PHY 142!
Assignment #3!
Summer 2013
Reading: Electrostatics 3.
Key concepts: Capacitance, energy storage, dielectrics, energy in the E-field.
1.!
2.!
3.!
4.!
Questions about charging and discharging capacitors.
a.!
When an uncharged capacitor is connected across a battery, why do equal
and opposite charges appear on the plates?
b.!
If the plates have different shapes or sizes, will the charges still be equal
and opposite? Explain.
c.!
Where does the energy come from when a capacitor is charged by a
battery? Where does it go if an isolated charged capacitor is discharged by
connecting a wire between the plates?
Questions about altering the capacitance of a parallel plate capacitor.
a.!
Why does moving the plates closer together increase C? [Consider an
isolated charged capacitor; what happens to ΔV ?]
b.!
If you suspend a copper sheet between the plates, parallel to them, what
happens to C? Suppose the copper sheet is resting on the bottom plate; is
the change different?
c.!
You have a copper sheet and a sheet of glass, the same size, that you can
insert between the capacitor plates. Which one will change C more? Why?
An isolated parallel plate capacitor has charge Q.
a.!
If the plates are moved closer together, the stored energy decreases. Where
does the “lost” energy go?
b.!
If a dielectric slab is inserted part way into the gap between the plates and
released, what happens? Explain.
c.!
If, instead of being isolated, the capacitor remained connected to the
charging battery, the energy increases when the plates are moved closer
together. Explain this carefully in terms of how much work is done by
what agent.
For some purposes, such as the effects of inserting slabs of material between
capacitor plates, a conductor can be regarded as a dielectric with an infinite value
of κ . Explain why this is so.
PHY 142!
5.!
Assignment #3!
Summer 2013
Two capacitors C1 and C2 connected in series or parallel can be replaced by a
single equivalent capacitor C, with charge Q and potential difference ΔV .
a.!
If the two capacitors are in series, which of the two quantities Q and ΔV is
the same for both and which is divided between them?
b.!
Answer the same question if the two capacitors are in parallel.
c.!
Use your answers to derive the rules giving C in terms of C1 and C2 for
the two cases.
6.!
An external E-field produces aligned dipole moments in a dielectric material; this
in turn results in the appearance of excess charge on the surface of the material. It
is not free to move around, so it is called “bound” charge, and its area density is
σ b . Consider a slab of dielectric filling the gap between the parallel plates of a
capacitor. The charge on the plates would produce a field in the gap E0 if the
dielectric were not there.
a.!
Because the dipoles are aligned with the external field, show that the
bound surface charge next to the positive plate of the capacitor is negative,
and vice versa. [Draw a picture.]
b.!
Express the net field in the slab, Eav , in terms of E0 and σ b .
c.!
Because the forces and torques that produce the aligned dipoles are
proportional to Eav , we can write σ b = ε 0 χ Eav , where the dimensionless
number χ is the electric susceptibilty of the material. Use this to show that
Eav = E0 /(1 + χ ) . [The dielectric constant is defined as κ = 1 + χ .]
7.!
Questions about energy in the E-field.
a.!
The formula for energy density in the absence of dielectrics is ue = 12 ε 0E2 ,
while the formula for dielectrics is ue = 12 κε 0E2 . The dielectric constant κ
is greater than 1, but inserting a dielectric into a region where there is an
E-field reduces the stored energy. Explain.
b.!
Refer to Question 3 (a). Explain the reduction of stored energy in terms of
energy in the E-field.
PHY 142!
8.!
9.!
10*.!
Assignment #3!
Summer 2013
More questions about energy in the E-field.
a.!
A balloon carries charge Q spread uniformly over its outer surface. If its
radius is increased by blowing it up, what happens to the total stored
energy in the E-field?
b.!
Two spheres, each of radius R, carry charge Q. Sphere A is a conductor,
while sphere B is a non-conductor and the charge is distributed uniformly
through its volume. For which one is the energy stored in the E-field
greater? Explain.
A charged parallel plate capacitor is slowly lowered into a liquid
dielectric material until its plates are half submerged. The
liquid level between the plates is observed to be higher than
that outside the plates, as shown.
+
a.!
Explain why the level rises between the plates.
b.!
How would you go about determining how much the
level rises? What principle would you use? What facts
about the capacitor and the dielectric material would you need?
A spherical capacitor consists of a pair of concentric
spherical conducting shells as shown. The inner shell, of
radius a, carries positive charge Q. The outer shell, of radius
b, is grounded.
−
Q
a.!
What is the potential at the surface of the inner shell?
⎛ 1 1⎞
Ans: V(a) = kQ ⎜ − ⎟ .
⎝ a b⎠
b.!
What is the capacitance of the system? Ans:
c.!
How much energy is stored in the E-field? Ans: kQ 2
d.!
What would your answers to the previous questions be if the gap between
the shells were filled with a dielectric of constant κ ?
1 ab
.
k b− a
b− a
.
2ab
PHY 142!
11.!
Assignment #3!
Referring to the previous problem:
a.!
What is the energy density ue in the gap between the shells at distance r
from the center? Ans:
12*.!
Summer 2013
kQ 2 1
.
8π r 4
b.!
Consider an infinitesimal volume dV consisting of a spherical shell of
radius r and thickness dr situated in the gap between the conductors. Call
the energy stored in this volume dU. Express dU in terms of Q and r.
c.!
Integrate over r from a to b to find the total energy in the field. Compare
your answer to Prob. 10(c).
d.!
Let b → ∞ to find the total stored energy in the field of a conducting
sphere of radius a carrying charge Q.
A 1 µF parallel plate capacitor is charged by a 100 V battery. The separation
between the plates is slowly reduced until the capacitance becomes 2 µF , while
the battery remains connected.
13*.!
a.!
How much has the capacitor’s stored energy increased? Ans: 5 × 10−3 J.
b.!
How much additional charge ΔQ has been put on the plates? Ans: 10−4 C.
c.!
How much additional energy was added to the system by the battery? [It
moves charge ΔQ through potential difference 100 V.] Ans: 10−2 J.
d.!
Account for the difference between your answers to (a) and (c)
Consider a system of three identical parallel plate capacitors
(numbered 1, 2, 3) each having a capacitance C = 1 µF and
connected as shown.
a.!
b.!
If 60 V is applied across the terminals of the system and
the capacitors become fully charged, how much energy
is stored in capacitor #1? [Find the charge on it.] Ans:
8 × 10−4 J.
2
1
3
The system is now isolated from the charging source, and a slab of
dielectric with κ = 2 is slipped between the plates of #1, filling the gap
completely. What is the new stored energy in #1? [What remains
constant?] Ans: 4 × 10−4 J.
c.!
What is the new potential difference across the terminals? Ans: 40 V.
PHY 142!
14.!
Assignment #3!
The capacitor shown is half filled with a dielectric of constant κ .
Its positive plate is at potential V0 above that of its negative plate.
Summer 2013
−
+
The plates have total area A and are separated by distance d.
a.!
What is the surface charge density σ on the bottom half of
the positive plate? [What is the E-field in the empty gap
between the plates?] Ans: σ = ε 0V0 /d .
b.!
The potential difference is the same in both halves, so the
total E-field must be the same in both halves. Because of the dielectric the
charge density σ ′ on the top half of the plates is larger than on the bottom
half. What is σ ′ ? Ans: σ ′ = κε 0V0 /d .
c.!
15.!
Find the total charge on the positive plate and the capacitance of the
ε A(1 + κ )
ε A(1 + κ )
system. Ans: Qtot = V0 ⋅ 0
, C= 0
.
2d
2d
Two capacitors of capacitance C1 and C2 are arranged as shown.
They are charged such that the total charge is Q. The switch
shown is closed and after a time all charges are at rest.
1
2
a.!
What fraction of the charge is now on capacitor #1? Ans:
Q1 /Q = C1 /(C1 + C2 ) .
b.!
What is the potential difference across them? Ans: Q/(C1 + C2 ) .!
c.!
If, before the switch was closed, all the charge was on capacitor #1, what is
the change in stored energy after the switch is closed? Ans:
C2
Q2
ΔU = −
.
C1 2(C1 + C2 )