electric potential of uniform charge distributions
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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. 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. - PowerPoint PPT PresentationTRANSCRIPT
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
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
Electric Potential Due to a Charged Rod on the Axis of the Rod
• 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
Electric Potential Due to a Charged Rod on the Axis of the Rod
• 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
Electric Potential Due to a Charged Rod on the Axis of the Rod
• 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).
Electric Potential Due to a Charged Rod on the Axis of the Rod
• For each point charge dq:
x
dxλkdV
x
dqkdV
dxλdqdx
dqλso
dx
dq
L
Qand
L
Qλ
x
dqkdVbecomes
x
QkV
Electric Potential Due to a Charged Rod on the Axis of the Rod
• 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
Electric Potential Due to a Charged Rod on the Axis of the Rod
• From a table of integrals (or your TI – 89):
• Completing the integral:
xlnx
dx
d
LdlnλkV
dlnLdlnλkV
xlnλkV Ld
d
Electric Potential Due to a Charged Rod on the Axis of the Rod
• 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
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
Electric Potential Off the Axis of a Finite Line of Charge
• 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 Off the Axis of a Finite Line of Charge
• Electric potential equation for a point charge:
• The electric potential contribution from each element of charge dq is designated dV:
rdqk
dV
rQk
V
Electric Potential Off the Axis of a Finite Line of Charge
• 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).
Electric Potential Off the Axis of a Finite Line of Charge
• 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.
Electric Potential Off the Axis of a Finite Line of Charge
• 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
Electric Potential Off the Axis of a Finite Line of Charge
• 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
Electric Potential Off the Axis of a Finite Line of Charge
• 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
Electric Potential Off the Axis of a Finite Line of Charge
• From a table of integration:
L
0 2122 yx
dxλkV
xyxln
yx
dx
uaulnau
du
21
22
21
22
21
22
21
22
Electric Potential Off the Axis of a Finite Line of Charge
• 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
Electric Potential Off the Axis of a Finite Line of Charge
• 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
Electric Potential Off the Axis of a Finite Line of Charge
Electric Potential Off the Axis of a Finite Line of Charge
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
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
Electric Potential of a Uniform Ring of Charge
• 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
Electric Potential of a Uniform Ring of Charge
• 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
Electric Potential of a Uniform Ring of Charge
• 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
Electric Potential of a Uniform Ring of Charge
• Right side of the integral:
• The electric potential at point P is:
21222
122 xa
QkQ
xa
k
Qdq
2122 xa
QkV