Electric Field of a Uniform Ring of Charge

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Electric Field Calculations for
Uniform Ring of Charge and
Uniformly Charged Disk
Montwood High School
AP Physics C
R. Casao
Electric Field of a Uniform
Ring of Charge
Montwood High School
AP Physics C
R. Casao

Consider the ring as a line of charge that
has been formed into a ring.
 Divide the ring into equal elements 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 field at
point P.


At point P, the electric field contribution
from each element of charge dq can be
resolved into an x component (Ex) and a
y component (Ey).
The Ey component for the electric field
from an element of charge dq on one
side of the ring is equal in magnitude but
opposite in direction to the Ey component
for the electric field produced by the
element of charge dq on the opposite
side of the ring (180º away). These Ey
components cancel each other.

The net electric field E lies completely
along the x-axis.
Ex
cos θ 
E

E x  E  cos θ
Each element of charge dq can be
considered as a point charge:
E
k Q
r
2
so : dE x 
becomes dE 
k  dq
r
2
 cos θ
k  dq
r
2

cos q can be expressed in terms of x and r:
x
cos θ 
r


k  dq x
dE x  2 
r
r
dE x 
k  x  dq
r
3
The total electric field can be found by adding
the x-components of the electric field
produced by each element of charge dq.
Integrate around the circumference of the ring:

dE x 

k  x  dq
r
3


is the symbol for integrating around

a closed surface.
Left side of the integral: adding up all the
little pieces of dEx around the
circumference gives us Ex (the total
electric field at the point).


dE x  E x
Right side of the integral: pull the
constants k, x, and r out in front of the
integral sign.



k  x  dq
r
3
dq  Q

kx
r
3
so :

 dq
k x Q
r
3
However, r can be expressed in terms of
the radius of the ring, a, and the position
on the x-axis, x.
r a x
2
2
2

r  a x  a x
2
2
2
2

1
2

Combining both sides of the integration
equation:
Ex 

k x Q

 2
2
a

x


p. 652 #31, 37.

1
2


3

a
k x Q
2
x
2

3
2
MIT Visualizations



URL:
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/electrost
atics/index.htm
The Charged Ring
Integrating Around a Ring of Charge
Electric Field 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.


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.
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
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 (see the ring problem).
 Point charge equation:

k Q
E 2
r

The distance from the point charge to the point
P (r) was labeled as L in the picture.
k Q
E 2
L

The contribution of each element of charge dq
to the net electric field at point P is:
k  dq
dE  2
L
k  2  π  σ  r  dr
dE 
2
L

At point P:
The y-components for each opposite
charge dq cancels; only the x-components
contribute to the net electric field at point
P.
 This is true for every ring.
 The net electric field is given by:

Ex
cos θ 
E

Substitute:
dE x 
E x  E  cos θ
2  π  k  σ  r  dr
2
L
 cos θ

Express the cos q in terms of the
variables x and r. L is the distance
from dq to point P.
L r x
x
cos θ 
L
2
2
2

L r x
L r x
x
cos θ 
2
2
r x
2
2
2
2

1
2
dE x 
dE x 
2  π  k  σ  r  dr
r x
2
2

r
x
2
x
2

1
2
2  π  k  σ  x  r  dr
r
2
x
2

3
2
Integrate with respect to the radius from
the center of the disk (r = 0) to the outer
rim of the disk (r = R).
 The 2, k, s, p, and x are constant and
can be pulled out in front of the integral.


R
0

dE x  2  π  k  σ  x 
r  dr
R
 r
0
2
x
2

3
2
Left side of the equation: adding all the
x-components together gives us the net
R
electric field, Ex.
dE x  E x

0

Right side of the equation: this integral
has to be solved by substitution (there is
no formula for this integral on the
integration table):

Substitution method:
Let u = r2 + x2
 Then du = 2·r dr + 0; du = 2·r dr.
2
 The derivative of x 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



3
3
3
2
2
2 2
2
2
2

u
u
r x




u
du
3
u
3
2
 du 
2
3  2
u 2 2
3
2
2
1
u 2
2
2
 1 
1
2
2
2
u
r x
2



1
2
Pull the ½ back into the equation:
1

2
r
2
2
x
2

1

2
r
1
2
x
2

1
2
2

So:
R
Ex
 1 
 2πk σx 
1 
2
2 2
r  x  0


1
Ex  2 π  k σ  x  
 2 2
 R x


1


1 
2
2 2
0 x

 
1
2



1
Ex  2 π  k σ  x  
 2 2
 R x



1
Ex  2 π  k σ  x  
 2 2
 R x


1 

1 
2 2
x

  
1

1
2
2

1

x


For problems in which x is very small in
comparison to the radius of the disk (x
<< R), called a near-field approximation:
Ex  2  π  k  σ
p. 652, #33, 34, 37 (2nd half)
 Homework 3, #5

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