

12e

q qU kr

=

More About U of Multiple Charges

If the two charges are the same sign, Uis positive and work must be done to bring the charges togetherIf the two charges have opposite signs, U is negative and work is done to keep the charges apart

U with Multiple Charges, finalIf there are more than two charges, then find U for each pair of charges and add themFor three charges:

The result is independent of the order of the charges

1 3 2 31 2

12 13 23e

q q q qq qU kr r r

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

Finding E From VAssume, to start, that E has only an xcomponent

Similar statements would apply to the y and zcomponentsEquipotential surfaces must always be perpendicular to the electric field lines passing through them

xdVEdx

= −

E and V for an Infinite Sheet of ChargeThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

E and V for a Point ChargeThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

E and V for a DipoleThe equipotential lines are the dashed blue linesThe electric field lines are the brown linesThe equipotential lines are everywhere perpendicular to the field lines

Electric Field from Potential, General

In general, the electric potential is a function of all three dimensionsGiven V (x, y, z) you can find Ex, Ey and Ez as partial derivatives

x y zV V VE E Ex y z

∂ ∂ ∂= − = − = −

∂ ∂ ∂

Equipotentials and ConductorsWe argued previously that the electric field must be perpendicular to the surface of a conductorThis is because otherwise the charges on the surface of the conductor would be movingIt then follows that the surface of a conductor is an equipotential

Potential GradientWe will now derive a fundamental relationship between potential and electric field

But

This must hold for any (a,b) pair and any path between the twoFor this to be true then

Write out the components

Then, in terms of components:

Suppose the displacement is in the x-directionThen dy=dz=0, and –dV=Exdx, or

Can do same thing for the other two components

Or in short-hand notation

is called the "gradient" or the "grad"

If we shift V V + Const. the E-field does not changeMakes sense, since V is only defined up-to arbitrary constantThe expression above for the gradient is in "Cartesian Coordinates"

Cartesian coordinates: x,y,z

One important result:If V is a function of r (and not of angle) then

Simple applications of gradient law

Point charge:

Infinite line of chargeSlide 16, this lecture

constant

Electric Potential for a Continuous Charge Distribution

Consider a small charge element dq

Treat it as a point charge

The potential at some point due to this charge element is

edqdV kr

=

V for a Continuous Charge Distribution, cont.

To find the total potential, you need to integrate to include the contributions from all the elements

This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions

edqV kr

= ∫

V for a Uniformly Charged RingP is located on the perpendicular central axis of the uniformly charged ring

The ring has a radius a and a total charge Q

= = =+ +

∫ ∫ ∫2 2 2 2e

e ekdq dqV k k dq

r x a x a

V for a Uniformly Charged Disk

The ring has a radiuses b and aand surface charge density of σ

( )( )−= = +

+∫ ∫

12 2 2

12 2 2

2 2a a

e eb b

rdrV πk σ πk σ x r rdrx r

( ) ( )= =

+ +1 1

2 2 2 22 2

2e ek dq k σ πrdrVx r x r

( ) ( )⎡ ⎤ ⎡ ⎤= + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 12 2 2 22 22 2e eV πk σ x a πk σ x b

V for a Uniformly Charged Disk

The ring has a radiuses a and band surface charge density of σ

( ) ( )= =

+ +1 1

2 2 2 22 2

2e ek dq k σ πrdrVx r x r

( )( )−= = +

+∫ ∫

12 2 2

12 2 2

2 2b b

e ea a

rdrV πk σ πk σ x r rdrx r

( ) ( )⎡ ⎤ ⎡ ⎤= + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 12 2 2 22 22 2e eV πk σ x b πk σ x a

V for a Uniformly Charged Disk, cont

The ring has a radius a and radius b=0. Surface charge density of σ

( )−= +∫1

2 2 2

0

2a

eV πk σ x r rdr

( )⎡ ⎤= + −⎢ ⎥⎣ ⎦

12 2 22 eV πk σ x a x

V for a Finite Line of Charge

A rod of line ℓ has a total charge of Q and a linear charge density of λ

2 2

lnek Q aVa

⎛ ⎞+ += ⎜ ⎟⎜ ⎟

⎝ ⎠

( )( )

( )

−= = + =

+

⎛ ⎞+ +⎜ ⎟⎝ ⎠

∫ ∫1

2 2 21

2 2 20 0

12 2 2ln

l l

e e

e

dx QV k λ k x r dxlx a

Qk x x al

V for a Uniformly Charged Sphere

A solid sphere of radius R and total charge QFor r > R, For r < R,

eQV kr

=

= 3e

rk QE rR

V for a Uniformly Charged Sphere

For r < R,

( )2 232

eD C

k QV V R rR

− = −

( )

=

− = − = =

−

∫ ∫

3

3

2 232

er

r re

D C rR R

e

k QE rR

k QV V E dr rdrR

k Q R rR

( ) ( )= + − = −2 2 2 23 3 3

2 2e e e

Dk Q k Q k QV R r R rR R R

= eC

k QVR

V for a Uniformly Charged Sphere, Graph

The curve for VD is for the potential inside the curve

It is parabolicIt joins smoothly with the curve for VB

The curve for VB is for the potential outside the sphere

It is a hyperbola

V Due to a Charged Conductor

Consider two points on the surface of the charged conductor as shownE is always perpendicular to the displacement dsTherefore, E · ds = 0Therefore, the potential difference between Aand B is also zero

V Due to a Charged Conductor, cont.

V is constant everywhere on the surface of a charged conductor in equilibrium

ΔV = 0 between any two points on the surfaceThe surface of any charged conductor in electrostatic equilibrium is an equipotential surfaceBecause the electric field is zero inside the conductor, we conclude that the electric potential is constant everywhere inside the conductor and equal to the value at the surface

E Compared to VThe electric potential is a function of rThe electric field is a function of r2

The effect of a charge on the space surrounding it:

The charge sets up a vector electric field which is related to the forceThe charge sets up a scalar potential which is related to the energy

Irregularly Shaped ObjectsThe charge density is high where the radius of curvature is small

And low where the radius of curvature is large

The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points

Irregularly Shaped Objects, cont.

The field lines are still perpendicular to the conducting surface at all pointsThe equipotential surfaces are perpendicular to the field lines everywhere

Cavity in a ConductorAssume an irregularly shaped cavity is inside a conductorAssume no charges are inside the cavityThe electric field inside the conductor must be zero

Cavity in a Conductor, contThe electric field inside does not depend on the charge distribution on the outside surface of the conductorFor all paths between A and B,

A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity

0B

B A AV V d− = − ⋅ =∫ E s

Corona DischargeIf the electric field near a conductor is sufficiently strong, electrons resulting from random ionizations of air molecules near the conductor accelerate away from their parent moleculesThese electrons can ionize additional molecules near the conductor

Corona Discharge, cont.This creates more free electronsThe corona discharge is the glow that results from the recombination of these free electrons with the ionized air moleculesThe ionization and corona discharge are most likely to occur near very sharp points

Millikan Oil-Drop Experiment –Experimental Set-Up

Millikan Oil-Drop ExperimentRobert Millikan measured e, the magnitude of the elementary charge on the electronHe also demonstrated the quantized nature of this chargeOil droplets pass through a small hole and are illuminated by a light

Oil-Drop Experiment, 2With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electronThe drop reaches terminal velocity with FD = mg

Oil-Drop Experiment, 3When an electric field is set up between the plates

The upper plate has a higher potential

The drop reaches a new terminal velocity when the electrical force equals the sum of the drag force and gravity

Oil-Drop Experiment, finalThe drop can be raised and allowed to fall numerous times by turning the electric field on and offAfter many experiments, Millikan determined:

q = ne where n = 1, 2, 3, …e = 1.60 x 10-19 C

Van de GraaffGenerator

Charge is delivered continuously to a high-potential electrode by means of a moving belt of insulating materialThe high-voltage electrode is a hollow metal dome mounted on an insulated columnLarge potentials can be developed by repeated trips of the beltProtons accelerated through such large potentials receive enough energy to initiate nuclear reactions

Electrostatic PrecipitatorAn application of electrical discharge in gases is the electrostatic precipitatorIt removes particulate matter from combustible gasesThe air to be cleaned enters the duct and moves near the wireAs the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles become chargedMost of the dirt particles are negatively charged and are drawn to the walls by the electric field

Application – Xerographic Copiers

The process of xerography is used for making photocopiesUses photoconductive materials

A photoconductive material is a poor conductor of electricity in the dark but becomes a good electric conductor when exposed to light

The Xerographic Process

Application – Laser PrinterThe steps for producing a document on a laser printer is similar to the steps in the xerographic process

Steps a, c, and d are the sameThe major difference is the way the image forms on the selenium-coated drum

A rotating mirror inside the printer causes the beam of the laser to sweep across the selenium-coated drumThe electrical signals form the desired letter in positive charges on the selenium-coated drumToner is applied and the process continues as in the xerographic process

Potentials Due to Various Charge Distributions

Application – Laser PrinterThe steps for producing a document on a laser printer is similar to the steps in the xerographic process

Steps a, c, and d are the sameThe major difference is the way the image forms on the selenium-coated drum

A rotating mirror inside the printer causes the beam of the laser to sweep across the selenium-coated drumThe electrical signals form the desired letter in positive charges on the selenium-coated drumToner is applied and the process continues as in the xerographic process

Problem, Method 1.

Alternative way to get term is to recognize that there are 4 side pairs and 2 face diagonal pairs.

4 sides of length s and 2 diagonals with length s√2

12 23 34 14

13 24

12 132 2

2 2

4 2

4 2 12

24 5.412

e e

e e

U U U UU UU U U

k Q k QU

s s

k Q k QU

s s

= = =

=

= +

= +

⎛ ⎞= + =⎜ ⎟

⎝ ⎠

Problem, Method 2.

Alternative way to get term is to recognize that there are 4 side pairs and 2 face diagonal pairs.

4 sides of length s and 2 diagonals with length s√2

( ) ( )1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 5.41

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

= + + +

= + + + + + +

⎛ ⎞ ⎛ ⎞= + + + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= + =⎜ ⎟⎝ ⎠

Problem.Find potential energy.

( ) ( )9 2 2 91

1

8.99 10 N m C 20 10 C4.50 kV

0.04 mek q

Vr

−× ⋅ ×= = =

( ) ( )9 3 512 2 1 10 10 C 4.5 10 V 4.50 10 JU q V − −= = × × = ×

( )

( )23 13 3 2 3 1 3 2 1

9 99 9 2 2

5 5

10 10 C 20 10 C20 10 C 8.99 10 N m C

0.04 m 0.08 m

4.50 10 J 4.50 10 J

U U q V q V q V V

− −−

− −

+ = + = +

⎛ ⎞× ×= − × × ⋅ +⎜ ⎟⎝ ⎠

= − × − ×5

12 23 13 4.50 10 JU U U −+ + = − ×

.

To place the 10-nC charge there we must put in energy

Next, to bring up the –20-nC charge requires energy

No energy is required for the 20-nC charge placed at its location

Problem, cont.Find 40 nC charge is released.

The three fixed charges create potential at the location where the fourth is released:

( )9 9 9

9 2 21 2 3 2 2

3

20 10 10 10 20 108.99 10 N m C C m

0.03 0.050.04 0.03

3.00 10 V

V V V V

V

− − −⎛ ⎞× × ×= + + = × ⋅ + −⎜ ⎟

⎝ ⎠+

= ×

Energy of the system of four charged objects is conserved as the fourth charge flies away:

( ) ( ) ( )( )

2 2

9 3 13 2

44

13

1 12 2

10 40 10 C 3.00 10 V 2.00 10 kg 0

2

2 1.20 10 J3.46 10 m s

2 10 kg

i f

m v qV m v qV

v

v

− −

−

−

⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ × × = × +

×= = ×

×

Quick Quiz 1

In the figure below, two points A and B are located within a region in which there is an electric field. The potential difference ΔV = VB – VA is

(a) positive

(b) negative

(c) zero

Answer: (b). When moving straight from A to B, E and dsboth point toward the right. Thus, the dot product E · ds is positive and ΔV is negative.

Quick Quiz 1

Quick Quiz 2

In this figure, a negative charge is placed at A and then moved to B. The change in potential energy of the charge–field system for this process is

(a) positive

(b) negative

(c) zero

Answer: (a). From ΔU = q0 ΔV, if a negative test charge is moved through a negative potential difference, the potential energy is positive. Work must be done to move the charge in the direction opposite to the electric force on it.

Quick Quiz 2

Quick Quiz 3

The labeled points of the figure below are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves along the following transitions. (a) A -> B, B -> C, C -> D, D -> E

(b) A -> B, D -> E, B -> C, C -> D

(c) B -> C, C -> D, A -> B, D -> E

(d) D -> E, C -> D, B -> C, A -> B

Answer: (c). Moving from B to C decreases the electric potential by 2 V, so the electric field performs 2 J of work on each coulomb of positive charge that moves. Moving from C to D decreases the electric potential by 1 V, so 1 J of work is done by the field. It takes no work to move the charge from A to B because the electric potential does not change. Moving from D to E increases the electric potential by 1 V, and thus the field does –1 J of work per unit of positive charge that moves.

Quick Quiz 3

Quick Quiz 4

For the equipotential surfaces in this figure, what is the approximate direction of the electric field? (a) Out of the page

(b) Into the page

(c) Toward the right edge of the page

(d) Toward the left edge of the page

(e) Toward the top of the page

(f) Toward the bottom of the page

Answer: (f). The electric field points in the direction of decreasing electric potential.

Quick Quiz 4

Quick Quiz 5a

A spherical balloon contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric potential at the surface of the balloon will

(a) increase

(b) decrease

(c) remain the same.

Answer: (b). The electric potential is inversely proportional to the radius (see Eq. 25.11).

Quick Quiz 5a

Quick Quiz 5b

Recall that the spherical balloon from part a) contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric flux through the surface of the balloon will

(a) increase

(b) decrease

(c) remain the same.

Answer: (c). Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant.

Quick Quiz 5b

Quick Quiz 6

In Figure 25.10a, take q1 to be a negative source charge and q2 to be the test charge. If q2 is initially positive and is changed to a charge of the same magnitude but negative, the potential at the position of q2 due to q1

(a) increases

(b) decreases

(c) remains the same

Answer: (c). The potential is established only by the source charge and is independent of the test charge.

Quick Quiz 6

Quick Quiz 7

Consider the situation in question 6 again. When q2 is changed from positive to negative, the potential energy of the two-charge system

(a) increases

(b) decreases

(c) remains the same

Answer: (a). The potential energy of the two-charge system is initially negative, due to the products of charges of opposite sign in Equation 25.13. When the sign of q2 is changed, both charges are negative, and the potential energy of the system is positive.

Quick Quiz 7

Quick Quiz 8

In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude that the x component of the electric field in this region is

(a) zero

(b) in the x direction

(c) in the –x direction.

Answer: (a). If the potential is constant (zero in this case), its derivative along this direction is zero.

Quick Quiz 8

Quick Quiz 9

In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is

(a) zero

(b) constant

(c) positive

(d) negative

Answer: (b). If the electric field is zero, there is no change in the electric potential and it must be constant. This constant value could be zero but does not have to be zero.

Quick Quiz 9

Quick Quiz 10

Consider starting at the center of the left-hand sphere (sphere 1, of radius a) in the figure below and moving to the far right of the diagram, passing through the center of the right-hand sphere (sphere 2, of radius c) along the way. The centers of the spheres are a distance b apart. Draw a graph of the electric potential as a function of position relative to the center of the left-hand sphere.

Answer: The graph would look like the sketch below. Notice the flat plateaus at each conductor, representing the constant electric potential inside a solid conductor.

Quick Quiz 10

Conceptual questionsfor practice

on chapter 23

The positive charge is the end view of a positively charged glass rod. A negatively charged particle moves in a circular arc around the glass rod. Is the work done on the charged particle by the rod’s electric field positive, negative or zero?

1. Positive 2. Negative3. Zero

1. Positive 2. Negative3. Zero

The positive charge is the end view of a positively charged glass rod. A negatively charged particle moves in a circular arc around the glass rod. Is the work done on the charged particle by the rod’s electric field positive, negative or zero?

Rank in order, from largest to smallest, the potential energies Ua to Ud of these four pairs of charges. Each + symbol represents the same amount of charge.

1. Ua = Ub > Uc = Ud2. Ua = Uc > Ub = Ud3. Ub = Ud > Ua = Uc4. Ud > Ub = Uc > Ua5. Ud > Uc > Ub > Ua

Rank in order, from largest to smallest, the potential energies Ua to Ud of these four pairs of charges. Each + symbol represents the same amount of charge.

1. Ua = Ub > Uc = Ud2. Ua = Uc > Ub = Ud3. Ub = Ud > Ua = Uc4. Ud > Ub = Uc > Ua5. Ud > Uc > Ub > Ua

A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton

1. moves toward A with an increasing speed. 2. moves toward A with a steady speed. 3. remains at rest at B. 4. moves toward C with a steady speed.5. moves toward C with an increasing speed.

1. moves toward A with an increasing speed.2. moves toward A with a steady speed. 3. remains at rest at B. 4. moves toward C with a steady speed.5. moves toward C with an increasing speed.

A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton

Rank in order, from largest to smallest, the potentials Va to Ve at the points a to e.

1. Va = Vb = Vc = Vd = Ve2. Va = Vb > Vc > Vd = Ve3. Vd = Ve > Vc > Va = Vb4. Vb = Vc = Ve > Va = Vd5. Va = Vb = Vd = Ve > Vc

Rank in order, from largest to smallest, the potentials Va to Ve at the points a to e.

1. Va = Vb = Vc = Vd = Ve2. Va = Vb > Vc > Vd = Ve3. Vd = Ve > Vc > Va = Vb4. Vb = Vc = Ve > Va = Vd5. Va = Vb = Vd = Ve > Vc

Rank in order, from largest to smallest, the potential differences ∆V12, ∆V13, and ∆V23 between points 1 and 2, points 1 and 3, and points 2 and 3.

1. ∆V12 > ∆V13 = ∆V232. ∆V13 > ∆V12 > ∆V233. ∆V13 > ∆V23 > ∆V124. ∆V13 = ∆V23 > ∆V125. ∆V23 > ∆V12 > ∆V13

Rank in order, from largest to smallest, the potential differences ∆V12, ∆V13, and ∆V23 between points 1 and 2, points 1 and 3, and points 2 and 3.

1. ∆V12 > ∆V13 = ∆V232. ∆V13 > ∆V12 > ∆V233. ∆V13 > ∆V23 > ∆V124. ∆V13 = ∆V23 > ∆V125. ∆V23 > ∆V12 > ∆V13

What are the units of potential difference?

1. Amperes2. Potentiometers3. Farads4. Volts5. Henrys

What are the units of potential difference?

1. Amperes2. Potentiometers3. Farads4. Volts5. Henrys

New units of the electric field were introduced in this chapter. They are:

1. V/C. 2. N/C. 3. V/m. 4. J/m2.5. W/m.

New units of the electric field were introduced in this chapter. They are:

1. V/C. 2. N/C. 3. V/m.4. J/m2.5. W/m.

The electric potential inside a capacitor

1. is constant.2. increases linearly from the negative to

the positive plate.3. decreases linearly from the negative to

the positive plate.4. decreases inversely with distance from

the negative plate.5. decreases inversely with the square of the

distance from the negative plate.

The electric potential inside a capacitor

1. is constant.2. increases linearly from the negative to

the positive plate.3. decreases linearly from the negative to

the positive plate.4. decreases inversely with distance from

the negative plate.5. decreases inversely with the square of the

distance from the negative plate.