Electricity and Magnetism: PHY-204.
3 October, 2014
Assignment 2: Electric Potential and Capacitors
Solution
1. Find the potential at P due to a uniformly charged triangular sheet of charge density
σ with dimensions shown. Not only set up the integral, but solve it as well.
h
θ
P
L
Fig. (1)
Answer
To solve this problem, we need to evaluate the potential at point P due to a trapezoidal
strip at distance x from P (as shown in figure) and then integrate from x = 0 to x = L.
h
dy y
0
r
P
y
β θ
x
dx
L
Potential due to the darkly shaded segment possessing small amount of charge dq, at
√
a distance r = x2 + y 2 from P , is given by
σdxdy
dq
√
=
·
4πε0 r
4πε0 x2 + y 2
The potential at P due to trapezoidal strip located at distance x becomes
∫
σdx y0
dy
√
·
4πε0 0
x2 + y 2
Due Date: October. 14, 2014, 5:00 pm
(1)
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Electricity and Magnetism: PHY-204.
3 October, 2014
For a definition of θmax , refer to the diagram depicting the geometry. Using the
geometry of figure;
y = x tan β
⇒ dy = x sec2 β,
√
x
⇒ x2 + y 2 = x cos β.
and cos β = √
x2 + y 2
By making these substitutions, the integral in Eq (1) becomes
∫
√
dy
x2 + y 2
∫
θ
=
θ=0
∫
x sec2 βdβ
x sec β
θ
=
sec βdβ
θ=0
= ln | sec θ + tan θ| − ln | sec 0◦ + tan 0◦ |
= ln(sec θ + tan θ) − ln(1) = ln(sec θ + tan θ).
Finally, the potential at P due to a uniformly charged triangular sheet is
∫ L
σdx
VPdue to triangle =
ln(sec θ + tan θ)
0 4πε0
∫
σ ln(sec θ + tan θ) L
σL
=
ln(sec θ + tan θ).
dx =
4πε0
4πε0
0
2. (a) Suppose we have a metallic, solid sphere of radius R. It carries a charge Q.
What is the potential energy of this sphere? Remember, as in class, we like to
“assemble” this charged body. Try bringing in tiny amounts of charge from far
away. Remember it’s a metal.
(b) Now consider two metallic spheres of radii R1 and R2 , far far apart—sufficiently
far away that they don’t electrically interact. We have a total amount of charge Q
that needs to be distributed on these spheres. Suppose f Q goes onto one sphere
and (1 − f )Q goes onto the other. For what fraction f , is the total potential
energy minimized? Let’s call this the electrostatic ground state. Draw a sketch
that will help you in solving the problem.
(c) What is the potential difference between the spheres in the electrostatic ground
state?
(d) What is the electric field along the axis joining the centres of the spheres?
Due Date: October. 14, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
3 October, 2014
(e) If the first sphere carries all the charge Q, and is then connected to the second
far away sphere by a metallic wire, will the charge redistribute? If yes, how?
Answer
(a) In a metallic sphere of radius R, all the charge is on the surface. Suppose we
have put some charge q already on the surface. The potential at the surface is
V =
1 q
·
4πε0 R
We bring in some extra charge dq from infinity to R. The amount of external
work required for this action is
dU = V (R)dq =
1 qdq
4πε0 r
all of this goes to increase the potential energy. For assembling a total charge Q,
the potential energy becomes,
∫ Q
1 Q2
Q2
1
qdq
=
=
·
U=
4πε0 0 R
4πε0 R 2
8πε0 R
(b) Suppose a sphere of radius R1 carries a charge f Q and a sphere of radius R2
carries a charge (1 − f )Q. The potential energies associated with them are
(f Q)2
U1 =
,
8πε0 R1
(
)2
(1 − f )Q
U2 =
·
8πε0 R2
d
fQ
(1-f)Q
R2
R1
The total potential energy associated with two spheres far far apart—sufficiently
far away that they don’t electrically interact, is
(
)
Q2 f 2
(1 − f )2
U = U1 + U2 =
+
.
8πε0 R1
R2
Due Date: October. 14, 2014, 5:00 pm
(2)
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Electricity and Magnetism: PHY-204.
3 October, 2014
The minimum of potential energy can be found by setting
dU
df
= 0.
(
)
Q2 2f
2(1 − f )(−1)
dU
=
+
=0
df
8πε0 R1
R2
2f
2(1 − f )
⇒
=
R1
R2
(1 − f )
R2
=
f
R1
R2
(1 − f ) = f
R
(1
)
R2
R1
1 = f 1+
,
⇒ fmin =
R1
R1 + R2
(3)
which is the fractional distribution for which total potential energy is minimum.
The minimum potential energy is
[ (
)2
(
)2 ]
Q2 1
R1
1
R2
Uf =fmin =
+
8πε0 R1 R1 + R2
R2 R1 + R2
[
]
2
1
Q
=
8πε0 R1 + R2
R1
R2
(c) Potentials V1 due to sphere of charge f Q and V1 due to sphere of charge (1 − f )Q
are
V1 =
1 fQ
,
4πε0 R1
V2 =
1 (1 − f )Q
·
4πε0
R2
So the potential difference between the spheres in the electrostatic ground state
is given by
∆V = V1 − V2
(
)
Q
f
(1 − f )
=
−
4πε0 R1
R2
(
)
Q
1
1
=
−
= 0,
4πε0 R1 + R2 R1 + R2
i.e., ground state corresponds to both spheres being at the same potential.
Due Date: October. 14, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
3 October, 2014
d
fQ
(1-f)Q
x
R1
R2
x
(d)
potential at x = Potential due to big sphere + potential due to small sphere
f Q 1 (1 − f )Q 1
Vx =
+
4πε0 x
4πε0 (d − x)
∂Vx
E at x = −
∂x (
)
(
)
fQ
1
(1 − f )Q
1
= −
− 2 +
−
4πε0
x
4πε0
(d − x)2
(
)
f
Q
(1 − f )
=
−
2
4πε0 x
(d − x)2
which is the field at any distance between the spheres.
(e) Yes. When the first sphere carrying all charge Q is connected to second sphere
then charge will continue to flow until equilibrium is established such that both
spheres are at the same potential V1 = V2 = V . This means
fQ
4πε0 R1
1−f
f
1
⇒ −1
f
1
f
=
=
=
=
f =
(1 − f )Q
4πε0 R2
R2
R1
R2
R1
R1 + R2
R1
R1
·
R1 + R2
(4)
Hence we have shown that in electrostatic equilibrium, the same charge distribution Eq (4) appears as in Eq (3). Hence equilibrium is really the lowering
of the total potential energy. This is in complete agreement with the laws of
thermodynamics.
3. A spherical shell with radius R and surface charge density σ is sandwiched between
two infinite sheets with surface charge densities −σ and +σ, as shown in Fig. (2). If
Due Date: October. 14, 2014, 5:00 pm
5
Electricity and Magnetism: PHY-204.
3 October, 2014
the potential far to the right at x = +∞ is taken to be zero, what is the potential
at the center of the sphere? At x = −∞? Before you start solving this problem, I
suggest you also find the potential at the centre of a uniformly charged ring of radius
R.
Fig. (2)
Answer
We start by finding as suggested, the potential at the center of a charged shell.
∫
1
dq
Vat point O =
4πε0
R
∫
σ
dA
=
4πε0 R
σ
σR
=
4πR2 =
·
4πε0 R
ε0
Note that the interior of the shell is equipotential and there is zero electric field inside
the shell.
Now let the sheets be at x = ±R and point O at x = 0. In the region outside the
sheets (x > R, x < −R), there is zero electric field because of the plates. However,
there is a field due to the shell, which is pointing radially outwards from the origin O.
(Use superposition). Hence we have
Region A: net field horizontal from right to left
Region B: some complicated field we don’t care about
Region C and D: radially outwards.
Due Date: October. 14, 2014, 5:00 pm
6
Electricity and Magnetism: PHY-204.
D
3 October, 2014
B
C
A
O
a
Field due to shell
x=o
x=-R
x=+R
Potential at r → ∞ is 0. The potential at point a is found by displacing a test charge
along the dashed line and will be the same as the potential due to a charged shell at
the surface of the shell. Hence
1 Q
4πε0 R
1
4πR2
σR
=
σ
=
·
4πε0
R
ε0
Vr =
Now as you move from point a to O, the only field seen is the uniform field due to the
plates. The potential drops by ER =
σ
R.
ε0
Hence potential at O is
σR
ε0
−
σR
ε0
= 0. By
symmetry, we can find the potential on the other side of the plate. In summary, the
graph of potential along the dashed line will look below.
V(r)
charge on shell
1 Q
4πεo R
-R
0
R
1 Q
4πεo R
4. A parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed
to slide in the direction parallel to the paltes. Let x be the distance of overlap, as
shown in Fig. (3). The separation between the plates is fixed.
Due Date: October. 14, 2014, 5:00 pm
7
Electricity and Magnetism: PHY-204.
3 October, 2014
(Movable)
(Fixed )
x
Fig. (3)
(a) Assume that the plates are electrically isolated, so that their charges ±Q are
constant. In terms of Q and the (variable) capacitance C, derive an expression
for the leftward force on the movable plate. HINT : Consider how the energy of
the system changes with x.
(b) Now assume that the plates are connected to a battery, so that the potential
difference ϕ is held constant. In terms of ϕ and the capacitance C, derive an
expression for the force.
(c) If the movable plate is held in place by an opposing force, then either of the above
two setups could be the relevant one, because nothing is moving. So the forces
in (a) and (b) should be equal. Verify that this is the case.
Answer
l
+++++++++++++++++
d
− − −−−− − −−− − −− − −− −
x
(a)
Total charge on each plate = Q
( )
x
overlapping charge = Q0 = Q
l
ε0 wx
,
capacitance = C(x) =
d
where w is the width of each plate.
Due Date: October. 14, 2014, 5:00 pm
8
Electricity and Magnetism: PHY-204.
3 October, 2014
(b) In this part, Q is fixed.
Q20
2C(x)
Q2 x2
=
2C(x)l2
dU
F = −
=
dx
U =
only Q0 matters
Q2 xd
·
2l2 ε0 w
−Q2 d
·
2l2 ε0 w
=
(5)
Let’s put this equation in terms of V , the resulting potential difference between
the plates.
Now
( )
x
⇒ Q0 = Q
= CV
l
( )
l
⇒ Q = CV
.
x
Q0
V =
C
By inserting Q above in Eq (5):
F = −
(
)
(l) 2
CV x
d
2l2 ε0 w
(
)
−C 2 V 2
d
=
2x
ε0 wx
2 2
−C V 1
CV 2
=
=−
·
2x C
2x
(6)
(c)
1
C(x)V 2
2
dU
Now F = −
dx
1 ε0 wV 2
CV 2
= −
=−
·
2 d
2x
U =
Now V is fixed. This is identical to the result in Eq (6).
(d) We will discuss in class.
5. Three conducting plates are placed parallel to one another as shown in Fig. (4). The
outer plates are connected by a wire. The inner plate is isolated and has net surface
charge density of σ (the combined value from the top and bottom faces of the plate).
What are the surface densities, σ1 and σ2 , on the top and bottom faces of the inner
plate?
Due Date: October. 14, 2014, 5:00 pm
9
Electricity and Magnetism: PHY-204.
3 October, 2014
Fig. (4)
Answer
Due Date: October. 14, 2014, 5:00 pm
10
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