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General Physics IISpring 2008

Electric Potential

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Electric Potential Energy• Potential energy changes when an

object that interacts with other objects via a conservative force changes its position.

• For example, the gravitational potential energy of the book and Earth increases when you (external agent) do positive work to lift the book against the conservative gravitational force (its weight).

• The electric force is conservative. Thus, one can change the electric potential energy of a system of charges by doing work on them.

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Electric Potential Energy• Imagine pushing a positive charge at

constant speed toward a fixed positive charge. You would do positive work (force in the same direction as motion), with the electric (repulsive) force opposing the motion. This positive work is equal to the increase in the electric potential energy of the system of two charges:

• Since the work done is proportional to the force and the (electric) force is proportional to the charge, then the work done, and hence the change inpotential energy, are proportional to the charge.

++

.elecW U=Δ

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Electric Potential Energy• Like the gravitational potential energy, only changes in the

electric potential energy are significant. Thus, one can choose the zero of electric potential energy at any convenient location. (Remember that you could choose the zero of gravitational potential energy at any convenient position, e.g., the floor, table top, etc.)

• Let us choose Uelec = 0 at some point A. Then the change in potential energy in moving a charge from point A to another point B is

• Thus, when we speak of the potential energy at a single point B, what we really mean is the potential energy at point B relative to the point where we have chosen the potential energy to be zero.

, , , .elec elec B elec A elec BU U U UΔ = − =

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Electric Potential• Recall that the electric force on a probe charge at a point is

proportional to the probe charge. When we divide the force by the charge, we get the electric field, which is independent of the probe charge. The field is only dependent on the source charges producing it.

• We saw that the electric potential energy is proportional to the probe (movable) charge. If we divide the potential energy by the probe charge, we obtain a quantity, the electric potential, which is independent of the probe charge:

• Like the electric field, the electric potential is only dependent on the source charges, not the probe charge. The electric potential exists at a point whether a probe charge is there or not!

, or, .elecelec

UV U qVq= =

Electric Potential

• For the source charges shown, the potential at A is less than the potential at B. For a positive charge that moves from A to B, the potential energy increases (Recall: U = qV).

• However, for a negative charge going from A to B, the potential energy decreases. (the potential energy is zero at A and negative at B.

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Conservation of Mechanical Energy• If a charge moves under the influence of electric forces

alone, then the total mechanical energy of the charge is conserved:

• This applies to any number of charges moving under the influence of electric forces.

, , , or,

.

initial elec initial final elec fin

i

l

f f

a

i

K U K U

K qV K qV

+ = +

+ = +

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Conservation of Mechanical Energy• It is also useful to restate the conservation of mechanical

energy in terms of differences: the change in the total mechanical energy is zero.

• Thus, if the potential energy increases, the kinetic energy must decrease by the same amount.

• For charges moving under the influence of electric forces,

• Note that

( ) 0, i.e., 0.K U K UΔ + = Δ +Δ =

( ) 0, or,0.

KK q V

qVΔ +Δ

Δ+

=Δ =

and . i if fK K K V V VΔ = − Δ = −

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Group Problem-SolvingA proton has a speed of 3.5 x 105 m/s at a point where the electrical potential is 600 V. It moves through a point where the electric potential is 1000 V. What is its speed at this second point?

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.

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Electric Potential and Parallel-Plate Capacitors

• Since the electric field inside the capacitor is uniform, the force on a probe charge q is constant ( ).

• If probe charge moves at constant velocity, the net force is zero. Thus, Fhand = Felec.

• If x is the displacement from the negative plate and the then the work done by the hand is

• W = Fhand x = Felec x = qEx.Now, W = ΔUelec, so ΔUelec = qEx.Further, ΔV = ΔUelec/q. Hence,

ΔV = Ex.

F qE=

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Electric Potential and Parallel-Plate Capacitors

• Now, ΔV = V(at x) – V(at 0). If we choose V(at 0) = 0, the we have V = Ex.

• Since E is constant, the potential is proportional to x. Every point that has the same value of x has the same potential. All points having the same value of x – and therefore the same potential – lie on a plane surface parallel to the plates of the capacitor. Such imaginary surfaces on which all points are at the same potential are called equipotentialsurfaces.

• Note that the potential increases as you go from the negative plate to the positive plate.

• Going from the negative plate to the positive plate is in the opposite direction of the electric field. In other words,thepotential always decreases when you move in the direction of the electric field.

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Electric Potential and Parallel-Plate Capacitors

• If the potential difference between the plates of the capacitor is ΔVC and the distance between the plates is d, then

ΔVC = Ed.

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Checking UnderstandingRank in order, from largest to smallest, the electric potentials at the numbered points.

a) 1 = 2, 3, 4 b) 3 = 4, 1 = 2c) 3, 2, 1, 4d) 1 = 2, 3 = 4e) 1, 2, 3, 4

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Potential Energy of Two Point Charges• A single point produces an electric field and electric potential

in all the space around it. Another point charge brought into this space will form a system that has potential energy. The potential of two point charges is given by

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Potential of a Point Charge• Consider the charge q to be the source charge and q’ the

probe charge. The potential at the position of the probe charge is given by V = Uelec/q’ = Kq/r. Note that the potential is independent of the probe charge and is dependent only on the source charge. Thus, the electric potential of a point charge q is

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Workbook: Chapter 21, Question 3

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Two probe charges are brought separately into the vicinity of a charge +Q. First, test charge +q is brought to a point a distance r from +Q. Then this charge is removed and test charge -q is brought to the same point. Which system has the greater electrostatic potential energy?

1. System A.2. System B.3. It is the same for both.

System A System B

Checking Understanding

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Potential of Many Point Charges• The potential at a point P in space due to many point

charges is just the sum of the individual potentials due to each point charge.

• In the above equation, q1 is the first point charge and r1 is the distance between q1 and the point P, etc.

• Potential is a scalar quantity and so adding different potentials together is as simple as adding numbers. The potential is negative for a negative charge and positive for a positive charge.

31 21 2 3

...qq qV K r r r⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= + + +

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Checking Understanding

Rank in order, from greatest (most positive) to smallest, the electric potentials at the numbered points.

1. 1 = 2, 3 2. 3, 1 = 23. 3, 2, 14. 1 = 2, 3 = 45. 1, 2, 3

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Textbook Chapter 21,Problem 48

P21.48. Prepare: The net potential is the sum of the scalar potentials due to each charge. Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this point, V1 + V2 = 0 V.

Solve: Using Equation 21.9:

14πε 0

3 .0 × 10−9 Cx

+−1.0 ×10 −9 C

x − 4.0 cm

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 0 V ⇒ −x + 3 x − 4.0 cm = 0 cm

Either −x + 3(x − 4.0 cm) = 0 cm, or−x + 3(4.0 cm − x) = 0 cm. In the first case, x = 6.0 cm and, in the second case, x = 3 cm. That is, we have two points on the x-axis where the potential is zero. Assess: Both values seem reasonable.

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Relating Electric Field and Potential• Consider an equipotential

surface. Since there is zero potential difference between any two points on the surface, there is no change in potential energy when a probe charge is moved between the two points. Thus, no work is done.

• Since no work is done, the electric force acting on the probe charge must be zero and so there cannot be an electric field parallel to the surface.

• It follows that the electric field must be perpendicular to an equipotential surface at every point on the surface.

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Relating Electric Field and Potential• We saw that for a capacitor,

ΔVC = Ed,where d is the distance between the plates.

• This relationship is also true for any two closely spaced equipotentialsurfaces. Thus, the electric field between the surfaces has a magnitude given by

where ΔV is the potential difference between the surfaces and d is the separation distance.

• The direction of the field is from higher to lower potential.

,VE dΔ=

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Conductors• Under electrostatic equilibrium, the electric field inside a

conductor is zero. Thus, no work is done in moving a probe charge between any two points within the volume of the conductor (including the surface). Thus, the change in potential energy is zero and hence the change in potential must be zero.

• It follows that the entire conductor is an equipotentialregion.

• Since the surface is also an equipotential surface, it follows that the electric field must be perpendicular to the surface (as we saw in the previous chapter).

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A. What is the potential at point (a)? At which point, (a), (b), or (c), does the electric field have its largest magnitude?

B. Is the magnitude of the electric field at (a) greater than, equal to, or less than at point (d)?

ExampleSource charges create the electric potential shown.

C. What is the approximate magnitude of the electric field at point (c)?

D. What is the approximate direction of the electric field at point (c)?

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Capacitors• A capacitor consists of two conducting

electrodes or plates, separated by an insulator. Initially, we will take the insulator to be vacuum.

• Assume the plates are neutral at first. Then charge is moved from one plate to the other. Thus, the plates always have charges of equal magnitude but opposite sign. This process is called charging the capacitor.

• As the magnitude of the charge on each plate increases, the magnitude of the electric field and the potential difference between the plates both increase.

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Charging a Capacitor• Start with uncharged

capacitor. Electrons from lower plate flow toward the higher potential positive terminal of the battery.They are then "pumped" through the battery and deposited on the upper plate.

• As more charge builds up, the incoming charges are repelled more and more by the charge already on the plate so the flow slows.

positive

Positive

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Charging a Capacitor• As the charge on each plate

increases, the potential difference increases. Eventually, the potential difference between the plates equals the potential difference of the battery. Charge flow stops and the capacitor is said to be fully charged.

• If the battery is disconnected, the charge will remain on the plates. Thus, a capacitor stores charge.

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Capacitance• The charge on each plate of a

capacitor is proportional to the potential difference. The ratio of charge to potential difference is called the capacitance of the capacitor:

.C

QC V=Δ

+Q -Q

+

+

+

+

-

-

-

-

ΔVC• The capacitance depends onlyon geometric factors such as the area of a plate and the distance between the plates. As the potential difference changes, the charge changes to keep the ratio (C) the same.

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Parallel-Plate Capacitor• For a parallel-plate capacitor, the capacitance is given by

0 .AC dε

=

• Recall that for a parallel-plate capacitor, the electric field is uniform and only depends on the charge per unit area on each plate (E=Q/(ε0A). Hence, for an isolatedcapacitor, the electric field remains the same if the distance between the plates changes.

• Since ΔVC = Ed, then for an isolated capacitor, ΔVC is proportional to d.

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Workbook, Chapter 21 Question 22

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Dielectrics• A dielectric is an insulating material. As we

saw in the previous chapter, insulators can be polarized by an electric field. This behavior can be used to increase the capacitance of a capacitor with vacuum between the plates.

• A dielectric is placed between the plates of a fully charged isolated capacitor. The electric field between the plates polarizes the dielectric. The separated charges produce an electric field (positive to negative) that opposes the original electric field. Hence, the net electric field between the plates is reduced. The factor by which the field is reduced is called the dielectric constant:

.EEκ =′

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Dielectrics• Since ΔVC = Ed, if E is reduced, then ΔVC is reduced by the

same factor (κ). Because the capacitor is isolated, the charge on it stays the same when the dielectric is inserted. (No where for it to go!). Thus, the charge does not change, but ΔVC decreases by a factor of κ. Thus, the capacitance (C = Q/ ΔVC) increases by a factor of κ:

00 . (parallel-plate capacitor)AC C d

κεκ= =

• If instead the battery remains connected to the capacitor, the potential difference ΔVC always stays the same. In this case, the charge on the capacitor plates increases when the dielectric is inserted, which results in the same increase in capacitance seen above.

• Textbook, Chapter 21, Problem 35.

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Energy in Capacitors• To charge a capacitor, positive charges must be

separated from negative charges. To do this, work must be done by an external agent, e.g., a battery. This work is appears as potential energy stored in the capacitor. This energy is calculated as

2)1 ( .2 CU C V= Δ

• Using Q = CΔVC, the energy stored can be rewritten as22)1 1 1( .2 2 2C C

QCU C V Q V= Δ = = Δ

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Energy in Capacitors• One can rewrite the energy stored in terms of the

electric field. One finds that2

0Energy Stored 1 .2VolumeEu Eκε= =

• For a capacitor with vacuum between the plates,2

0Energy Stored 1 .2VolumeEu Eε= =

• Even though the volume between the plates is empty of material, there is an electric field present. Thus, one can think of the energy as being stored in the electric field.

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Workbook: Chapter 21, Question 23