ECE 230B, HW1, Winter 2010
Solutions
1.
3D Gauss’ law is obtained after a volume integration of 3D Poisson’s equation and
takes the form of
E
⋅
=
∫∫
dS
Q
si
S
ε
,
where the LHS is an integral of the normal electric field over a closed surface
S
, and
Q
is the net
charge enclosed within
S
.
Use it to derive the electric field at a distance
r
from a point charge
Q
(Coulomb’s law).
What is the electric potential in this case?
Solution:
With the point charge
Q
at the center, construct a closed spherical surface
S
with radius
r
.
By symmetry, the electric field at every point on
S
has the same magnitude and points outward
perpendicular to the surface.
Therefore,
E
⋅
=
∫∫
dS
4
2
π
r
S
E
,
where
E
is the magnitude of the electric field on
S
.
3D Gauss’s law then gives
E
=
Q
r
si
4
2
πε
,
which is Coulomb’s law.
Since
E
=

dV
/
dr
, the electric potential at a point on the sphere is
V
Q
r
si
=
4
πε
,
if one defines the potential to be zero at infinity.
1
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2.
(a) Use Gauss’ law to show that the electric field at a point above a uniformlycharged
sheet of charge density
Q
s
per unit area is
Q
s
/2
ε
, where
ε
is the permittivity of the medium.
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 '09
 Electrostatics, Electric charge, negatively charged sheet, Vapp

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