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Physics 322 Homework 3-sol
Assigned: Wed. 9/21/16
Due: Wed. 9/28/16
1. Griffiths 2.27 (exercise in integrating the charge density for the potential). Leave the expression for V in terms of
an integral of the form
Z L/2
ρ
dz0 f (z − z0 ).
V (0, 0, z) =
2ε0 −L/2
i.e. Find f (z − z0 ) and the electric field at point (0, 0, z).
ans
2. Griffiths 2.30 b) and c) (boundary matching condition exercise). The “long” description means the edges are far
away from where you are evaluating the ~E field.
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ans
3. The field at the surface of a charged conductor can be written as ~E patch + ~Eother , as described in lecture. Just
outside the conductor ~E patch (+) and ~Eother (+) both have magnitudes σ /(2ε0 ) where σ is the local surface
charge density. Calculate directly ~Eother (+) for a uniformly charged sphere of radius R with charge density σ ,
assuming that the patch is a small disk of radius a. Show that the result becomes exact as a/R → 0 (e.g. show
that lima/R→0 ~Eother (+) = σ n̂/(2ε0 ) where n̂ is the normal to the surface element – which in the picture below
is ẑ in this limit; note that the angle ε in the figure is related to a and R through sin ε = a/R; compute the electric
field at the north pole).
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ans
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4. A hollow conducting sphere (i.e. a spherical shell) of radius R carries total charge Q. The sphere is cut along the
equator. Find the force with which the two hemispheres repel each other.
ans
Based on the result of lecture 7, we can write
~F =
Z
da~f
hemisphere
where da is the area element and
2
~f = σ r̂
2ε0
Q
.
4πR2
By the symmetry of the problem, we expect only the ẑ component (if the north pole is in the ẑ direction) to survive.
Evaluating the integral, we find
σ=
Z 1
ẑ · ~F
=
−1
Z 2π
d cos θ
= 2πR2
=
dφ R2
0
σ2
σ2
r̂ · ẑ
2ε0
Z 1
d cos θ cos θ
2ε0
2
πR σ 2
.
2ε0
0
Hence, we conclude the northern hemisphere is pushed with a force
~F =
Q2
ẑ.
32πε0 R2
By symmetry, the southern hemisphere has the same magnitude but with direction in the −ẑ direction.
5. Using Gauss’ theorem, show that
a) Any net charge on a conductor must lie on its surface.
ans
Conductor is defined to be at an equipotential, which means that ~E = 0 inside the conductor. Hence, if we take a
volume integral of ~∇ · ~E = ρ/ε0 over any volume v inside the conductor, it is zero since the left hand side vanishes:
Z
v
d 3 x~∇ · ~E =
Z
v
d 3 xρ/ε0 = 0
Since v can be arbitrarily small, we find
ρ =0
inside the conductor. As soon as the surface of the volume is not inside the conductor, the electric field need not vanish
and ρ 6= 0 is possible. This means that any net charge of the conductor must reside at the surface.
b) A closed hollow conductor shields its interior from fields due to exterior charge, but does not shield the exterior
from the charge on the interior. [hint: Use the properties of the solution to ∇2 Φ = 0 for part of the problem.]
ans
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Call the hollow region R. In region R, the field equation is
∇2 Φ = 0
with the conductor boundary condition Φ|∂ R = constant on the boundary of R. A solution to this problem is Φ =
constant in R. By uniqueness, this is the only solutions. ∴ −~∇Φ = 0 in R. =⇒ ~E = 0 in R. To see outside fields are
not shield from the charges from the inside, place a point charge Q at the center of the cavity. Gauss’s law gives
Q
d~a · ~E =
6 0
=
ε
S
0
Z
where S is a closed surface that encloses the cavity but lies outside of the conductor.
c) The electric field at the surface of a conductor has zero tangential component, and a normal component of σ /ε0
where σ is the charge density. (For the tangential component, you will need ~∇ × ~E = 0.)
ans
For the tangential component, we can draw a loop as in page 6 of lecture 6 and use ~∇ × ~E = 0 to conclude that the
tangential component is continuous across the boundary of the conductor. Since ~E = 0 inside the conductor, Ek = 0 .
For the normal component, apply Gauss’s law with a small pill box of cross sectional area ∆a at the surface of the
conductor as shown where h → 0:
where the surface ∆a is parallel to the surface of the
conductor.
Gauss’s law states
Z
σ ∆a.
d~a · ~E =
ε0
pill box sides
Since h → 0, the flux integral on the sides of the pill box only picks up the Ek |sur f ace ) contribution which is zero (even
if Ek |sur f ace 6= 0 were finite it vanishes since h → 0). Since ~E = 0 inside the conductor, Gauss’ law gives
E⊥ ∆a =
σ ∆a
.
ε0
Cancelling ∆a from both sides, we obtain
E⊥ =
σ
.
ε0
6. A capacitor is composed of two concentric cylinders of radii a and b, with b > a. The tubes are long and protrude
from one another. The length of the overlapping region is l. Assume b − a l length of cylinders
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a) Calculate the capacitance as a function of l neglecting the edge effects (i.e. accounting only for the overlapping
region capacitance).
ans
Accounting only for the overlapping region, we have
Er 2πrl =
Vout −Vin
= −
Z b
a
=
=
Qin
ε0
drEr = −
Z b
dr
a
Qin
2πrlε0
Qin
a
ln
2πlε0 b
Qout
b
ln .
2πlε0 a
Hence, we conclude
C
=
=
Qout
Vout −Vin
2πε0 l
ln ba
b) Suppose the potential difference between the tubes is V (assume that the outer cylinder is at a higher potential).
Based on an intuitive qualitative guess of the charge distribution, draw a clear sketch of the elctric field lines in
a plane through the axis. Your drawing should qualitatively show how a force originates along the axis.
ans
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