Physics tutorial #15

1. Show that the equivalent capacitance C

eq

of a combination of capacitors C

1

and C

2

satisfies relation

(a) C

−1

eq

= C

−1

1

+ C

−1

2

when they are connected in series,

(b) C

eq

= C

1

+ C

2

when they are connected in parallel.

2. The capacitor C

1

in figure below initially has a charge Q

0

, and the voltage between its plates is V

0

.

After the switch is closed, the capacitor C

2

becomes charged. Find the ratio V /V

0

where V is the

final voltage across the capacitors, and evaluate the ratio U/U

0

of the final to the initial stored energy.

0

C

1

0

Q

V

C

2

3. Two concentric spherical conducting shells are separated by vacuum. The inner shell has total charge

+Q and outer radius r

a

= 9.5 cm, and the outer shell has charge −Q and inner radius r

a

= 10.5 cm.

(a) Use the Gauss’s law to find the capacitance of this spherical capacitor.
(b) Find the electric potential energy using the capacitance from (a).
(c) Find the electric potential energy by integrating the electric-field energy density.

4. Is it possible to find two capacitors which connected in parallel give the same equivalent capacity as

when connected in series? Prove your answer.

5. An air capacitor is made by using two flat plates, each with area A, separated by a distance d. Then

a metal slab having thickness a (less than d) and the same shape and size as the plates is inserted
between them, parallel to the plates and not touching either plate. (a) What is the capacitance of
this arrangement? (b) Express the capacitance as a multiple of the capacitance C

0

when the metal

slab is not present. (c) Discuss what happens to the capacitance in the limits a → 0 and a → d.

6. A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric, one with

constant K

1

and one with constant K

2

. Each slab has thickness d/2, where d is the plate separation.

Show that the capacitance is

C =

2

0

A

d



K

1

K

2

K

1

+ K

2



7. A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric, one

with constant K

1

and one with constant K

2

. The thickness of each slab is the same as the plate

separation d, and each slab fills half of the volume between the plates. Show that the capacitance is

C =



0

A

2d

(K

1

+ K

2

)

Nivas Babu Selvaraj, Maciej Wo loszyn
http://fatcat.ftj.agh.edu.pl/~woloszyn/phys/