PHYS 3327 PRELIM 1 Prof. Itai Cohen, Fall 2016 Wednesday, 10/12/16 Name: Read all of the following information before starting the exam: • Put your name on the exam now. • Show all work, clearly and in order, if you want to get full credit. • Box or otherwise indicate your final answers. • Question 5(g) is a extra credit question worth 5 points. The total exam score cannot exceed 100, but the extra credit can help you make up points lost elsewhere. Problem # Score 1 /12 2 /08 3 /20 4 /25 5 /35 5(g) Extra credit /05 Total /100 • It is your responsibility to make sure that you have all of the pages! • Good luck!
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PHYS 3327 PRELIM 1
Prof. Itai Cohen, Fall 2016
Wednesday, 10/12/16 Name:
Read all of the following information before starting the exam:
• Put your name on the exam now.
• Show all work, clearly and in order, if you want to get full credit.
• Box or otherwise indicate your final answers.
• Question 5(g) is a extra credit question worth 5 points. The total exam score cannot
exceed 100, but the extra credit can help you make up points lost elsewhere.
Problem # Score
1 /12
2 /08
3 /20
4 /25
5 /35
5(g) Extra credit /05
Total /100
• It is your responsibility to make sure that you have all of the pages!
A uniformly charged sphere of radius a with total charge −Q is surrounded by a concentric
spherical shell of radius b, which has a total charge +Q distributed uniformly on its surface.
(a) [6 points] Find the net electric field produced by the charge configuration, ~E(~r), in
each of the following regions:
(i) outside the shell, r > b.
Ans. By symmetry, the electric field must point in the radial direction, and is a function
of r alone. Taking a spherical Gaussian surface at radius r, we get
E(4πr2) = 4πQencl = 0 . =⇒ E = 0.
(ii) between the sphere and the shell, a < r < b.
Ans. Here Qencl = −Q. Hence, E(4πr2) = −4πQ , =⇒ ~E(~r) = −Qr2r .
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(iii) inside the sphere, r < a.
Ans. The charge density inside the inner sphere is ρ = −Q/(43πa3). Hence the total charge
inside radius r is Qencl = ρ(43πr3) = −Q(r3/a3). Therefore,
E(4πr2) = −4πQ(r3/a3) , =⇒ ~E(~r) = −Qra3
r .
(b) [6 points] Sketch graphs of the radial component of the electric field, E(r), and the
electric potential Φ(r) as a function of r, assuming Φ(r →∞) = 0.
You don’t need to find an expression for Φ(r). However, pay attention to qualitative
features such as slope, curvature, and relative magnitudes in your plots, particularly
at r = 0, r = a, and r = b.
Ans. The plots are shown below. The electric field is given by the negative gradient of the
potential, E(r) = −dΦdr
. Since the electric field is zero for r > b, Φ(r > b) = Φ(r →∞) = 0. Note that Φ(r) is continuous everywhere, and smooth at r = a. Also, it has
zero slope at r = 0, since the electric field there vanishes.
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Question 2: Current loop in a magnetic field [8 points]
A current-carrying loop of arbitrary shape is partially immersed in a uniform magnetic field~B. The magnetic field is pointing into the page, perpendicular to the plane of the loop.
(a) [4 points] Write an expression for the magnetic force, d~F , on a small segment d~l of the
loop inside the field region. Sketch d~l and d~F on the figure.
Ans. From the Lorentz force law, we get d~F = Id~l× ~B. The vectors are shown in the figure.
(b) [4 points] Integrate your expression to find the net magnetic force on the loop, in terms
of I, a, B, x and y (see figure). Hint: I and ~B are constants!
Ans. To find the net force, we integrate d~F over the part of the loop inside the field region.
Both I and ~B can be taken out of the integral since they are both constants. Thus we
get
~F = I
(∫d~l
)× ~B .
However,∫d~l is just summing over all the little vectors from the left end to the right
Consider an infinitely long dielectric cylinder of radius R. It has a frozen-in polarization~P (~r) = −k r, where r denotes the distance from its axis, and k is a constant.
(a) [4 points] Find the bound charge density inside the cylinder, ρb(~r).
Ans. The volume charge density is given by
ρb(~r) = −~∇. ~P
= −1
r
∂(rPr)
∂r− 1
r
∂Pφ∂φ− ∂Pz
∂z
= −1
r
∂(−k r)∂r
=k
r.
(b) [4 points] Find the bound charge density on the surface of the cylinder, σb(~r).
Ans. The surface charge density is given by σb(~r) = ~P .n, where n denotes the outward
normal to the surface. Here, n = r. Hence, σb(~r) = −k r.r = −k.
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(c) [6 points] Find the net electric field, ~E(~r), in each of the following regions:
(i) inside the cylinder, r < R
Ans. By symmetry, the electric field must be a function of r alone, and in the radially
outward direction. Thus we apply Gauss’ law to a closed concentric cylinder of
length l and radius r < R. The total enclosed charge is given by
Qencl =
∫ r
0
(k/r′)(2πr′ldr′) = 2πkrl .
Hence, E(2πrl) = 4π(2πkrl), or E = 4πk. =⇒ ~E(~r) = 4πk r.
(ii) outside the cylinder, r > R
Ans. Here we consider a cylinder of radius r > R. The total enclosed charge is given
by
Qencl = 2πkRl + (−k)(2πRl) = 0 .
Hence, E = 0.
Note: The result Qencl = 0 for a closed surface enclosing the dielectric is true
more generally, as the dielectric as a whole is electrically neutral.
(d) [6 points] Plot the magnitudes of the electric field ( ~E) and the displacement ( ~D) as a
function of r. Show that they satisfy the required boundary conditions at r = R.
Ans. By definition, the displacement is given by ~D = ~E + 4π ~P , which is zero both inside
and outside the cylinder. The plots are shown below.
They satisfy the boundary conditions ~E⊥r→R+ − ~E⊥r→R− = 4πσ = −4πk, and ~D⊥r→R+ −~D⊥r→R− = 0. Note that the parallel components are identically zero.
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Question 4: Dipolar hemispheres [25 points]
A conducting sphere of radius a is made up of two hemispherical shells separated by a thin
insulating ring. The two hemispheres are held at potentials +V and −V (see figure).
(a) [4 points] Roughly sketch the electric field lines you would expect for this system.
Focus on qualitative features, such as the direction of the field lines, and the angle
they make with the spherical surface.
y
z
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Suppose Φin(r, θ) and Φout(r, θ) describe the electric potential inside and outside the sphere,
where r denotes the distance from the origin, and θ denotes the angle from the z axis.
(b) [5 points] Write the boundary conditions on Φin and Φout at r = 0, r = a, and r →∞.
Ans. The boundary conditions are given by
(i) Φin(r, θ) is finite at r = 0.
(ii) Φout(r, θ)→ 0 as r →∞.
(iii) Φin(a, θ) = Φout(a, θ) =
+V θ < π/2
−V θ > π/2.
(c) [5 points] Write down the most general solution for Φin(r, θ) and Φout(r, θ) which satisfy
the boundary conditions at r = 0 and r →∞ respectively.
Ans. The most general solution to Laplace’s equation consistent with the boundary condi-
tions (i) and (ii) are given by
Φin(r, θ) =∞∑l=0
AlrlPl(cos θ) ,
Φout(r, θ) =∞∑l=0
Bl
rl+1Pl(cos θ) .
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(d) [8 points] Apply the boundary conditions at r = a to determine Φin(r, θ) and Φout(r, θ).
You may need the following property of Legendre polynomials: Pl(−x) = (−1)lPl(x),
and you may express your answer in terms of Cl ≡∫ 1
0Pl(x)dx.
Hint: If you’re stuck, try the substitution x = cos θ.
Ans. The boundary conditions at r = a read
∞∑l=0
Bl
al+1Pl(cos θ) =
∞∑l=0
AlalPl(cos θ) =
+V θ < π/2
−V θ > π/2
From the first equality we get Bl = a2l+1Al. Next we apply the orthogonality of
Legendre polynomials on the second equality to obtain the coefficients Al. To this
end, we multiply both sides by dθ sin θPl′(cos θ) and integrate over θ from 0 to π,
yielding
∞∑l=0
Alal
∫ π
0
dθ sin θPl(cos θ)Pl′(cos θ) =
∫ π
0
dθ sin θPl′(cos θ)×
+V θ < π/2
−V θ > π/2
or,2Al′a
l′
2l′ + 1= V
[ ∫ 1
0
dxPl′(x)−∫ 0
−1
dxPl′(x)
].
Here we have used the orthogonality relations of the Legendre polynomials in the left-
hand side, and made the substitution x = cos θ in the right-hand side. The right-hand
side can be further simplified as∫ 1
0
dxPl′(x)−∫ 0
−1
dxPl′(x) =
∫ 1
0
dx(Pl′(x)− Pl′(−x)) = (1− (−1)l′)
∫ 1
0
dxPl′(x) ,
which vanishes for all even values of l′. For odd l′, we get
2Al′al′
2l′ + 1= 2V
∫ 1
0
dxPl′(x) = 2V(−1)
l′−12
l′(l′ + 1)
l′!!
(l′ − 1)!!
or, Al = (−1)l−12
2l + 1
l(l + 1)
l!!
(l − 1)!!
V
al.
Therefore,
Φin(r, θ) = V∑odd l
(−1)l−12
2l + 1
l(l + 1)
l!!
(l − 1)!!
(ra
)lPl(cos θ) ,
Φout(r, θ) = V∑odd l
(−1)l−12
2l + 1
l(l + 1)
l!!
(l − 1)!!
(ar
)l+1
Pl(cos θ) . [using Bl = a2l+1Al.]
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(e) [3 points] Show that the potential Φ(~r) has the symmetry Φ(x, y,−z) = −Φ(x, y, z),