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Homework 6 Solutions
Problem 6.1: Square loop (based on Griffiths 2.4) Consider a square loop
with each side a length a carrying a uniform linear charge density λ.
a) Find the electric field at the center of the square.
Solution: The electric field to to each of the opposite sides will be in an
opposite direction, so the total electric field at the center of the square is
zero.
b) Find the work needed to bring a charge in from infinity to the center of
the square. (You may start with the electrostatic potential due to a single
finite line of charge.)
Solution: The work needed to bring a charge Q in from infinity to the
origin is just the charge times the electrostatic potential at the origin.
U = QV (0)
(1)
The electrostatic potential due to a finite line charge, in the plane that
bisects the line, is
!
√
a2 + 4r2 + a
(2)
V (r) = kλ ln √
a2 + 4r2 − a
!
√
2+1
V (l/2) = kλ ln √
(3)
2−1
Since there are four sides, we just end up with
U = 4kQλ ln
!
√
2+1
√
2−1
(4)
Problem 6.2: Three charges (Griffiths 2.32) Three charges are situated at
the corners of a square (side s). Two have charge −q and are located on opposite
corners. The third has charge +q and is opposite an empty corner.
a) How much work does it take to bring in another charge, +q, from far away
and place it at the fourth corner?
1
Solution: To find the work needed, I need to calculate the potential at
the final location of the fourth charge due to the presence of the other
charges. (The subscripts denote the charge that is creating the potential.)
W
=
q V (~r)
=
q (V1 + V2 + V3 )
k(−q) k(q) k(−q)
q
+√ +
s
s
2s
√ !
2
−k q 2
2−
s
2
=
=
The negative sign indicates that the charge is losing energy is being placed
at the corner of the square.
b) How much work does it take to assemble the whole configuration of four
charges?
Solution:
4
W =
1X
qi V (~ri )
2 i=1
The potential for charges on opposite corners is the same (here, the subscripts denote the total potential at the location of the charge due to the
presence of the other four charges):
VT 1 = VT 3
k q2
=
s
VT 2 = VT 4 =
W
=
=
−k q 2
s
√ !
2
2−
2
√ !
2
2−
2
1
[2(−q)V1 + 2(q)V2 ]
2
√ !
−(q) k q 2
(−q) k q 2
2
2−
+
s
2
s
k q2
= −2
s
√ !
2
2−
2
√ !
2
2−
2
The negative sign indicates that the system loses some energy in order to
be placed in the square.
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Problem 6.3: A thin spherical shell
radius R and total charge Q.
Consider a thin spherical shell with
a) What is the surface charge density of the shell?
b) Find the electrostatic potential everywhere within the shell.
c) Compute the gradient of the electrostatic potential to find the electric field
within the shell.
d) Find the electrostatic potential everywhere outside the shell.
e) Compute the gradient of the electrostatic potential to find the electric field
outside the shell.
f) Comment on your results.
Solution:
a) The charge over the area is
σ=
Q
4πR2
b) The potential is clearly spherically symmetric by symmetry, so I only need
to work out V (r). We can get this by just adding up the potential due to
all the little bits of charge.
Z
kdQ0
V (r) =
(5)
|~r − ~r0 |
In this case, we’re talking about a surface, so dQ0 = σdA0
Z
dA0
V (r) = kσ
|~r − ~r0 |
(6)
The obvious set of coordinates is spherical, and since the only thing that
isn’t spherically symmetric is the point ~r we’re curious about, let’s choose
our coordinate system so that that point is on the z axis, to maintain as
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much symmetry as we can.
ZZ 2
R dϕ sin θdθ
V (r) = kσ
(7)
|~r − ~r0 |
ZZ
1
q
dϕ sin θdθ
(8)
= kσR2
(R cos θ − r)2 + R2 sin2 θ
Z π
1
q
= 2πkσR2
sin θdθ
(9)
0
(R cos θ − r)2 + R2 sin2 θ
Z π
1
p
= 2πkσR2
sin θdθ (10)
2
2
2
0
R cos θ + r − 2R cos θr + R2 sin2 θ
Z π
1
2
√
sin θdθ
(11)
= 2πkσR
2
2
R + r − 2Rr cos θ
0
Z 1
1
√
= 2πkσR2
d cos θ
(12)
2
2
R + r − 2Rr cos θ
−1
Here we just define
ξ = R2 + r2 − 2Rr cos θ
V (r) = 2πkσR
Z
2
R2 +r 2 −2Rr
R2 +r 2 +2Rr
kπσR
=
r
Z
kπσR
=
r
Z
dξ = −2Rrd cos θ
R2 +r 2 +2Rr
R2 +r 2 −2Rr
R2 +r 2 +2Rr
R2 +r 2 +−Rr
1
√
ξ
1
dξ
−
2Rr
(14)
1
√ dξ
ξ
(15)
1
√ dξ
ξ
(16)
2
2
kπσR p R +r +2Rr
2 ξ 2 2
r
R +r −2Rr
p
kπσR p 2
=
2
R + r2 − 2Rr − R2 + r2 + 2Rr
r
p
kπσR p
=
2
(R + r)2 − (R − r)2
r
kπσR
=
2 (R + r − (R − r))
r
kπσR
=
4r
r
= k4πσR
kQ
=
R
=
(13)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
Note that we assumed
that we are talking about the inside of the shell,
p
when converting (R − r)2 to R − r.
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c) The electric field is the gradient of the potential. Since the potential is
independent of ~r (it involve r, θ or ϕ), its gradient is zero and the electric
field is zero inside the shell.
d) The outside is found in a manner precisely analogous to the inside.
Z R2 +r2 −2Rr
1
1
√
V (r) = 2πkσR2
−
dξ
2Rr
ξ
R2 +r 2 +2Rr
Z R2 +r2 +2Rr
1
kπσR
√ dξ
=
r
ξ
R2 +r 2 −2Rr
Z R2 +r2 +2Rr
1
kπσR
√ dξ
=
r
2
2
ξ
R +r −2Rr
2
2
kπσR p R +r +2Rr
=
2 ξ
r
R2 +r 2 −2Rr
p
p
kπσR
=
2
R2 + r2 + 2Rr − R2 + r2 − 2Rr
r
p
kπσR p
=
2
(R + r)2 − (R − r)2
r
kπσR
=
2 (r + R − (r − R))
r
kπσR
=
4R
r
= k4πσR
kQ
=
r
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
e) To find the electric field outside the shell, we just need to compute the
gradient of kQ
r . This is
~ r ) = −∇
~ kQ
E(~
r
kQ
∂
~ r) = − ∂ r̂ + 1 ∂ θ̂ + 1
E(~
ϕ̂
∂r
r ∂θ
r sin θ ∂ϕ
r
∂ kQ
= −r̂
∂r r
kQ
= r̂ 2
r
(34)
(35)
(36)
(37)
f) The two answers look quite similar. However, there is a huge difference
between R and r. The former is the radius of the shell, while the latter
is the distance between the center of the shell and the position whose
electrostatic potential we want to know.
The answer outside the shell
kQ
r
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is the same as the electrostatic potential due to a point charge! So outside
a spherical shell of charge, it behaves as if it were simply a point charge.
In contrast, the potential inside the shell
kQ
R
is completely independent of position, which means that there is no work
required to move a point charge around within the charged shell, and there
is no electric field inside the shell. These are interesting and deep results.
Problem 6.4: Classical Electron Radius Consider a solid sphere of charge
of uniform charge density, with total charge Q, and radius R. We will compute
the total energy required to create this charge distribution.
The process for computing this total energy will to build up the sphere by
bringing in charge from r = ∞, keeping track of how much work we have done
as we go. We will bring in one infinitesimally thin spherical shell at a time.
You may use the property that the electrostatic potential due to a spherically
symmetric charge distribution outside that charge distribution is the same as if
it were a point charge.
a) Begin by assuming we have already built the sphere up to a radius r with
the same charge density as our final sphere. How much work is needed to
bring a charge dQ to the surface of this sphere?
b) Express the charge dQ of a thin spherical shell with thickness dr (together
with the quantities given above: Q and R.
c) Integrate the quantity from part (a) from r = 0 to r = R to find the total
work needed to construct a solid sphere of charge with total charge Q and
radius R.
d) You have probably heard of Einstein’s famous formula
E = mc2
(38)
which relates the energy of a particle at rest to its mass and the speed of
light. Use this relation together with the energy of a sphere of charge that
you just computed to find what the radius of an electron must be, if its
mass arises purely from the electrostatic energy needed to put it together.
This is closely related to the so-called classical electron radius (which is
not related to the actual size of an electron), which is obtained by ignoring
any dimensionless factors of order one in your answer.
Please provide an actual numerical answer.
Solution:
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a) I’ll use U for the total energy, so dU will be the small amount of work
needed to bring in the charge dQ. This will be the same as its potential
energy, so I can just use the electrostatic potential, which requires that I
find the charge of the sphere that I’m bringing the dQ up to.
4π 3
r
3
4π
Q = ρ R3
3
r3
q(r) = Q 3
R
kq(r)dQ
dU =
r
r2
= kQ 3 dQ
R
q(r) = ρ
(39)
(40)
(41)
(42)
(43)
b)
dQ = ρdτ
(44)
2
= ρ4πr dr
Q
= 4π 3 4πr2 dr
3 R
= 3Q
r2
dr
R3
(45)
(46)
(47)
(48)
c)
Z
U=
dU
r2
kQ 3 dQ
R
Z
kQ
r2
2
r
3Q
dr
R3
R3
Z
3kQ2 R 4
r dr
R6 0
3kQ2 R5
R6 5
3kQ2
5R
(49)
Z
=
=
=
=
=
7
(50)
(51)
(52)
(53)
(54)
d)
U = me c2
(55)
2
3ke
5R
3ke2
R=
5me c2
(56)
=
=
(57)
2
1.6 × 10−19 C
2
5 (9.1 × 10−31 kg) 3 × 108 ms
J
3 9 × 109 mC
= 1.7 × 10−15 m
(58)
(59)
which is pretty small. Amazingly enough, we know from experiment that
electrons are smaller than this. As far as we can tell, they are true point
charges—unlike, for instance, protons, which have a radius that can be
measured by pummeling them with electrons. In an interesting coincidence, the classical radius of an electron is not far from the actual radius
of a proton.
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