22.10
What are the horizontal and vertical components of the resultant electrostatic force
on the charge in the lower left corner of the square if q = 1.0 ∗ 10−7 C and a = 5.0cm ?
+q
-q
a
ϑ
a
+2q
-q
F =
F12 =
k(q)(2q)
a2
F13 =
kq1 q2
r2
k(−q)(2q)
cosθx̂ + sinθŷ
2a2
tanθ =
a
a
θ=
F14 =
k(−2q)(2q)
a2
π
4
Sum the forces to get the total force on the +2q particle
X
22.31
F = F12 + F13 + F14 =
q2
[4.707x̂ − 1.293ŷ]
4π0 a2
An electron is in a vacuum near the surface of Earth. Where should a second electron be
placed so that the electrostatic force it exerts on the first electron balances the weight of
the first electron?
-e
r
-e
Earth
Fg = mg − ŷ
Fe =
Sum the forces to get the total force on the electron
1
e2
ŷ
4π0 r 2
X
s
r=
22.40
2
e2
=
4π0 mg
s
2
F = Fg + Fe =
e2
− mgŷ = 0
4π0 r 2
(9 ∗ 109 F/m)(1.6 ∗ 10−19 C)2
= 5.02m below the electron
(9.31 ∗ 10−31 kg)(9.8m/s2 )
In the radioactive decay of U-238, the center of the emerging He particle is, at
a certain instant, 9.0 ∗ 10−15 m from the center of the daughter nucleus Th-234. At
this instant, (a) what is the magnitude of the electrostatic force on the He particle, and
(b) what is that particle’s acceleration?
Fe =
q1 q2
where q1 = 90 ∗ 1.6 ∗ 10−19C and q2 = 2 ∗ 1.6 ∗ 10−19 C
4π0 r 2
Fe = 512N
mHe = 6.68 ∗ 10−27 kg
F = ma
a=
23.25
512N
F
=
= 7.66 ∗ 1028 m/s2
m
6.68 ∗ 10−27kg
Find the magnitude and direction of the electric field at point P due to the electric
dipole in Fig. 23-37. P is located at a distance r > d along the perpendicular bisector of
the line joining the charges. Express your answer in terms of the magnitude and direction of
the electric dipole moment p.
The electric field at point P is the sum of the fields from the positive and negative charges
of the dipole, so E = E+ + E−. The electric field from one point charge in the dipole is
given by E = 4πq0 R2
+q
r+
d/2
ϑ
r
ϑ
E-
d/2
P
ϑ
E+
r-q
kq
E+ = 2 r̂+
r+
r
where r+ =
r
cosθ = q
2
r 2 + ( d2 )2
so,
E+ =
2
d
r 2 + ( )2
2
and r̂+ = cosθx̂ − sinθŷ
d
and
sinθ = q 2
2
r 2 + ( d2 )2
kq
d
ŷ)
3 (−r x̂ −
2
(r 2 + ( d2 )2 ) 2
2
r
kq
E− = 2 r̂−
r−
2
where r− =
d
r 2 + ( )2
2
d
r
cosθ = q
2
2
r + ( d2 )2
again,
so,
E− =
−kp
3
(r 2 + ( d2 )2 ) 2
ŷ
sinθ = q 2
2
r 2 + ( d2 )2
and
kq
(r 2
+
3
( d2 )2 ) 2
E = E+ + E− =
N ow,
E=
and r̂− = −cosθx̂ − sinθŷ
where
d
(r x̂ − ŷ)
2
−kqd
3 ŷ
(r 2 + ( d2 )2 ) 2
p = dipole moment
For, r >> d then r 2 + ( d2 )2 → r 2 so,
E→
23.34
(a) λ =
(b)
−kp
ŷ
r3
In Fig. 23-42, a nonconducting rod of length L has charge -q uniformly distributed along its length. (a)
What is the linear charge density of the rod? (b) What is the electric field at point P, a distance a from
the rod? (c) If P were very far from the rod compared to L, the rod would look like a point charge. Show that
your answer to (b) reduces to the electric field of a point charge for a >> L.
−q
L
dE =
kdq
x2
with
giving
Z
integrating gives
E=
a
a+L
dE =
dq =
λ
dx
kλ
dx
x2
1
−kq
−kλ a+L
1
kλ
|
)=
x̂
dx =
= kλ( −
x2
x a
a a+L
a(a + L)
(c) for a >> L, a + L → a, so
E→
23.54
−kq
x̂
a2
At some instant the velocity components of an electron moving between two charged parallel plates are
vx = 1.5 ∗ 105 m/s and vy = 3.0 ∗ 103 m/s. Suppose that the electric field between the plates is
given by E = (120N/C)j. (a) What is the acceleration of the electron? (b) What will be the
velocity of the electron after its x coordinate has changed by 2.0 cm?
F = ma = qE
a=
−1.6 ∗ 10−19 C(120N/C ĵ)
qE
=
= −2.1 ∗ 1013 m/s2 ĵ
m
9.11 ∗ 10−31 kg
There is no acceleration in the x direction, so x − x0 = v0xt giving
t=
0.02m
δx
= 1.3 ∗ 10−7 s
=
v0x
1.5 ∗ 105 m/s
3
vy = v0y + at = 3.0 ∗ 103 m/s + (−2.1 ∗ 1013 m/s2 ĵ)(1.3 ∗ 10−7 s) = −2.8 ∗ 10−6 m/s
v = (1.5 ∗ 105 î − 2.7 ∗ 106 ĵ)m/s
23.59
An electric dipole consists of charges of +2e and -2e separated by 0.78nm. It is in an electric field
of strength 3.4 ∗ 106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment
is (a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.
(a)
pkE
τ = p×E = 0
(b)
p⊥E
τ = p × E = pE = (2.5 ∗ 10−28C · m)(3.4 ∗ 106 N/C) = 8.8 ∗ 10−22N · m
(c)
p k −E
τ = p×E = 0
24.12
Find the net flux through the cube of Exercise 3 and Fig 24-28 if the electric field is given by
(a) E = 3.00yj and (b) E = −4.00i + (6.00 + 3.00y)j. E is in N/C, and y is in meters
(c) In each case, how much charge is enclosed by the cube?
(a)
I
E · dA
Φ=
Only sides of the cube with area vectors parallel to the E field will contribute to the flux. For part (a), the
E field is going in the y direction, so the two sides with area vectors in the y direction are those perpendicular
to the y-axis.
Z
1.4 Z 1.4
Φ=
Z
1.4 Z 1.4
Φ=
0
0
0
0
Z
Edxdzy=0 −̂j +
Z
(3.00)(0m)dxdz − ĵ +
1.4 Z 1.4
0
Edxdzy=1.4 ĵ
0
1.4 Z 1.4
(3.00)(1.4m)dxdz ĵ = 8.23
0
0
N · m3
C
Φ = qenc0
so,
qenc = (8.23
N · m3
)(8.85 ∗ 10−12 F/m) = 7.28 ∗ 10−11 C
C
(b)
1.4 Z 1.4
Z
Φ=
0
Z
0
1.4 Z 1.4
+
Z
0
0
1.4 Z 1.4
Φ=
0
0
Z
(−4.00î + (6.00 + 3.00y)ĵ) · dxdzy=0 (−ĵ) +
0
Z
(−4.00î + (6.00 + 3.00y)ĵ) · dydzx=0 (−î) +
Z
−(6.00 + 3.00(0))dxdz +
1.4 Z 1.4
1.4 Z 1.4
0
0
1.4 Z 1.4
0
(−4.00î + (6.00 + 3.00y)ĵ) · dxdzy=1.4(ĵ)
(−4.00î + (6.00 + 3.00y)ĵ) · dydzx=1.4 (î)
Z
(6.00 + 3.00(1.4))dxdz +
0
0
1.4 Z 1.4
4.00dydz +
0
4
Z
1.4 Z 1.4
0
0
0
−4.00dydz
Φ = −(6.00)(1.4)(1.4) + (6.00 + (3.00)(1.4))(1.4)(1.4) + (4.00)(1.4)(1.4) − (4.00)(1.4)(1.4)
N · m3
C
Φ = qenc0
Φ = 8.23
so,
qenc = (8.23
24.19
N · m3
)(8.85 ∗ 10−12 F/m) = 7.28 ∗ 10−11 C
C
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
metallic satellite 1.3m in diameter accumulates 2.4µC of charge in one orbital revolution. (a) Find
the resulting surface charge density. (b) Calculate the magnitude of the resulting electric field just
outside the surface of the satellite due to the surface charge.
(a)
σ=
2.4µC
C
q
=
= 4.52 ∗ 10−7 2
Area
4π(0.65m)2
m
(b)
I
E · dA =
E(4πR2 =
24.24
2.4µC
0
qenc
0
E = 5.1 ∗ 104 N/C
giving
Figure 24-32 shows a section of a long, thin-walled metal tube of radius R, carrying a charge per unit
length λ on its surface. Derive expressions for E in terms of distance R from the tube axis, considering
both (a) r > R and (b) r < R. Plot your results for the r = 0 to r = 5.0cm, assuming that λ = 2.0 ∗ 10−8 C/m//and R =
(a) for r > R
I
E · dA =
qenc
0
with
E(2πrL) =
Er>R =
λL
0
λ
2πr0
(b) for r < R, qenc = 0, so
Er<R = 0
5
qenc = λL
24.52
A solid nonconducting sphere of radius R has a nonuniform charge distribution of volume charge density
ρ = ρs r/R, where ρs is a constant and r is the distance from the center of the sphere.
Show that (a) the total charge on the sphere is Q = πρs R3 and (b) the electric field inside the
sphere has a magnitude given by:
E=
(a)
Z
Z
ρdv =
Q=
1 Q 2
r
4π0 R64
Z
2π
R
sinθdθ
0
v
Z
π
dφ
0
0
r2
ρs r
dr
R
4πρs r 4 R
| = πρs R3
Q=
R 4 0
(b)
I
E · dA =
qenc
0
with
qenc =
E(4πr 2 ) =
4πρs r 04 r πρs 4
| =
r
R 4 0
R
πρs 4
r
R0
so,
E=
Q
ρs 2
r =
r2
4R0
4πR4 0
6