Download Jackson 2.9 Homework Solution

Jackson 2.9 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
An insulated, spherical, conducting shell of radius a is in a uniform electric field E0. If the sphere is cut
into two hemispheres by a plane perpendicular to the field, find the force required to prevent the
hemispheres from separating
(a) if the shell is uncharged;
(b) if the total charge on the shell is Q.
SOLUTION:
(a) Align the z axis with the direction of the electric field. Find the potential outside a sphere at the
origin in a uniform field by placing charges at z = -R and z = +R with charges +Q and -Q and letting R
and Q approach infinity with Q/R2 constant. The response of the sphere can be represented by placing
two image charges -Qa/R and +Qa/R in the sphere at -a2/R and +a2/R. The potential outside an
uncharged conductor in a uniform field is therefore the potential of these four charges:
Φ=
Q
4 π ϵ0
[
1
1
− 2
+
2
√ r +R +2 r R cos θ √ r +R −2 r Rcos θ
2
2
a /R
√
r2 +
4
a
r
−2 a 2 cos θ
2
R
R
−
a/R
√
r2 +
4
a
r
+2 a 2 cos θ
2
R
R
In the limit R >> r, this becomes:
Φ=−E 0 r cos θ+E 0
a3
cos θ
r2
where E0 was recognized as 2Q/4πε0R2
The first term is just the potential due to the applied field in spherical coordinates. The second term is
the potential of a perfect dipole. The sphere there has an induced charge distribution that acts as a
perfect dipole.
The electric field is therefore:
E=−∇ Φ
3
̂ a (2 cos θ r̂ +sin θ θ)]
̂
E= E 0 [cos θ r̂ −sin θ θ+
3
r
The electric field at the surface of the sphere is:
E( r=a)=E 0 3 cos θ r̂
]
The charge distribution on the sphere's surface is found using:
σ=ϵ 0 r̂⋅E(r =a)
σ=3 ϵ0 E 0 cos θ
If the sphere is now cut into hemispheres at the polar angle θ = π/2, the bottom hemisphere will feel a
total force:
F=∫ σ( x) E(x) d a
We have to be careful and not include the force of the bottom hemisphere on itself. We do this by using
the relation E= 2σϵ which gives us just the electric field at the surface of a conductor due to non-self
contributions. Using this, we have:
0
F=
1
σ2 r̂ d a
2 ϵ0 ∫
F=k̂
2π π
a2
∫ ∫ (3 ϵ0 E 0 cos θ)2 cos θ sin θ d θ d ϕ
2 ϵ0 0 π/ 2
F=k̂ 9 ϵ0 π E 0 a
2
2
π
∫ cos 3 θ sin θd θ
π /2
9
2 2
F=− π ϵ 0 E 0 a k̂
4
The force needed to keep the bottom hemisphere in place would therefore have to be equal and in the
opposite direction:
9
2 2
F= π ϵ 0 E 0 a k̂
4
Due the symmetry, the force needed to keep the other hemisphere in place would be equal and
opposite.
(b) If the sphere has a total charge of Q, it will just spread out uniformly on the sphere as an additional
charge to the induced one.
σ=3 ϵ0 E 0 cos θ+
Q
4 π a2
The total force on the bottom hemisphere will therefore be:
F=
1
2
σ r̂ d a
∫
2 ϵ0
F=k̂
2 2π π
a
∫ ∫ (3 ϵ0 E 0 cos θ+ 4 πQa 2 )(3 ϵ0 E 0 cos θ+ 4 πQa2 )cos θ sin θ d θ d ϕ
2 ϵ0 0 π/ 2
2π π
F=
a2 ̂
Q
Q2
k ∫ ∫ (9 ϵ20 E 20 cos 2 θ+6 ϵ 0 E 0 cos θ
+
)cos θsin θ d θ d ϕ
2 ϵ0 0 π/ 2
4 π a 2 16 π 2 a 4
The first term represents the force on the induced charges due the external field and the field from the
induced charges. The second term represents the force on the charge Q due the external field. The third
term represent the force Q on itself. Note that the force of the external field on the point-charge-like
charge Q will just tend to shift it and not separate it. Because we just want forces that will separate the
two hemispheres, we must drop the middle term:
2π π
a2 ̂
Q2
F=
k ∫ ∫ (9 ϵ20 E 20 cos 2 θ+
) cos θ sin θ d θ d ϕ
2 ϵ0 0 π/ 2
16 π2 a 4
F=−k̂
[
9
Q2
π a 2 ϵ0 E 20+
4
32 π ϵ 0 a 2
]
The force needed to keep the bottom hemisphere touching the upper sphere is therefore:
F=k̂
[
9
Q2
π a2 ϵ 0 E 20 +
4
32 π ϵ0 a 2
]