# Homework Assignment #2

Due: Thursday 4 February 1999

Chapters 25 and 26.

### QUESTIONS

Chapter 25 - 1, 2, 4, 6, 8, 10, 14, 15, 16.
Chapter 26 - 1, 4, 10, 11.

### PROBLEMS

Chapter 25 - 2, 6, 12, 23, 24, 26, 32, 39, 42, 45, 50, 53, 54A, 54, Review Problem.
Chapter 26 - 3, 6, 8, 10, 15, 20, 24, 28, 29, 33, 46, 53, 58.

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to copy these and turn them in as homework. You must show your work.

Q 25-1) They are fundamentally different quantities; electric potential (or voltage) is the electric potential energy per unit charge. The MKS unit of potential is the volt (=joule/coulomb) and the MKS unit of potential energy is the joule.

Q 25-2) The potential energy U increases. It moves to a position of lower potential V.
Q 25-4) In a direction perpendicular to the electric field (along the y or z directions).
Q 25-6)
1. concentric cylinders
2. concentric spheres

Q 25-8) No, it implies only that the potential V is a constant.
Q 25-10) The smaller sphere has the greater charge density. (Although the larger sphere has the greater total charge.)
Q 25-14) For a given voltage V applied to a conductor, the electric field E is larger in the vicinity of sharp corners. If the electric field grows large enough, sparks may emerge from the corners. The charge may then drain away from the conductor. Notice that the collecting sphere at the top of the van de Graaff generator is very smooth.
Q 25-15) Put the circuit to be shielded in a closed conducting box. Such a device is called a "Faraday cage". If there are no charges inside the box, then there is no electric field inside the box. (See pages 722-3.)
Q 25-16) (See previous question.)
Q 26-1) The charge is also doubled. Q=CV --> 2Q=C(2V)
Q 26-4) The pair connected in parallel is more dangerous because it holds 4 times the charge of the pair in series (and twice as much charge as a single capacitor).
Q 26-10) The factor of 1/2 comes from integrating the infinitesimal work dW required to move an infinitesimal charge dq from one plate to the other. (See page 750.)
Q 26-11) By a factor of four. energy stored = 1/2 C V2
P 25-2) 7.8 x 10-4 J = 4.88 x 1015 eV
P 25-6) 0.502 V
P 25-12) 6.67 x 104 V/m (volt/meter = newton/coulomb)
P 25-23) q = 1.19 x 10-7 C; r = 2.67 m
P 25-24)
1. 0
2. 0
3. 45,000 V

P 25-26)
1. -4.83 m
2. +0.67 m and -2 m

P 25-32) -3.96 J
P 25-39)
Ex = -5 + 6xy
Ey = 3x2 - 2z2
Ez = -4yz
E=7.08 V/m
P 25-42)
1. 0

P 25-45) 0.553 kQ/R
P 25-50) k lambda [pi + 2 ln(3)]
P 25-53)
1. E=0; V=1.67 x 106 V
2. E=5.85 x 106 V/m; V=1.17 x 106 V
3. E=11.9 x 106 V/m; V=1.67 x 106 V

P 25-54A) none on the inner sphere, Q on the outer sphere
P 25-54) none on the inner sphere, 10 microcoulombs on the outer sphere
P 25-Review Problem)
1. kQ/r2
2. kQr/R3
3. kQ/r
4. kQ/R + kQ/2R3(R2 - r2)
5. (Graphs required)
6. kQq/r2 toward the center of the sphere
7. -kQq/r
8. sqrt[2kQq/m (1/R - 1/r)]

P 26-3)
2. 100 V

P 26-6) 31/3 = 1.44
P 26-8)
1. 5 microcoulombs on the larger sphere; 2 microcoulombs on the smaller sphere
2. 90,000 V

P 26-10) 43.6 square miles
P 26-15)
1. 1.11 x 104 V/m
2. 9.83 x 10-8 C/m2
3. 3.74 x 10-12 F
4. 74.8 x 10-12 C

P 26-20) 1.07 x 10-2 m
P 26-24) 7.08 x 10-4 F (corrected by Angela Watson)
P 26-28) 6 pF and 3 pF
P 26-29)
2. On the 2 microfarad capacitor: 12 V and 24 microcoulombs
On the 3 microfarad capacitor: 8 V and 24 microcoulombs
On the 6 microfarad capacitor: 4 V and 24 microcoulombs

P 26-33)
initial charge on C1 is 120 microcoulombs
final charge on C1 is 80 microcoulombs
final charge on C2 is 40 microcoulombs

P 26-46) both Q and V are doubled.
P 26-53) (Proof required.)
P 26-58) 21.9 pF

Please report any corrections to Professor Scalise.