-Electric Potential due to
Continuous Charge Distributions
AP Physics C
Mrs. Coyle
Electric Potential –What we used so far!

Electric Potential

Potential Difference

Potential for a point
charge

Potential for multiple
point charges
U
V
qo
B
U
V 
   E  ds
A
qo
q
V  ke
r
qi
V  ke 
i ri
Remember:

V is a scalar quantity

Keep the signs of the charges in the equations,
so V is positive for positive charges.

You need a reference V because it is changes in
electric potential that are significant. When
dealing with point charges and charge
distributions the reference is V=0 when r 
Electric Potential Due to a
Continuous Charge Distribution
How would you calculate
the V at point P?
Two Ways to Calculate Electric Potential
Due to a Continuous Charge Distribution

It can be calculated in
two ways:

Method 1: Divide the
surface into infinitesimal
elements dq

Method 2:If E is known
(from Gauss’s Law)
   E  dA  q /  o
dq
V  ke 
r
B
U
V 
   E  ds
A
qo
Method 1

Consider an infinitesimal
charge element dq and
treat it as a point charge

The potential at point P
due to dq
dq
dV  ke
r
Method 1 Cont’d

For the total potential, integrate to include the
contributions from all the dq elements
dq
V  ke 
r

Note: reference of V = 0 is when P is an
infinite distance from the charge distribution.
Ex 25.5 : a) V at a point on the perpendicular
central axis of a Uniformly Charged Ring
Assume that the total
charge of the ring is Q.
Show that:
dq
V  ke 

r
keQ
x 2  a2
Ex 25.5: b) Find the expression for the
magnitude of the electric field at P

Start with V 
and
dV
E x 
dx
kQx
Ans: E x  2
( x  a 2 )3/ 2
Note that at the center of the ring E=0.
How else had we calculated this result?
k eQ
x 2  a2
Ex 25.6: Find a)V and b) E at a point
along the central perpendicular axis of a
Uniformly Charged Disk

Assume radius a and
surface charge density of
σ. Assume that a disk is
a series of many rings
with width dr.

Ans.:V  2πkeσ  x 2  a 2



1
2

 x

Ex 25.6: Find a)V and b) E at a point
along the central perpendicular axis of a
Uniformly Charged Disk

Start with V  2πkeσ  x 2  a2



Ans: E x  2 k (1 
x
1
2

 x

x2  a2
)
Ex25.7: Find V at a point P a distance a
from a Finite Line of Charge


Assume the total charge of
the rod is Q, length l and a
linear charge density of λ.
Hint:  dx  ln( x  x 2  a 2 )
x2  a2
Ans:V 
keQ
 
ln 


 a2 


a

2
Method 2 for Calculating V for a
Continuous Charge Distribution:

If E is known (from Gauss’s Law)
   E  dA  q /  o

Then use:
B
U
V 
   E  ds
A
qo
Ex 25.8: Find V for a Uniformly Charged
Sphere (Hint: Use Gauss’s Law to find E)


Assume a solid
insulating sphere of
radius R and total
charge Q
For r > R,
Q
Ans : V  ke
r
Ex 25.8: Find V for a Uniformly Charged
Sphere


A solid sphere of
radius R and total
charge Q
For r < R,

keQ 2
2
Ans : VD  VC 
R

r
2R 3

Ex 25.8:V for a Uniformly Charged
Sphere, Graph


The curve for inside
the sphere is
parabolic
The curve for
outside the sphere is
a hyperbola