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Set 2. Chapters 22 and 23 – Due Thurs., Mar. 18, 2010
Page 601. HWR 7th edition (2005). 22: 41, 51. 23: 2, 17, 21, 43
Chapter 22.
41.
Beams of high-speed protons can be produced in “guns” using electric
fields to accelerate the protons. (a) What acceleration would a proton
experience if the gun’s electric field were 2.00 x 104 N/C? (b) What
speed would the proton attain if the field accelerated the proton
through a distance of 1.00 cm?
(a) The magnitude of the force acting on the proton is F = eE, where E is the magnitude
of the electric field. According to Newton’s second law, the acceleration of the proton is
a = F/m = eE/m, where m is the mass of the proton. Thus,
c1.60 × 10 Chc2.00 × 10
a=
−19
167
. × 10
−27
kg
4
NC
h = 1.92 × 10
12
2
ms .
(b) We assume the proton starts from rest and use the kinematic equation v 2 = v02 + 2ax
1
(or else x = at 2 and v = at) to show that
2
d
v = 2ax = 2 192
. × 1012 m s
51.
2
. × 10 m s.
ib0.0100 mg = 196
5
An electric dipole consisting of charges of magnitude 1.50 nC
separated by 6.20 µm is in an electric field of strength 1100 N/C.
What are (a) the magnitude of the electric dipole moment and (b) the
difference between the potential
energies for dipole orientations
r
parallel and antiparallel to E ?
(a) The magnitude of the dipole moment is
c
hc
h
p = qd = 150
. × 10 −9 C 6.20 × 10−6 m = 9.30 × 10 −15 C ⋅ m.
(b) Following the solution to part (c) of Sample Problem 22-6, we find
b g bg
c
hb g
U 180° − U 0 = 2 pE = 2 9.30 × 10−15 1100 = 2.05 × 10 −11 J.
Chapter 23.
2.
The cube in Fig. 23-27 has edge length 1.40 m and is oriented as
shown in a region of uniform electric field. Find the electric flux Φ
through the right face (which points in the y-direction) if the electric
field, in newtons per coulomb, is given by (a) 6.00 î , (b) –2.00 ĵ , and
(c) –3.00 î + 4.00 k̂ . (d) What is the total flux through the cube for
each field?
b
g
r r
r
2
We use Φ = E ⋅ A , where A = A$j = 140
. m $j .
2
(a) Φ = ( 6.00 N C ) ˆi ⋅ (1.40 m ) ˆj = 0.
2
(b) Φ = ( −2.00 N C ) ˆj ⋅ (1.40 m ) ˆj = −3.92 N ⋅ m 2 C.
2
(c) Φ = ( −3.00 N C ) ˆi + ( 400 N C ) kˆ  ⋅ (1.40 m ) ˆj = 0 .


(d) The total flux of a uniform field through a closed surface is always zero.
17.
Space vehicles traveling through Earth’s radiation belts can intercept a
significant number of electrons. The resulting charge buildup can
damage electronic components and disrupt operations. Suppose a
spherical metal satellite 1.3 m in diameter accumulates 2.4 µC of
charge in one orbital revolution. (a) Find the resulting surface charge
density. (b) Calculate the magnitude of the electric field just outside
the surface of the satellite, due to the surface charge.
(a) The area of a sphere may be written 4πR2= πD2. Thus,
q
2.4 × 10−6 C
σ=
=
= 4.5 × 10−7 C/m 2 .
2
2
πD
π (1.3 m )
(b) Eq. 23-11 gives
σ
4.5 × 10−7 C/m 2
E= =
= 5.1× 104 N/C.
−12
2
2
ε 0 8.85 ×10 C / N.m
21.
An infinite line of charge produces a field of magnitude 4.5 x 104 N/C
at a distance of 2.0 m. Calculate the linear charge density.
The magnitude of the electric field produced by a uniformly charged infinite line is E =
λ/2πε0r, where λ is the linear charge density and r is the distance from the line to the
point where the field is measured. See Eq. 23-12. Thus,
(
λ = 2πε 0 Er = 2π 8.85 × 10−12 C 2 / N ⋅ m 2
43.
)( 4.5 ×10
)
N/C ( 2.0 m ) = 5.0 ×10 −6 C/m.
4
An unknown charge sits on a conducting solid sphere of radius 10 cm.
If the electric field 15 cm from the center of the sphere has the
magnitude 3.0 x 103 N/C and is directed radially inward, what is the
net charge on the sphere?
Charge is distributed uniformly over the surface of the sphere and the electric field it
produces at points outside the sphere is like the field of a point particle with charge equal
to the net charge on the sphere. That is, the magnitude of the field is given by E =
q/4πε0r2, where q is the magnitude of the charge on the sphere and r is the distance from
the center of the sphere to the point where the field is measured. Thus,
2
q = 4πε 0 r E =
( 0.15 m )
2
( 3.0 ×10
9
2
3
N/C
8.99 × 10 N ⋅ m / C
2
) = 7.5 ×10
−9
C.
The field points inward, toward the sphere center, so the charge is negative: –7.5 ×10–9 C.