Download Chapter 2 - Omar A. Ashour

Omar Ashour
Chapter 2 Review
September 13, 2015
Problem 1
University Physics, problem 67
A region in space contains a total positive charge that is distributed spherically such that
the volume charge density ρ (r) is given by
for r ≤ R/2
ρ (r) = 3αr/ (2R)
2
ρ (r) = α[1 − (r/R) ]
for R/2 ≤ r ≤ R
for r ≥ R
ρ (r) = 0
Here α is a positive constant having units of C/m3
a. Determine α in terms of Q and R
b. Using Gausss law, derive an expression for the magnitude of the electric field as a function
of r. Do this separately for all three regions. Express your answers in terms of the total
charge.
c. What fraction of the total charge is contained within the region R/2 ≤ r ≤ R
~ at r = R/2
d. What is the magnitude of E
e. If an electron with charge q 0 = −e is released from rest at any point in any of the three
regions, the resulting motion will be oscillatory but not simple harmonic. Why?
Solution
Part (a)
ρ=
dQ
=⇒ Q =
dV
Z
b
ρ dV
a
dV = Adr = 4πr2 dr
Z
3αr
6πα R/2 3
6πα 1 R
3
2
Q1 =
4πr dr =
r dr =
= παR3
2R
R 0
R 4 16
32
0
Z R
r 2
7
31
47
Q2 =
α 1−
4πr2 dr = 4παR3
−
=
παR3
R
24
160
120
R/2
Z
R/2 (1)
(2)
1
Omar Ashour
Chapter 2 Review
September 13, 2015
From (1) and (2)
Q = Q1 + Q2 =
47
3
233
παR3 + παR3 =
παR3
120
32
480
α=
480Q
233πR3
Part (b)
For the region where r ≤ R/2
Z
~ · dA
~ = Qen , the enclosed charge is a fraction of Q1
E
0
S
r
Z
Qen =
0
r
3α
4πr2 dr =
2R
Z
r
3
0
480Q
233πR3
r 720Q
4πr2 dr =
·4
2R
233R4
E1 · 4πr2 =
E1 =
Z
r
r3 dr =
0
720Qr4
233R4
720Qr4
233R4 0
180Qr2
233π0 R4
For the region where R/2 ≤ r ≤ R
Z
~ · dA
~ = Qen , the enclosed charge is the sum of Q1 and a fraction of Q2
E
0
S
0 2 !
03
Z r 0
r
r04
r
r05 r
02
02
0
Qen = Q1 +
α 1−
4πr dr = Q1 + 4πα
r − 2 dr = Q1 + 4πα
−
R
R
3
5R2 R/2
R/2
R/2
3
3
r
R3
r5
R3
3
1
1 r5
1
3
3 1 r
Qen = Q1 + 4πα
−
−
+
= παR + 4παR
−
−
+
3
24
5R2 160
32
3 R3 24 5 R5 160
3
1 r 3 1 r 5
17
480Q
1 r 3 1 r 5
23
3
3
Qen = 4παR
+
−
−
= 4πR
−
−
128 3 R
5 R
480
233πR3
3 R
5 R
1920
3
1 Qen
4πR
480Q
1 r 3 1 r 5
23
E2 =
=
−
−
2
2
3
4π0 r
4π0 r
233πR
3 R
5 R
1920
Z
r
2
Omar Ashour
Chapter 2 Review
480Q
E2 =
233π0 r2
September 13, 2015
1 r 3 1 r 5
23
−
−
3 R
5 R
1920
For the region where r ≥ R
The enclosed charge is simply equal to the total charge Q.
1
Q
E3 =
· 2
4π0 r
Part (c)
The fraction of enclosed charge is simply the ratio between Q2 and Q.
Q2
47
233
=
παR3 ÷
παR3 = 0.807
Q
120
480
Part (d)
Using either E2 or E3 for r = R/2
E=
180 Q
233 4π0 R2
Part (e)
~ = qE
~
F
The condition for simple harmonic motion is to have F ∝ r which is not the case here as F ∝
1
r2
3
Omar Ashour
Chapter 2 Review
September 13, 2015
Problem 2
University Physics, problem 48 (exam problem)
A small conducting spherical shell with inner radius a and outer radius b is concentric with
a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has
total charge +2q the outer shell has charge −2q.
a. Calculate the electric field (magnitude and direction) in terms of q and the distance r
from the common center of the two shells for
i. r < a
ii. a < r < b
iii. b < r < c
iv. c < r < d
v. r > d
b. What is the total charge on the
i. inner surface of the small shell
ii. outer surface of the small shell
iii. inner surface of the large shell
iv. outer surface of the large shell
Solution
Part (a)
i. E1 = 0; charge enclosed is 0.
ii. E2 = 0; inside a conductor.
iii. E3 =
1 2q
1 q
=
4π0 r2
2π0 r2
iv. E4 = 0; inside a conductor.
v. E5 = 0; charge enclosed is −2q + 2q = 0.
Part (b)
To solve this part, we take a Gaussian surface in the regions where we want to find the charge.
i. We take a Gaussian surface with radius r where a < r < b. Electric field is zero thus the
charge enclosed is zero.
ii. We take a Gaussian surface with radius r where b < r < c. Electric field in this region is
the same as that of a point charge +2q, the charge on the surface is +2q
iii. We take a Gaussian surface with radius r where c < r < d. Electric field in this region
is zero, which means that the enclosed charge is zero. The enclose charge Qen = +2q +
qinner surface = 0 =⇒ qinner surface = −2q
iv. This spherical shell overall has a charge of −2q which is already present on the inner
surface, thus the charge on the outer surface is zero.
4