Electric Potential of Uniform Charge Distributions

Electric Potential of Uniform Charge Distributions

AP Physics C

Montwood High School

R. Casao

Electric Potential Due to a Charged Rod on the Axis of the Rod


• Divide the length of the rod L into small elements of length dx. – This also divides the total charge Q on

the rod into small elements of charge dq.

– Over the length L of the rod, the charge on each small piece of length dx is dq.

• The charge density on the rod is:

L

Qλ


• Because the charge is uniformly distributed over the length of the rod, we can set up a proportion relating the total charge per unit length to the charge per unit length for each small piece of the rod.

• Each piece of length dx contains a charge dq and can be considered to be a point charge.

dx

dq

L

Q


• Each element of charge dq contributes to the net electric potential V at point P1.

• Since the charge on the rod is positive, the net electric potential V at point P1 is positive.

• The distance from point P1 to each piece dx is x.

• Electric potential equation for a point charge:

xQk

VxrrQk

V


• The electric potential contribution from each element of charge dq is designated dV.

• To determine the total electric potential V at point P1, add the electric potential contribution from each element of charge dq from the left end of the rod to the right end of the rod.– Integrate from d at the left end of the rod to

(d + L) at the right end of the rod. In other words, the electric field contributions begin at d and end at (d + L).

– This means that dq needs to be expressed in terms of dx because we will integrate along the x-axis from d to (d + L).


• For each point charge dq:

x

dxλkdV

x

dqkdV

dxλdqdx

dqλso

dx

dq

L

Qand

L

Qλ

x

dqkdVbecomes

x

QkV


• Integrate both sides of the equation from the left end of the rod at x = d to the right end of the rod at d + L.

• On the left side of the equation: the sum of the contributions to the electric potential dV at point P1 is the electric potential V.

Ld

d

Ld

d xdxλk

dV

Ld

d

Ld

d xdx

λkx

dxλkV


• From a table of integrals (or your TI – 89):

• Completing the integral:

xlnx

dx

d

LdlnλkV

dlnLdlnλkV

xlnλkV Ld

d


• If Q is positive, will be positive and V will be positive.

• If Q is negative, will be negative and V will be negative.

Electric Potential Off the Axis of a Finite Line of Charge


• Divide the length of the rod L into small elements of length dx.– This also divides the total charge Q on

the rod into small elements of charge dq.

– Over the length L of the rod, the charge on each small element of length dx is dq.

• The charge density on the rod is:

L

Qλ


• Because the charge is uniformly distributed over the length of the rod, we can set up a proportion relating the total charge per unit length to the charge per unit length for each small piece of the rod.

• Each element of length dx contains a charge dq that can be considered to be a point charge.

• Each element of charge dq contributes to the net electric potential V at point P2.

dxdq

λdxdq

LQ


• Electric potential equation for a point charge:

• The electric potential contribution from each element of charge dq is designated dV:

rdqk

dV

rQk

V


• At point P2, the net electric potential V will be the sum of the electric potential contribution dV of each element of charge dq from the left end of the rod (x = 0) to the right end of the rod (x = L).


• For each element of charge dq from x = 0 to x = L, the values of V, r, and all change as x changes. The relationships between these variables must be determined.– Because electric potential is a scalar

quantity and not a vector quantity, we do not have to worry about the angle for the direction. We only have to concern ourselves with the distance r from element of charge dq to point P2.


• Integrate from 0 at the left end of the rod to L at the right end of the rod. In other words, the electric potential contributions begin at 0 and end at L.

• This means that dq needs to be expressed in terms of dx because we will integrate along the x axis as the distance x changes from 0 to L.

2122

22222

yx:asrewrite

yxryxr


• Integrate both sides of the equation from 0 to L:

21

22 yx

dxλkdV

rdxλk

dVbecomesrdqk

dV

dxλdqanddxdq

λ

L

0 2122

L

0yx

dxλkdV


• On the left side of the integral: adding each contribution dV from the left end of the rod to the right end of the rod gives us the total potential V:

• Pull the constants k and out in front of the integral sign and integrate. The y2 term cannot be removed from the integration because it is being added to the x2 term.

VdVL

0


• From a table of integration:

L

0 2122 yx

dxλkV

xyxln

yx

dx

uaulnau

du

21

22

21

22

21

22

21

22


• Replacing this in the equation:

yLyL

lnλkV

ylnLyLlnλkV

ylnLyLlnλkV

0y0nlLyLlnλkV

xyxlnλkV

21

22

21

22

21

221

22

21

2221

22

L

0

21

22


• Should the point P lie along a line that is a perpendicular bisector of the line of charge as shown, you would integrate from the left end of the rod at x = -x to the right end of the rod at x = x.

x

x- 2122

x

x-yx

dxλkdV


xyx

xyxlnλkV

yxlnxyxlnλkV

yxlnxyxlnλkV

-yx-nlxyxlnλkV

xyxlnλkV

2122

2122

21222

122

21222

122

21

222122

x

x-

2122

x

x

x

Electric Potential of a Uniform Ring of Charge


• Consider the ring as a line of charge that has been formed into a ring.– Divide the ring into equal elements of length dx

containing an element of charge dq; each element of charge dq is the same distance r from point P.

– Each element of charge dq can be considered as a point charge which contributes to the net electric potential at point P.

– Each element of charge dq is the same distance r from point P.

212222

222

xaxar

xar


• Because electric potential is a scalar quantity, we do not need to consider the vector components from the elements of charge dq on the opposite side of the ring (180° away) as is required when determining the electric field.

• For each element of charge dq:

2122 xa

dqkdV

r

dqkdVbecomes

r

QkV


• The total electric potential V can be found by adding the contribution to the electric potential by each element of charge dq.

• Integrate around the circumference of the ring:

• K, a, and x are constants and are pulled out in front of the integral sign.

2122 xa

dqkdV


• Left side of the integral: adding up all the contributions to the electric potential from each element of charge dq gives us the total electric potential.

dqxa

kdV

2122

VdV


• Right side of the integral:

• The electric potential at point P is:

21222

122 xa

QkQ

xa

k

Qdq

2122 xa

QkV

Electric Potential of Uniform Charge Distributions

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