Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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11.3 Solutions 529
The potential drop across C
1
plus that across C
4
must be equal to the
potential drop across C
2
plus that across C
3
.
V
1
+ V
4
= V
2
+ V
3
= V (1)
∴
q
1
C
1
+
q
1
−q
2
C
4
=
q − q
1
C
2
+
q − q
1
+q
2
C
3
(2)
By problem C
1
= C
2
= C
3
= C
4
.
∴ 2q
1
−q
2
= C
1
V (3)
2q − 2q
1
+q
2
= C
1
V (4)
Adding (3) and (4)
2q = 2C
1
V → C =
q
V
= C
1
= 2 μF
11.93 Applying the loop theorem to the circuit, traversing clockwise from the neg-
ative terminal of the battery we have the equation
ξ −iR−
q
C
= 0(1)
where ξ is the emf of the battery and the second and the third terms represent
the potential drop across the resistor and the capacitor.
Now i =
dq
dt
(2)
Using (2) in (1)
R
dq
dt
+
q
C
= ξ (3)
This differential equation describes the time variation of the charge on the
capacitor. Re-arranging (3)
dq
Cξ −q
=
dt
RC
(4)
Integrating (4) − ln (Cξ −q) =
t
RC
+ A (5)
where A is the constant of integration which can be determined from the
initial condition.
At t = 0, q = 0 since the capacitor was uncharged.
∴ A =−ln ξ C (6)
530 11 Electrostatics
Using (6) in (5) and re-arranging
q = Cξ(1 − e
−t/RC
)
(a)
i =
dq
dt
=
ξ
R
e
−t/RC
(b)
P = iξ =
ξ
2
R
e
−t/RC
(c)
H = i
2
R =
ξ
2
R
e
−2t/RC
(d)
dU
dt
=
d
dt
1
2
q
2
e
=
1
2
d
dt
C
2
ξ
2
1 − e
−t/RC
2
=
C
R
ξ
2
e
−t/RC
11.94
(i) By prob. (11.93)
q = Cξ(1 − e
−t/RC
)
∴ i =
dq
dt
=
ξ
R
e
−t/RC
R = (1200 + 600) × 10
3
= 1.8 ×10
6
RC = 1.8 × 10
6
× 2.5 × 10
−6
= 4.5
At t = 0, i =
ξ
R
=
50
1.8 × 10
6
= 27.8 ×10
−6
A = 27.8 μA
(ii) At t =∞, i = 0
(iii) i =−
ξ e
−t/RC
R
=−
50 × e
−t/4.5
1.8 × 10
6
=−27.8e
−0.222t
μA
11.95
(i) Time constant, RC = 200 × 10
3
× 500 × 10
−6
= 100
When the switch is closed there is no emf in the circuit, and (3) in prob.
(11.93) reduces to
(ii)
R
dq
dt
+
q
C
= 0(1)
or
dq
q
=−
dt
RC
(2)
Integrating, ln q =−t/RC + A
where A is the constant of integration. When t = 0, q = q
0
. Therefore
A = ln q
0
.
11.3 Solutions 531
∴ ln
q
q
0
=−
t
RC
∴ q = q
0
e
−t/RC
(3)
q
q
0
=
1
2
= e
−t/RC
∴ t = RC ln 2 = 100 ln 2 = 69.3s.
(iii) In (3) put t = 0. Then
q = q
0
= Cξ = 500 × 10
−6
× 200 × 10
3
= 100 C
(iv) Differentiating ( 3) with respect to time
i =
dq
dt
=−
ξ
R
e
−t/RC
(4)
The negative sign shows that the current in the discharging process
flows opposite to that in the charging process. At t = 0
i =−
ξ
R
=−
900
200 × 10
3
=−4.5 × 10
−3
A
(v) From (4) V = iR =−ξ e
−t/RC
At t = 25 s, V =−900 e
−25/100
=−701 V.
11.96 (i) By (4), prob. (11.50) the electric field inside the sphere is given by
E =
qr
4πε
0
R
3
(1)
∴ div E =
q
4πε
0
R
3
div r (2)
Now div r =
ˆ
i
∂
∂x
+
ˆ
j
∂
∂y
+
ˆ
k
∂
∂z
· (
ˆ
ix +
ˆ
jy+
ˆ
kz)
=
∂x
∂x
+
∂y
∂y
+
∂z
∂z
= 1 +1 +1 = 3
∴ div r =
3q
4πε
0
R
3
=
3 × 10
−9
× 9 × 10
9
1
3
= 27
(ii)
F = QE =
Qqr
4πε
0
R
3
= 1.6 ×10
−19
× 10
−9
× 9 × 10
9
×
0.8
1
3
= 1.152 ×10
−18
N
532 11 Electrostatics
(iii) By prob. (11.52), V (r) =
q
8πε
0
R
3 −
r
2
R
2
. The potential energy of
the proton at r will be U (r) = QV (r) =
Qq
8πε
0
R
3 −
r
2
R
2
.
∴ U(r = 0.8m) =
1.6 × 10
−19
× 10
−9
× 9 × 10
9
2 × 1.0
3 −
0.8
1.0
2
= 1.7 ×10
−19
J
Since U(r =∞) = 0, work done = 1.7 × 10
−19
J.
11.97 (a) If the dielectric is present, Gauss’ law gives
ε
0
E · ds = ε
0
EA= q −q
(Integral form)
or E =
1
ε
0
A
(q −q
)
where q is the free charge and −q
the induced charge.
∇·E = ρ/ε (Differential form)
(b) The displacement vector
D =
q
A
(1)
where A is the area
E =
E
0
K
=
q
K ε
0
A
(2)
Combining (1) and (2)
D = Kε
0
E (3)
As E is uniform in a parallel plate capacitor, D will be also uniform
via (3)
(c) By Gauss law
ε
0
K (x) E(x) ·ds = q = ε
0
E
0
· ds
∴ E(x) =
E
0
K (x)
(4)
E
0
=
q
ε
0
A
=
σ A
ε
0
A
=
σ
ε
0
(
∵ σ = q/ A
)
(5)
11.3 Solutions 533
(d)
V =
E(x)dx = E
0
d
0
dx
ax + b
(6)
Put y = ax + b,dx = dy/a. Then (6) becomes
V =
E
0
a
ad+b
b
dy
y
=
E
0
a
ln y
|
ad+b
b
=
σ
0
ε
0
a
ln
1 +
ad
b
(7)
where we have used (5).
The capacitance C =
q
V
=
Aε
0
a
ln
1 +
ad
b
(8)
where we have used (7) and q = σ A.
(e) Vacuum polarization charge density
P(x) = ε
0
(k(x) −1)E =
ε
0
E
0
k(x)
(k(x) − 1)
= ε
0
E
0
1 −
1
ax + b
(9)
11.98 Newton’s law of gravitation is
F =−
GMm
r
2
ˆe
r
= m g
The Gauss’ law for gravitation may be written as
g · ds =−
GM
r
2
4πr
2
=−4π GM
The divergence theorem gives
∇·g d
3
r =
g · ds
∴ ∇·g
4
3
πr
3
=−4π GM
∴ ∇·g =−4π Gρm
This is analogous to the law for electric field
∇·E =
ρ
q
ε
0
Chapter 12
Electric Circuits
Abstract Chapter 12 is mainly concerned with the analysis of electric network
employing Kirchhoff’s laws. Problems are solved under resistivity, Joule heating,
emf, internal resistance, arrangement of cells, electric instruments such as ammeter,
voltmeter, potentiometer and Wheatstone bridge.
12.1 Basic Concepts and Formulae
Electric current (i) is defined as the rate at which the net charge q passes through a
cross-section of a conductor
i = q/t (12.1)
The instantaneous current is defined by
i = dq/dt
The current density ( j) is given by
j = i/ A (12.2)
where A is the cross-sectional area.
The Drift Velocity
v
d
= i/nAe = j/ne (12.3)
where n is the number of electrons per unit volume
Resistance (R) and Resistivity (ρ)
R = ρ L/A (12.4)
535
536 12 Electric Circuits
where L is the length of the conductor. Unit of resistivity is m.
Electrical Conductance (σ )
σ = 1/ρ (12.5)
The units of conductance are mho/m.
Variation of Resistance and Resistivity with Temperature
R
t
= R
0
(1 + αT ) (12.6)
ρ
t
= ρ
0
(1 + αT ) (12.7)
where α is the temperature coefficient of resistance or resistivity.
Resistors in Series
R =
n
R
n
(12.8)
Resistors in Parallel
1
R
=
n
1
R
n
(12.9)
Joule’s law: The power (P) developed is given by
P = i
2
R = iV = V
2
/R (12.10)
Cells
Cells in series:Ifn cells each of emf ξ and internal resistor r are connected in series
then the current in the circuit is given by
i =
nξ
R + nr
(12.11)
where R is the external resistance.
Cells in parallel: In this case the total emf is that of a single cell ξ. As the internal
resistances of the cells are in parallel, the equivalent internal resistance is r/n, with
the external resistance R in series. The current is
i =
ξ
R +r/n
(12.12)
12.1 Basic Concepts and Formulae 537
Mixed Grouping of Cells
Let there be m rows of cells, with each row containing n cells in series. The emf for
each row of cells would be nξ and the equivalent internal resistance nr. The effective
emf for m rows would again be nξ, but since the rows are in parallel, the effective
internal resistance would be nr/m. The total resistance then becomes R + (nr /m)
(Fig. 12.1):
i =
nξ
R +
nr
m
=
mnξ
mR +nr
=
Nξ
mR +nr
(12.13)
where N = m × n = total number of cells.
Instruments
Potentiometer may be used to measure the internal resistance of a cell: B =
battery, E =cell of internal resistance r,S=resistor, R =resistor, G = galvanome-
ter, AC =potentiometer, AX =l =balancing length.
If l
1
= balancing length with K
1
open and K
2
closed and l
2
with K
1
closed,
r = R
(l
1
−l
2
)
l
2
(12.14)
Fig. 12.1
Wheatstone bridge consists of a network of four resistors P, Q, R and S, battery
E and galvanometer G, Fig. 12.2. Of the four resistors P, R and S are known whose
values are adjustable while Q is unknown. When the bridge is balanced, i.e. the
galvanometer shows null deflection:
P
Q
=
R
S
or Q =
PS
R
(12.15)
538 12 Electric Circuits
Fig. 12.2
Kirchhoff’s Laws
1. Junction theorem: At any junction of an electric network (branched circuit), the
algebraic sum of the currents flowing towards that junction is zero, i.e. the total
current flowing towards the junction is equal to the total current flowing away
from it:
i = 0 (12.16)
2. The loop theorem: The sum of the changes in the potential, encountered in
traversing a loop (closed circuit) in a particular direction (clockwise or coun-
terclockwise), is zero.
(i) If a resistor is traversed in the direction of current, the change in the potential
is −iR, while in the opposite direction it is +iR.
(ii) If a seat of emf is traversed in the direction of the emf, the change in potential
is +ξ, while in the opposite direction it is −ξ.
12.2 Problems
12.2.1 Resistance, EMF, Current, Power
12.1 All resistors in Fig. 12.3 are in ohms. Find the effective resistance between the
points A and B.
[Indian Institute of Technology 1979]
Fig. 12.3
12.2 Problems 539
12.2 If a copper wire is stretched to make it 0.1% longer, what is the percentage
change in its resistance?
[Indian Institute of Technology 1978]
12.3 The equivalent resistance of the series combination of two resistors is p. When
they are joined in parallel, the equivalent resistance is q.Ifp = nq , find the
minimum possible value of n.
12.4 Five resistors are connected as in Fig. 12.4. Find the equivalent resistance
between A and C.
Fig. 12.4
12.5 Five resistors are arranged as in Fig. 12.5. Find the effective resistance
between A and B.
Fig. 12.5
12.6 Each of the resistances in the network, Fig. 12.6, is equal to R. Find the resis-
tance between the terminals A and B.
Fig. 12.6