Electric Potential of
Uniform Charge
Distributions
Part II
AP Physics C
Montwood High School
R. Casao
Potential Difference Due to a
Circular Arc of Charge





Consider a plastic rod having a uniformly
distributed charge –Q that is bent into a circular
arc of radius r.
The x-axis passes through the center of the
circular arc and the point P lies at the center of
curvature of the circular arc.
We will determine the potential difference V due
to the charged rod at point P.
The equation for arc length is: s = r·.
Divide the circular arc into small, equal pieces of
length ds.
Potential Difference Due to a
Circular Arc of Charge
Potential Difference Due to a
Circular Arc of Charge

Each length ds will contain an equal amount of
charge dq.
Q
λ
L

Q
Ls λ
s
Uniform charge density allows us to set up a
proportional relationship between Q, s, dq, and
ds:
Q dq

s ds
dq
λ
ds
Potential Difference Due to a
Circular Arc of Charge

Each length ds containing charge dq contributes
to the net electric potential at point P and can
be considered as a point charge:
k Q
V
r

becomes
k  dq
dV 
r
Keep the sign of the charge in the problem
because electric potential is a scalar quantity.
There is no direction associated with the electric
potential.
Potential Difference Due to a Circular Arc
of Charge
The distance of any element of charge dq to point P
is a constant r (the radius).
 The angle  with respect to the x-axis is different
for each element of charge dq.

s  rθ

ds  r  dθ
For each element of charge dq:
k  λ  ds
dq  λ  ds
dV 
r
k  λ  r  dθ
dV 
dV  k  λ  dθ
r
Potential Difference Due to a
Circular Arc of Charge

The net electric potential at point P is the
sum of the contribution for each element
of charge dq from one end of the circular
arc ( = - p/3) to the other end ( = p/3).
π
π
dV

k

λ

dθ


3
π
3
3
π
3
Potential Difference Due to a Circular Arc of
Charge

On the left side of the equation: the sum of the
contributions to the electric potential dV at point
P is the electric potential V.
π
 dV  V
3
π
3

On the right side of the equation:
π
π
 k  λ  dθ  k  λ   dθ
3
π
3
 k λ θ
3
π
3
π
3
π
3
Potential Difference Due to a Circular Arc of
Charge
π -π
V  k λ θ
 k λ  

3 
3
2π
π π
V  k λ     k  λ 
3
3 3
2πk  λ
V
3
π
3
π
3


Because the charge on the plastic circular arc is negative,
the electric potential V will be negative.
If the charge on the circular arc was positive, the electric
potential V would be positive.
Electric Potential of a Uniformly Charged
Disk
Electric Potential of a Uniformly Charged
Disk

Surface charge density:
Q
σ
A
Divide the disk into concentric rings which
will increase in size from the center of the
disk to the outer rim of the disk.
 r is the distance from the center of the
disk to a particular ring.
 Each ring will have a different charge,
radius, and area.

Electric Potential of a Uniformly Charged
Disk
Electric Potential of a Uniformly Charged Disk

For each ring, as the radius changes from
the center of the disk to the ring location,
so does the amount of charge on the ring
and the area of the ring.
Electric Potential of a Uniformly Charged Disk
A  πr
2

r  radius of ring

   π  2  r  dr
dA  d π  r  π  d r
dA  2  π  r  dr

2
2
For each ring:
Q dq
dq

σ
A dA
dA
dq  σ  dA dq  σ  2  π  r  dr
dq  2  π  σ  r  dr
Electric Potential of a Uniformly Charged Disk
dq is expressed in terms of dr because the
radius of each ring will vary from the center of
the disk to the rim of the disk.
 The charge within each ring can be divided into
equal elements of charge dq, which can then be
treated as point charges which contribute to the
electric field at point P.
 R is the distance from dq to point P.

Electric Potential of a Uniformly Charged Disk

For each element of charge dq:
k Q
k Q
V
becomes V 
r
R
k  dq
k  2  π  σ  r  dr
dV 
dV 
R
R
R r x
2
2
2
dV 
R  r x
2
R  r  x
2
2
2  π  k  σ  r  dr
r
2
x
2

1
2
2

1
2
Electric Potential of a Uniformly Charged Disk
To determine the electric potential at point
P, add each contribution from every
element of charge dq on the disk.
 Integrate with respect to the radius from
the center of the disk (r = 0) to the outer
rim of the disk (r = R).

dV

0
0
R
R
2  π  k  σ  r  dr
r
2
x
2

1
2
Electric Potential of a Uniformly Charged Disk

On the left side of the equation: the sum of the
contributions to the electric potential dV at point
P is the electric potential V.
 dV  V
R
0

On the right side of the equation: the 2, k, s, p,
and x are constant and can be pulled out in
front of the integral.
2  π  k  σ  0
r  dr
R
r
2
x
2

1
2
Electric Potential of a Uniformly Charged Disk

This integral has to be solved by substitution
(there is no formula for this integral on the
integration table):
– Let u = r2 + x2
– Then du = 2·r dr + 0; du = 2·r dr.
– The derivative of x2 is 0 because it is a constant and
the derivative of a constant is 0; r is a quantity that
changes.
du
du  2  r  dr
r  dr 
2
r  dr
du
1 du
  1
1
 2 2 12  
r  x 
2u 2 2 u 2
Electric Potential of a Uniformly Charged Disk
1  2
2
2
1
1
du
1
1
u
2
 1  u
 du 

2
2
2 1  2
u 2
2
2
1
2

1
1 u
1

  2  u 2  r2  x 2
2 1
2
2


1
2
The electric potential V at point P is:

V  2  π  k  σ  r  x
2
2

1

R
2
0
Electric Potential of a Uniformly Charged Disk

The electric potential is:

V  2  π  k  σ  R
V  2  π  k  σ  R
V  2  π  k  σ  R  x
2
2
x
2
2
x
2
2



1
1
1
2
 0  x
2
 x
2
x
2
2


1
2

2

1
2

Application of Gauss’s Law to
Charged Insulators

The electric field E:
E = 0 N/C at the middle of
the charged insulator.
E increases from the
middle of the charged
insulator towards the
surface.
E reaches its maximum
value at the surface of the
charged insulator.
Application of Gauss’s Law to
Charged Insulators

To move a positive test
charge from the middle of
the charged insulator to
the outer surface, the
work required increases
as the test charge moves
from the middle of the
charged insulator to the
outer surface.
Application of Gauss’s Law to
Charged Insulators
The maximum amount of
work is performed as the
test charge reaches the
outer surface.
 No additional work is
required to move the test
charge beyond the outer
surface.

Application of Gauss’s Law to
Charged Insulators

Inside the sphere:
Low voltage occurs in
center of the sphere.
High voltage occurs at the
outer surface of the sphere.
Electric field lines inside the
uniformly charged sphere
are directed radially inward
from the outer high voltage
surface to the low voltage
center.
Application of Gauss’s Law to
Charged Insulators

Electric field lines extend
radially outward from the high
voltage surface to 0 J/C at .

From  to the surface, the
work needed to move a
positive test charge from
 to the surface increases
from 0 J to a maximum
value at the surface of the
sphere.