Simple Nature - Light and Matter

j/Applying Gauss’ law to an

infinite charged surface.

shown in the figure. The length,L, of the surface is irrelevant.

The field is parallel to the surface on the end caps, and therefore

perpendicular to the end caps’ area vectors, so there is no contribu-

tion to the flux. On the long, thin strips that make up the rest of the

surface, the field is perpendicular to the surface, and therefore paral-

lel to the area vector of each strip, so that the dot product occurring

in the definition of the flux isEj·Aj=|Ej||Aj||cos 0◦=|Ej||Aj|.

Gauss’ law gives

4 πkqin=

∑

Ej·Aj

4 πkλL=

∑

|Ej||Aj|.

The magnitude of the field is the same on every strip, so we can
take it outside the sum.

4 πkλL=|E|

∑

|Aj|

In the limit where the strips are infinitely narrow, the surface be-

comes a cylinder, with (area)=(circumference)(length)=2πRL.

4 πkλL=|E|× 2 πRL

|E|=

2 kλ

R

Field near a surface charge
As claimed earlier, the resultE = 2πkσfor the field near a
charged surface is a special case of Gauss’ law. We choose a Gaussian
surface of the shape shown in figure j, known as a Gaussian pillbox.
The exact shape of the flat end caps is unimportant.

The symmetry of the charge distribution tells us that the field

points directly away from the surface, and is equally strong on both

sides of the surface. This means that the end caps contribute equally

to the flux, and the curved sides have zero flux through them. If the

area of each end cap isA, then

4 πkqin=E 1 ·A 1 +E 2 ·A 2 ,

where the subscripts 1 and 2 refer to the two end caps. We have
A 2 =−A 1 , so

4 πkqin=E 1 ·A 1 −E 2 ·A 1

4 πkqin= (E 1 −E 2 )·A 1 ,

and by symmetry the magnitudes of the two fields are equal, so

2 |E|A= 4πkσA

|E|= 2πkσ

Section 10.6 Fields by Gauss’ law 649