Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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11.3 Solutions 509
Thus for r < R, E varies linearly with r. Note that at r = R, both (1)
and (4) give the same value as they should.
11.51 The electric field on the surface of the original sphere is
E
R
=
q
4πε
0
R
2
After creating the cavity, the charge in the remaining part of the sphere will
be
q
=
⎡
⎢
⎢
⎢
⎣
4
3
π R
3
ρ −
4
3
π
R
2
3
ρ
4
3
π R
3
ρ
⎤
⎥
⎥
⎥
⎦
q =
7
8
q
∴ The electric field on the surface is now
E
R
=
q
4πε
0
R
2
=
7
8
q
4πε
0
R
2
=
7
8
E
11.52
(a)
E =
q
4 πε
0
r
r
R
3
V (r) =−
Edr =−
q
4πε
0
R
3
r dr =−
qr
2
8πε
0
R
3
+ C
Now the potential at the surface (r = R)is
V (R) =
q
4πε
0
R
=−
q
8πε
0
R
+C
∴ C =
3q
8πε
0
R
∴ V (r) =
q
8πε
0
R
3 −
r
2
R
2
(b) At the centre r = 0. From the result of (a)
V (0) =
3q
8πε
0
R
=
3
2
V (R)
11.53
(a) E = 0 when r < b as no charge exists in this region.
(b) Region b < r < a: Consider a Gaussian surface, a sphere of radius
r, where b < r < a. Charge at distance a between b and r only will
contribute to the field. The charge residing in the shell of radii b and r is
510 11 Electrostatics
Q
=
4
3
π(r
3
− b
3
)Q
4π
3
(a
3
− b
3
)
=
(r
3
− b
3
)
(a
3
− b
3
)
Q
E =
Q
4πε
0
r
2
=
Q
4πε
0
r
2
(r
3
− b
3
)
(a
3
− b
3
)
(c) Region r > a : E =
Q
4πε
0
r
2
11.54
(a) Construct a Gaussian surface in the form of a cylinder of radius r > R
and height h. By Gauss’ law
(ε
0
E)(2πrh) = Q = ρπ R
2
h
∴ E =
1
2
ρ
ε
0
R
2
r
(b) Construct a Gaussian surface in the f orm of a cylinder of height h and
radius r < R coaxial with the cylinder of radius R and height h. By
Gauss’ law
(ε
0
E)(2πrh) = Q
= ρπ R
2
h
∴ E =
1
2
ρ
ε
0
R
2
r
(r < R)
11.55
(a) E =
q
4 πε
0
r
2
∴ q = 4 πε
0
r
2
E =
1
9 × 10
9
× (0.5)
2
× 800 = 2.22 × 10
−8
C
(b) E =
σ
ε
0
∴ σ = Eε
0
= 120 ×8.85 × 10
−12
= 1.062 ×10
−9
C/m
2
(c) q = ε
0
φ
φ = EA= 120 × (100 × 75) = 9 ×10
5
Nm
2
/C
11.56
(a)
E
0
E · dA = q (Gauss’ law) (1)
Consider an isolated positive point charge. Construct a Gaussian surface,
a sphere of radius r centred at the charge. At every point on the spherical
11.3 Solutions 511
surface, the field is perpendicular to the surface. As both E and dA are
directed radially outwards, the angle θ between them is zero so that the
dot product E · dA reduces to EdA and (1) can be written as
ε
0
E·dA = ε
0
EdA = q (2)
As E is constant, it can be factored out of the integral:
ε
0
E
dA = (ε
0
E)(4π r
2
) = q
∴ E =
q
4πε
0
r
2
(Coulomb’s law) (3)
(b) (i) r < R
By prob. (11.50), E =
Qr
4πε
0
R
3
φ = EA=
Qr
4πε
0
R
3
4π r
2
=
Qr
3
ε
0
R
3
(ii) r > R
E =
Q
4πε
0
r
2
φ = EA=
Q
4πε
0
r
2
4πr
2
=
Q
ε
0
11.57 φ =
5
E · dA, q = ε
0
A
(a) Construct a Gaussian surface, a sphere of radius r
1
concentric with the
spherical shells. Since no charge is enclosed by the Gaussian surface
with r < R
1
, E = 0.
(b) Here the net charge enclosed by the Gaussian surface is Q
1
.AsE is
normal to the spherical surface by Gauss’ law
(ε
0
E)(4π r
2
) = Q
1
or E =
Q
1
4πε
0
r
2
(c) Here the net charge enclosed by the Gaussian surface is Q
1
+ Q
2
and E
is normal to the spherical surface (Fig. 11.44). By Gauss’ law
(ε
0
E)(4πr
2
) = Q
1
+ Q
2
or E =
Q
1
+ Q
2
4πε
0
r
2
In (c) if Q
1
+ Q
2
= 0, then Q
1
/Q
2
=−1.
512 11 Electrostatics
Fig. 11.44
11.58 Let the charges be q
1
= q
2
on the two spheres before contact.
Force F =
q
1
q
2
4πε
0
r
2
=
q
2
1
4πε
0
r
2
(1)
When the spheres are brought into contact and separated they are at a com-
mon potential. Let the new charges be q
1
and q
2
, respectively
q
1
+q
2
= q
1
+q
2
= 2q
1
(from charge conservation) (2)
Also V =
q
1
4πε
0
R
1
=
q
2
4πε
0
R
2
(3)
or
q
1
q
2
=
R
1
R
2
(4)
Solving (2) and (4)
q
1
=
2q
1
R
1
R
1
+ R
2
, q
2
=
2q
1
R
2
R
1
+ R
2
(5)
After separation, force F
=
q
1
q
2
4πε
0
r
2
(6)
F
F
=
q
2
1
q
1
q
2
=
(r
1
+r
2
)
2
4r
1
r
2
=
(1 + 3)
2
4 × 1 × 3
=
4
3
(7)
11.59
V =
q
4πε
0
r
∴ q = 4 πε
0
rV=
(5)(2 × 10
4
)
9 × 10
9
= 1.11 ×10
−5
C
11.60 (a) V =
q
4πε
0
a
∴ Capacitance C =
q
V
= 4πε
0
a
11.3 Solutions 513
(b) q
1
= 4πε
0
aV
1
, q
2
= 4πε
0
aV
2
U
1
=
1
2
q
1
V
1
=
1
2
q
2
1
4πε
0
a
, U
2
=
1
2
q
2
2
4πε
0
a
Initially total energy U = U
1
+U
2
=
q
2
1
+q
2
2
8πε
0
a
(1)
When the spheres are connected they reach a common potential and
when disconnected let the charges be q
1
and q
2
.
q
1
= 4πε
0
aV, q
2
= 4πε
0
aV
∴ q
2
= q
1
Final energy in the spheres
U
1
=
1
2
q
1
V =
q
2
1
8πε
0
a
, U
2
=
q
2
2
8πε
0
a
Total final energy U
= U
1
+U
2
=
q
2
1
+q
2
2
8πε
0
a
(2)
q
1
+q
2
= q
1
+q
2
= 2q
1
(charge conservation) (3)
U = U
1
−U
2
=
1
8πε
0
a
q
2
1
+q
2
2
− 2q
2
1
(4)
Eliminating q
1
between (3) and (4) and simplifying
U =
1
16πε
0
a
(q
1
−q
2
)
2
(5)
This energy is dissipated in Joule heating of the wire.
11.61 When the spheres are brought into contact they reach a common potential,
say V . If the charges on them are now Q
1
and Q
2
V =
Q
1
4πε
0
R
1
=
Q
2
4πε
0
R
2
(1)
σ
1
=
Q
1
4π R
2
1
=
ε
0
V
R
1
(2)
wherewehaveused(1).
514 11 Electrostatics
Similarly
σ
2
=
ε
0
V
R
2
(3)
Thus σ
α
1
R
.
11.62
(a) V = Er = 5 × 10
6
× 0.1 = 5 × 10
5
V
(b) Pressure p =
1
2
σ
2
ε
0
=
1
2ε
0
(Eε
0
)
2
=
1
2
ε
0
E
2
=
1
2
× 8.85 × 10
−12
× (5 × 10
6
)
2
= 110.6N/m
2
(c) If q is the charge and C the capacitance then the electrostatic energy
U =
1
2
q
2
C
=
1
2
q
2
4πε
0
r
Now q = 4πε
0
r
2
E =
(0.1)
2
× 5 × 10
6
9 × 10
9
= 5.55 ×10
−6
C
∴ U =
1
2
× 9 ×
10
9
0.1
× (5.55 × 10
−6
)
2
= 1.386 J
11.63 Work done in the isobaric expansion (constant pressure) is
W = Pv =
4π
3
R
3
2
− R
3
1
p (1)
where we have written υ for volume.
Increase in electrostatic energy
U =
1
2
q
2
C
1
−
1
2
q
2
C
2
=
q
2
2
1
4πε
0
1
R
1
−
1
R
2
(2)
where C is the capacitance of the spherical bubble. Equating (1) and (2) and
simplifying, we obtain
q =
32
3
π
2
ε
0
pR
1
R
2
R
2
1
+ R
1
R
2
+ R
2
2
1/2
(3)
11.3 Solutions 515
11.64
q =
r
2
r
1
ρ(r) dυ =
A
r
· 4πr
2
dr = 2π A
r
2
2
−r
2
1
(1)
Region (i), 0 ≤ r < r
1
, ρ(r) = 0
As no charge is enclosed, E = 0(2)
Region (ii), r
1
≤ r ≤ r
2
, ρ(r) =
A
r
By Gauss’ law
ε
0
E · dS =
r
r
1
ρ(r)dυ =
r
r
1
A
r
4πr
2
dr = 2π A
r
2
−r
2
1
∴ (ε
0
E)(4πr
2
) = 2π A
r
2
2
−r
2
1
∴ E =
A
2ε
0
r
2
r
2
−r
2
1
(3)
Region (iii), r > r
2
, ρ(r) = 0
ε
0
E · dS = q = 2π A
r
2
2
−r
2
1
by (1)
∴ (ε
0
E)(4πr
2
) = 2π A
r
2
2
−r
2
1
∴ E =
A
2ε
0
r
2
r
2
−r
2
1
(4)
11.65
(a) In order to show that the electric field is conservative, it is sufficient to
establish the existence of a potential. Now, if potential V exists, it must
be such that
F · dr =−dV
where F = f (r)e
r
is the central force and e
r
is the unit vector along the
radius vector r
F · dr = f (r)e
r
· dr = f (r) dr
∴ −dV = f (r) dr
or V =−
f (r)dr
We conclude that the field is conservative and V represents the potential
given by the above relation.
516 11 Electrostatics
(b) Initial electrostatic energy
U
1
=
1
2
qV =
1
2
CV
2
=
1
2
4πε
0
r
1
V
2
=
1
2
×
1
9 × 10
9
× 0.01 × 10
2
= 5.55 ×10
−11
J
Final electrostatic energy
U
2
=
1
2
4πε
0
r
2
V
2
=
1
2
×
1
9 × 10
9
× 0.001 × 10
2
= 5.55 ×10
−12
J
∴ Energy decrease =U
1
−U
2
= (5.55 −0.555)×10
−11
= 5×10
−11
J
11.66 Construct a Gaussian cylindrical surface of radius r and length L coaxial
with the cylinder.
(i) r > a. The charge enclosed is q = πa
2
Lρ. By Gauss’ law
ε
0
E · dA = q
∴ ε
0
E (2πrL) = π a
2
Lρ
∴ E =
a
2
ρ
2ε
0
r
(ii) 0 < r < a. By Gauss’ law
ε
0
E · dA = q
(ε
0
E)(2πrL) = π(a
2
−r
2
)Lρ
∴ E =
(a
2
−r
2
)ρ
2ε
0
r
Centripetal force = Electric force
mv
2
r
= Ee =
a
2
ρe
2ε
0
r
∴ v =
a
2
ρe
2ε
0
m
11.3 Solutions 517
11.67 For an infinite non-conducting charge sheet E = σ/2ε
0
∴ σ = 2Eε
0
= 2 ×200 ×8.85 × 10
−12
= 3.54 ×10
−9
C/m
2
The electric field is independent of the distance.
For an infinite conducting sheet
σ = Eε
0
= 200 ×8.85 × 10
−12
= 1.77 ×10
−9
C/m
2
11.3.3 Capacitors
11.68 The field strength E
0
between the plates of a parallel plate capacitor in vac-
uum is
E
0
=
V
d
=
σ
ε
0
(1)
where V is the applied voltage, σ the charge density (charge per unit area)
and ε
0
the permittivity in vacuum. Now σ = q/A. Therefore the capacitance
in air or vacuum will be
C
0
=
q
V
=
Aε
0
d
(2)
With the introduction of the dielectric slab the electric field in the slab will
be E
0
/K , and the potential across the capacitor becomes
V = E
0
(d − K) +
E
0
t
K
= E
0
(d − t) +
t
K
=
q
Aε
0
(d − t) +
t
K
∴ C =
q
V
=
ε
0
A
d − t
1 −
1
K
(3)
If a metal of thickness t is to be introduced, the effective distance between
the capacitor plates is reduced and the capacitance becomes
C =
ε
0
A
d − t
(4)
a result which is obtained by putting K =∞in (3).
518 11 Electrostatics
11.69
C
1
+C
2
= 9 (parallel) (1)
C
1
C
2
C
1
+C
2
= 2 (series) (2)
(C
1
−C
2
)
2
= (C
1
+C
2
)
2
− 4C
1
C
2
= (C
1
+C
2
)
2
− 8(C
1
+ C
2
) = 9
where we have used (1)
∴ C
1
−C
2
= 3(3)
Solving (1) and (3), C
1
= 6 μF and C
2
= 3 μF.
11.70
(a)
U
1
=
1
2
C
1
V
2
=
1
2
× 2 × 10
−6
× (100)
2
= 0.01 J
U
2
=
1
2
C
2
V
2
=
1
2
× 4 × 10
−6
× (100)
2
= 0.02 J
(b)
C =
C
1
C
2
C
1
+ C
2
=
2 × 10
−6
× 4 × 10
−6
(2 + 4) × 10
−6
=
4
3
× 10
−6
F
U =
1
2
CV
2
=
1
2
×
4
3
× 10
−6
× (100)
2
= 0.0067 F
(c)
C = C
1
+ C
2
= (2 +4) ×10
−6
= 6 ×10
−6
F
U =
1
2
CV
2
=
1
2
× 6 × 10
−6
× (100)
2
= 0.03 J
11.71
C
0
=
ε
0
A
d
=
8.85 × 10
−12
× 1
0.01
= 8.85 ×10
−10
F
U
0
=
1
2
Q
2
C
0
=
1
2
×
(10
−6
)
2
8.85 × 10
−10
= 5.65 ×10
−4
J
U =
1
2
Q
2
C
=
1
2
Q
2
C
0
K
=
U
0
K
=
5.65 × 10
−4
2
= 2.83 ×10
−4
J
The energy is decreased by U = U
0
− U = (5.65 − 2.83) × 10
−4
=
2.82 × 10
−4
J, that is, by a factor of 2.
11.72
C =
ε
0
KA
d
∴ K =
Cd
ε
0
A
=
0.1 × 10
−6
× 0.001
8.85 × 10
−12
× 1.0
= 11.3