Physics 1B: EField from a PointSource Integral
EField of a Plane of Charge
1
Problem
Using the remotepointsource expression for the electric ﬁeld (rather than Gauss’s law), solve for
the electric ﬁeld everywhere surrounding an inﬁnite plane with surface charge density
σ
.
2
Quick Solution
By symmetry,
E
only points toward or away from the plane, and it can only depend on
z
.
E
(
z
) =
K
e
Z
distr.
d
q
0
R
2
ˆ
R
=
K
e
Z
distr.
d
q
0
R
3
R
=
K
e
Z
plane
d
q
0
R
3
R
z
ˆ
z
=
K
e
Z
∞
0
σ
2
πs
0
d
s
0
h
s
0
2
+
z
2
i
3
/
2
z
ˆ
z
= 2
πK
e
σz
ˆ
z
Z
∞
0
s
0
d
s
0
h
s
0
2
+
z
2
i
3
/
2
= 2
πK
e
σz
ˆ
z
Z
∞
z
2
1
2
d
u
u
3
/
2
= 2
πK
e
σz
ˆ
z
±

1
u
1
/
2
²
∞
z
2
= 2
πK
e
σz
ˆ
z
"
0

³

1

z

´
#
=
±
2
πK
e
σ
ˆ
z
(+ sign for
z >
0
,

sign for
z <
0)
=
±
2
π
³
1
4
πε
0
´
σ
ˆ
z
=
±
σ
2
ε
0
ˆ
z
We see that the electric ﬁeld in fact does not depend on
z
after all.
E
=
σ
2
ε
0
ˆn
, where
ˆn
points away from the plane,
E
=
0
inside the plane.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document3
Explained Solutions
3.1
Qualitative Analysis
As with every problem, we should make a qualitative analysis before getting quantitative; we can get
a sense of what the answer should be and potentially simplify the calculation with our conceptual
ideas. A good way to start is to draw a picture:
We can arbitrarily choose to draw the plane in a horizontal orientation. Let’s pick a random
point
P
above the plane and see if we can reason what direction the electric ﬁeld (“Eﬁeld”)
E
P
at this point is pointing. Let’s assume for now that the charge density is positive,
σ >
0. If we
examine a small patch on the plane, it will have a small positive amount of charge d
q
1
and that will
contribute a small amount of Eﬁeld d
E
1
P
at the point
P
that points directly away from the patch.
(See the illustration on the left, below.) If we look at another patch that is symmetrically placed
on the other side of
P
, it will contribute its own small amount of Eﬁeld d
E
2
P
, and the magnitudes
of these two contributions are the same, d
E
1
P
= d
E
2
P
, since the source charges d
q
1
and d
q
2
are the
same distance from point
P
. The geometry of this situation is such that the horizontal components
of these two Eﬁeld vectors cancel, but the vertical components add to point away from the plane.
Since an inﬁnite plane can be completely constructed from these pairs of patches, it follows
that the total electric ﬁeld at point
P
should have no horizontal component and only a vertical
component pointing away from the plane. This same logic allows us to conclude that for negative
charge density,
σ <
0, the electric ﬁeld must point directly toward the plane, with no component
parallel to the plane. (See the illustration on the right, above.)
Another kind of argument gives us this same result. This other argument is a proof by contra
diction. Here it is: Since the source charge distribution is symmetrical with respect to translation
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Corbin
 Electrostatics, Electric charge, Fundamental physics concepts

Click to edit the document details