MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2008
Problem Set 3 Solutions
Problem 1: Closed Surfaces
Four closed surfaces,
S
1
through
S
4
, together with the charges
–2
Q
,
Q
, and –
Q
are sketched in the figure at right. The
colored lines are the intersections of the surfaces with the
page.
Find the electric flux through each surface.
By Gauss’s Law, the flux through the closed surfaces is equal
to the charge enclosed over
ε
0
.
So,
1
2
3
0
0
2
;
0;
;
S
S
S
Q
Q
4
0
S
ε
ε
Φ
= −
Φ
=
Φ
= −
Φ
=
Problem 2:
Gauss’s Law Warm Up – Cylinder
Consider a cylinder of radius
R
and linear charge density
λ
.
(a) What is the electric field outside the cylinder as a function of distance
r
from its axis?
We use a Gaussian cylinder of length
L
and radius
r
>
R
.
Then,
0
0
0
0
ˆ
2
2
2
enc
d
E
rL
Q
L
E
r
r
π
ε
λ
ε
λ
πε
λ
πε
⋅
=
⋅
=
=
⇒
=
⇒
=
∫∫
E
A
E
r
G
G
G
(b) What is the electric potential difference between a point on its surface and a point a
distance
r
>
R
from its central axis?
To get the potential difference we just integrate, from the outer edge of the cylinder to
r
:
( )
(
)
(
)
(
)
(
)
0
0
0
'
2
'
'
2
ln
'
2
ln
r
r
R
r
R
V
V
r
V
R
d
r dr
r
R r
λ
πε
λ
πε
λ
πε
=
∆
=
−
= −
⋅
= −
=
−
=
∫
∫
E
s
G
G
What is the sign of this?
r > R, so the natural log is negative, and thus so is the potential
difference.
This should make sense, since we are going downhill as we move away from
the charged cylinder.
PS03 Solution  1
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Problem 3: A potential for finding charge
The electric potential
V(r)
for a distribution of charge with
spherical symmetry is:
2
0
0
2
0
0
Region I:
( )=
for
0
2
Region II:
( )
for
2
2
R egion III:
( )
0
for
2
V r
V r
V
r
R
R
V R
V
V r
R
r
R
r
V r
r
R
−
+
<
<
=
−
<
<
=
>
Where
V
0
is a constant of appropriate units.
This function is
plotted above.
(a)
What is the electric field
( )
r
E
G
for this problem?
Give your answer analytically for
the three regions above and also plot the magnitude.
Be sure to indicate the maximum
value on the vertical axis for E.
( )
ˆ
r
r
E
=
E
G
r
where
E
r
is given below and plotted above at right.
Region I:
0
r
R
<
<
0
2
r
V
V r
E
r
R
∂
= −
=
∂
Region II:
2
R
r
R
<
<
0
2
r
V
V R
E
r
r
∂
= −
=
∂
Region III:
2
r
R
>
0
r
V
E
r
∂
= −
=
∂
(b)
Use Gauss’s Law arguments to determine the distribution of charges that produces
the above electric potential and electric field.
When making a Gauss’s Law argument,
state the Gaussian surface you are using and what you conclude about the charge
distribution by applying Gauss’s Law to that surface.
First take a Gaussian sphere with radius
r > 2R
.
Since the electric field is zero on the
surface of the sphere, the total charge enclosed by this sphere must be zero.
Now take a
Gaussian sphere with any radius
R < r < 2R
.
We have
2
0
0
2
4
4
inside
o
closed surface
Q
V R
r
V
r
π
π
ε
⋅
=
=
=
∫∫
E dA
R
G
G
w
So for any
r
in the range
R < r < 2R
0
4
inside
o
Q
V R
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 Fall '08
 SCIOLLA
 Electrostatics, High Voltage, Electric charge, charge density

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