# Solution - Solar Physics Group ```–1–
Solution to Final Exam
~ &times; f~ = −k(ẑ + x̂) 6= ~0. So the force is rotational.
1. (a) We first find the curl of the force to be ∇
~ E
~ = ~0, the electrostatic field is nonrotational,
According to the property of the electrostatic field ∇&times;
so the force cannot be electrostatic force.
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(b) The angle θ is given by tanθ1 = E1 /E1⊥ , and tanθ2 = E2 /E2⊥ . So tanθ1 /tanθ2 = (E1 /E2 )(E2⊥ /E1⊥ ).
From the boundary condition, we know that across the boundary, the tangential component of the
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electrostatic field is continuous, or E2 = E1 ; and with no free charge, the normal component of the
displacement field is continuous D1⊥ = D2⊥ , or ǫ1 E1⊥ = ǫ2 E2⊥ . These lead to tanθ1 /tanθ2 = ǫ1 /ǫ2 .
(c) The capacitor without the dielectric material in between carries a surface charge σ = Q0 /A.
The electric field is E = σ/ǫ0 = Q0 /(Aǫ0 ), and the potential difference is V0 = Ed = Q0 d/(Aǫ0 ). So
the capacitance is C0 = Q0 /V0 = Aǫ0 /d. The energy in the capacitor is W0 = 21 Q20 /C0 = 21 C0 V02 .
When it’s filled with dielectric material, the capacitance is increased to C = Aǫ0 ǫr /d = 2C0 . In the
first situation when the battery is disconnected, the free charge on the plates is constant, so the
energy of the capacitor is W1 = 12 Q20 /C = 12 W0 . The energy is decreased by one half. In the second
situation, when the battery is connected and maintains the potential difference V0 , the energy in
the capacitor is W2 = 12 CV02 = 2W0 , the energy is increased by a factor of 2. The energy in the first
case is deduced because the electrostatic force does the work to “pull” in the dielectric material; in
the second case, to maintain the constant potential difference, the battery does work to transport
more free charges to the capacitor and increase the total energy in the capacitor.
2 (a) We first find the displacement field using Gauss’s law. The Gaussian surface is a coaxial
~ = Ds ŝ.
cylindrical shell with radius s. Given the symmetry, the displacement field must be radial D
H
R
~ = Ds 2πsL = ρdτ = ρπs2 L, leading to Ds = ρs/2. The electric
~ &middot; da
Inside the cylinder, s &lt; a, D
~ = ρs/(2ǫ0 )ŝ.
field is Es = Ds /ǫ0 , or E
H
~ = Ds 2πsL = Qenc = ρπa2 L, leading to
~ &middot; da
Inside the dielectric material a &lt; s &lt; 2a, D
~ = ρa2 /(4ǫ0 s)ŝ.
Ds = ρa2 /2s. The electric field is Es = Ds /(ǫ0 ǫr ) = ρa2 /(4ǫ0 s), or E
~ = ~0.
Outside the cable, s &gt; 2a, the total enclosed free charge is zero, so Ds = 0, and E
(b) The work done on the point charge is
Z
Z
Z 0
~ =q
~ &middot; dl
Es ds = q(
W = qE
s=2a
a
s=2a
ρa2
ds +
4ǫ0 s
Z
0
s=a
ρs
qρa2
ds) = −
(ln2 + 1)
2ǫ0
4ǫ0
The electric field does negative work by moving the charge from low potential to high potential,
and the charge gains energy.
~ = ǫ0 E
~ = ρa2 /(4s)ŝ The volume bound charge ρb =
(c) In a linear material, P~ = (ǫr − 1)ǫ0 E
~ &middot; P~ = 0, and the surface bound charge is σb = P~ &middot; n̂. At the inner surface, n̂ = −ŝ, σb (s = a) =
−∇
ρa2 /(4s)ŝ &middot; (−ŝ)|s=a = −ρa/4. At the outer surface, n̂ = ŝ, σb (s = 2a) = ρa2 /(4s)ŝ &middot; ŝ|s=2a = ρa/8.
The total bound charge Qb = σb (s = a)2πaL + σb (s = 2a)2π2aL = 0.
–2–
⊥
⊥
= 0 − ρa2 /(4ǫ0 2a) = −ρa/(8ǫ0 ). The total surface charge
− Ebelow
(d) At the outer surface, Eabove
includes the bound charge and the free charge. The free charge is determined by σf 2π2aL =
−ρπa2 L, or σf = −ρa/4. And the total surface charge is σ = σf + σb = −ρa/8. The boundary
⊥
⊥
= −ρa/(8ǫ0 ) = σ/ǫ0 .
− Ebelow
condition is therefore satisfied Eabove
3 (a) In this problem, by solving the Laplace’s equation in spherical coordinates, we get: inside the
sphere, Vin = ΣAl r l Pl (cosθ), and outside the sphere, Vout = ΣBl r −l−1 Pl (cosθ). The continuity at
the surface gives Bl = Al R2l+1 , and the second type boundary condition gives ∂Vin /∂r−∂Vout /∂r =
Σ(2l + 1)Al Rl−1 Pl (cosθ) = σ/ǫ0 = 3kcosθ. Comparing terms, we get Al = Bl = 0 for l 6= 1, and
A1 = k, and B1 = kR3 . So Vin = krcosθ, and Vout = kR3 cosθ/r 2 .
(b) It is seen that the charge on the top half of the spherical shell is positive, and the charge on the
bottom half is negative, so the total charge is zero. Therefore the monopole term of the potential is
zero. For the dipole term, we first find the dipole moment. Given the symmetry, the dipole moment
R
Rπ
must be in z direction only. We find p~ = pẑ = zẑσda = ẑ θ=0 Rcosθ3ǫ0 kcosθR2 sinθdθdφ =
~&middot;r̂
1 p
3
2
ẑ2π3kǫ0 R3 13 (−cosθ 3 )|π0 = 4πkǫ0 R3 ẑ. The dipole term is given by Vdip = 4πǫ
2 = kR cosθ/r , the
0 r
same as in (a). So for such a charge distribution, all higher order terms are zero.
(c) If the charge is induced charge by the external field, the electric field inside the sphere by the
induced charge should completely cancel the external field. The field by this charge inside the
~ ind = −∇V
~ in = −∇(kz)
~
~ ind + E~0 = ~0, we get −kẑ + E0 ẑ = ~0, so
sphere is E
= −kẑ. To have E
k = E0 .
(d) If a point charge q is placed in a cavity inside the metal sphere, an additional charge q will be
induced on the surface, which will be uniformly distributed on the sphere surface, so the charge on
the surface becomes σ = 3kǫ0 cosθ + Q/(4πR2 ). The effect of such is that an additional electric field
1 q
is applied outside the sphere Eq = 4πǫ
2 , in addition to the dipole field by the induced charge and
0 r
the original external field. This new charge distribution generates a zero electric field inside the
sphere (where there is no cavity), as should be. Inside the cavity, the electric field line starts from
the point charge and ends at the induced charge (of total amount -q) on the cavitiy wall. The net
electric field by the point charge and the induced charge on the cavity wall is zero in the metal.
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