SERWAY 23.26
The E-field along the axis of a uniformly charged disk of radius R and total charge Q was calculated in Example 23.8. Show that the E-field at distances x that are large compared with R approaches that of a particle with charge Q =
σπR 2
.
Suggestion: First show that x
 x
2 
R
2

1
2

1
   n 
1
 n

when
 
1 .



1

R
2 x
2


 1
2 and use the binomial expansion
First, what’s the point of this exercise? It’s simple. The farther you get from a charged disk, the more it starts to appear as a point charge. So, when you get far away, the Efield you perceive should look like the E-field of a point charge. Since the disk has a uniform surface charge density of σ, and an area of πR 2 , then the total charge on the disk is just given by:
Q
 
A
 
R
2 .
So, when we get far away from the disk on the x axis, the E-field should look like:
(1) E
Point Charge
 k e
Q r 2
 k e

R x 2
In Example 23.8, we showed that for points on the x axis, the E-field of a charged disk looks like:
(2) E x

2
 k e




1
 x
 x
2 
R
2

1
2



We’re going to make some approximations to show that Eqn. (2) reduces to Eqn. (1) when we’re really far down the x axis.
Let’s start with the nasty part of Eqn. (2) first… x
 x
2 
R
2

1
2

 x
2


1
 x
R
2 x
2



1
2
 x x


 1

R
2 x
2


1
2



1

1
R
2 x
2


1
2



1

R
2 x
2


 1
2

So, Eqn. (2) can be rewritten as:
(2) E x

2
 k e




1


 1

R
2 x
2


 1
2



Great. Now, let’s apply an approximation by using the binomial expansion. The binomial expansion says that, as long as
 
1 , then

1
   n 
1
 n

.
In our case, we’re dealing with


1

R x down the x axis), and n = -1/2. Thus:
2
2


 1
2
, so
 
R
2 x
2

1 (because we’re really far


1

R
2 x
2


 1
2
R
2

1

2 x
2
for x >> R.
Plugging this back into (2), we obtain:
E x

2
 k e



1

1

R
2
2 x
2



2
 k e

R
2
2 x
2
E x
 k e

R
2 x
2
, which is what we set out to prove.