# Electric Potential Difference for a Uniformly Charged Slab

```MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Electric Potential Difference for a Uniformly Charged Slab
Consider a semi-infinite slab of charge with charge density ! that extends from x = !d to x = d .
Find the electric potential difference V (x A ) ! V (0) where x A &gt; d .
Solution: Considering symmetry, we note that the electric field at the center of the slab must be zero.
To see this, imagine putting a test charge right at the center of the slab. It will feel no net force (it
would be pushed to the right by the charge to the left exactly as much as it would be pushed to the left
by the charge to the right), so the electric field there must be zero.
The symmetry is planar so we will use Gaussian pillboxes (cylinders of cross-section
! ! A and height x)
and will place one end of the pillbox at x = 0 to take advantage of the fact that E = 0 there.
There are two distinct regions of space, inside and outside of the slab. By symmetry the magnitude of
the field will be the same on the left as on the right of the slab, but will point in the opposite direction.
We will only calculate explicitly for x &gt; 0 .
(1) Calculate the electric field for each region.
Region 1 outside the slab ( x &gt; d ): We show the Gaussian surface in the figure below.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
The charge within this pillbox is Qenc = !Venc = ! Ad . The flux (integral of the electric field over this
pillbox) is zero on the sides (because E is perpendicular to the area normal there) and zero on the left
end (because E is zero there). Thus:
!
!
! !
E! dA +
&quot;
&quot;&quot; E ! d A = &quot;&quot;
left end-cap
&quot;&quot;
! !
E ! d A = 0 + EA
right end-cap
Applying Gauss’s Law:
!
!
&quot;
&quot;&quot; E ! dA = EA =
\$d
=
%E=
#0
#0
#0
Region 2 inside the slab ( 0 &lt; x ! d ): We show the Gaussian surface in the figure below.
The charge within this pillbox is Qenc = !Venc = ! Ax . As in region 1, the flux is given by:
!
!
! !
E! dA +
&quot;
&quot;&quot; E ! d A = &quot;&quot;
left end-cap
&quot;&quot;
! !
E ! d A = 0 + EA .
right end-cap
Applying Gauss’s Law:
! !
Q
\$ Ax
\$x
E
&quot;
&quot;&quot; ! dA = EA = #enc0 = # 0 % E = # 0
Summarizing (and using symmetry to get E for x &lt; 0):
&amp;
(
(
! ((
E='
(
(
(
()
!d
î
&quot;o
for x # d
!x
î
&quot;o
for \$ d &lt; x &lt; d
\$
!d
î for x % d
&quot;o
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Note that we explicitly insert the negative sign for x outside the slab on the left, but inside the slab on
the left the negative sign of x itself takes care of the direction. Ignoring these signs is a common
source of problems – always check a few concrete cases to make sure that the field as written points in
the direction you think it should.
In order to calculate the electric potential difference V (x A ) ! V (0) , where x A &gt; d , we need to integrate
the electric field across two regions: 0 ! x ! d and d ! x ! x A . Thus
V (x A ) ! V (0) = !
x&quot;= d
\$
x &quot; =0
x &quot; = xA
!
!
!
!
E( x &quot; ) # d s&quot; + ! \$ E( x &quot; ) # d s&quot; .
x&quot;= d
!
For both regions we choose d s! = dx ! î and we use our results above to find that
x# =d
&quot; x#
V (x A ) ! V (0) = ! &amp;
î % dx# î !
\$o
x# =0
x# =x A
&amp;
x# =d
*
&quot; d
&quot;d
&quot;d ' d
=!
!
xA ! d ) =
! xA ,
(
)
\$o 2 \$o
\$o ( 2
+
2
x# =d
&quot;d
&quot; x#
î % dx# î = ! &amp;
dx# !
\$o
\$o
x# =0
x# =x A
&amp;
x# =d
&quot;d
dx#
\$o
.
```