Set 3. Chap. 24 – Due at 11:00 pm Thurs., Mar. 25, 2010
Page 648. HWR 7th edition (2005). 24: 10, 15, 31, 39 and 57.
Chapter 24.
10.
As a space shuttle moves through the dilute ionized gas of Earth’s
ionosphere, the shuttle’s potential is typically changed by –1.0 V
during one revolution. Assuming the shuttle is a sphere of radius 10
m, estimate the amount of charge it collects.
The voltage on the sphere is V =
q=
15.
kq
so the charge is
R
RV
(10m) (−1.0V)
=
= −1.11×10−9 C.
9
2
2
k
8.99 ×10 N ⋅ m /C
In Fig. 24-33, what is the net electric potential at point P due to the
four particles if V = 0 at infinity, q = 5.00 fC, and d = 4.00 cm?
A charge –5q is a distance 2d from P, a charge –5q is a distance d from P, and two
charges +5q are each a distance d from P, so the electric potential at P is (in SI units)
9
−15
 1 1 1 1  kq (8.99 ×10 )(5.00 ×10 )
V = kq  −
− + + =
=
= 5.62 ×10−4 V.
−2
2(4.00 ×10 )
 2d d d d  2 d
The zero of the electric potential was taken to be at infinity.
31.
The electric potential at points in an xy plane is given by
V = (2 x 2 − 3 y 2 )V/m 2
In unit-vector notation, what is the electric field at the point (3.0 m,
2.0 m)?
We use Eq. 24-41:
c
c
h
h
∂V
∂
=−
(2.0V / m2 ) x 2 − 3.0V / m2 ) y 2 = −2(2.0V / m2 ) x;
∂x
∂x
∂V
∂
E y ( x, y) = −
=−
(2.0V / m2 ) x 2 − 3.0V / m2 ) y 2 = 2(3.0V / m2 ) y .
∂y
∂y
Ex ( x, y) = −
We evaluate at x = 3.0 m and y = 2.0 m to obtain
r
E = (−12 V/m)iˆ + (12 V/m)jˆ .
39.
(a) What is the electric potential energy of two electrons separated by
2.00 nm? (b) If the separation increases, does the potential energy
increase or decrease?
(a) We use Eq. 24-43 with q1 = q2 = –e and r = 2.00 nm:
d
i
8.99 × 109 NC⋅m2 (160
. × 10 −19 C) 2
q1q2
e2
U =k
=k
=
= 115
. × 10−19 J.
r
r
2.00 × 10−9 m
2
(b) Since U > 0 and U ∝ r–1 the potential energy U decreases as r increases.
57.
A metal sphere of radius 15 cm has a net charge of 3.0 x 10-8 C. (a)
What is the electric field at the sphere’s surface? (b) If V = 0 at
infinity, what is the electric potential at the sphere’s surface? (c) At
what distance from the sphere’s surface has the electric potential
decreased by 500 V?
(a) The magnitude of the electric field is
( 3.0 ×10 C )  8.99 ×10
−8
E=
σ kq
=
=
ε 0 R2

2
( 0.15 m )
9
N ⋅ m2 

C2 
= 1.20 × 104 N C.
(b) V = RE = (0.15 m)(1.20 × 104 N/C) = 1.80 × 103 V.
(c) Let the distance be x. Then
bg
∆V = V x − V =
FG
H
IJ
K
q
1
1
−
= −500 V,
4 πε 0 R + x R
which gives
x=
b
gb
g
015
. m −500 V
R∆V
=
= 58
. × 10−2 m.
−V − ∆V −1800 V + 500 V