12. ELECTROSTATICS
A 600 µF capacitor is charged by 250 V
battery.
i)
How much electrostatic energy is
stored in capacitor ?
ii)
The capacitor is disconnected from
battery and connected to another
600 µ F capacitor. What is the ∴
energy of system ?
Given :
2.
C
=
600 µF
=
600 × 10–6 F
=
6 × 10–4 F
V
=
250 V
To Find :
i)
U1
=
?
ii)
U2
=
?
Formula :
1.
U
1
CV2
2
=
2
=
=
U2
=
1
( 0.15 )
×
2
12 × 10 –4
0.0225
× 103
2 × 1.2
9.375 J
A network of four capacitors 5 µF each
are connected to a 240 V supply as shown
in figure
5 µF
5 µF
5 µF
5 µF
240V
Determine
i)
the equivalent capacitance of
network
1
2
ii)
the charge on each capacitor.
U
=
CV
2
Given :
1
C1
=
C2
=
C3
=
C4
U1
=
× 6 × 10–4 × (250)2
–6
=
5
×
10
F
2
V
=
240 V
=
3 × 250 × 250 × 10–4
To Find :
U1
=
18.75 J
i) Ceq = ?
Effective capacity of capacitor is given by
ii) Q1
= ?
iii) Q2
= ?
C
=
C1 + C2
iv) Q3
= ?
–4
–4
=
6 × 10 + 6 × 10
v)
Q
= ?
4
=
12 × 10–4 F
Solution :
The total charge Q will remain constant
The equivalent network for the given
by law of conservation of charge.
circuit is as follows
Q
=
CV =
6 × 10–4 × 250
C 1= 5 µ F C 2= 5 µ F C 3= 5 µ F
=
0.15 V
Energy stored becomes,
Solution :
i)
∴
ii)
∴
∴
C 4= 5 µ F
2
U2
=
1 Q
.
2 C
240 V
Electrostatics
MAHESH TUTORIALS SCIENCE
.. 68
Equivalent capacity of three capacitors is 3.
given by
1
CS
=
=
∴
∴
∴
∴
∴
∴
∴
∴
∴
∴
1
CS
=
1
1
1
+
+
C3
C2
C1
1
5 × 10
–6
3
5 × 10–6
+
1
5 × 10
–6
+
1
5 × 10 –6
A 10 µ F capacitor is connected to a
100 V battery. What is the electrostatic
energy stored ?
Given :
C
=
10 µF
=
10 × 10–6 F = 10–5 F
V
=
100 V
To Find :
U
=
?
Formula :
5
× 10–6 F
3
U
=
C 4 is connected across the parallel
Solution :
combination of C1, C2, C3
Cp
=
C4 + CS
U
=
5
=
5 × 10–6 +
× 10–6
3
CS
=
20
Cp
=
× 10–6
3
Cp
=
6.667 µF
Equivalent capacitance
∴
Ceq =
Cp
=
6.667 µF
Potential acrosss C1, C2, C3 is given by
4.
V1, V2, V3 respectively
Also C1
=
C2
=
C3
then V1
=
V2
=
V3
U
U
1
CV2
2
1
CV2
2
=
1
× 10–5 × (100)2
2
=
1
× 10–5 × 10000
2
=
0.05 J
A metal sphere of radius 1 cm is charged
with 3.14 µ C/m. Find the electric
intensity at a distance 1 m from centre
of metal sphere.
Given
:
V
240
V1
=
=
= 80 V
R
=
1 cm
3
3
=
1 × 10–2 m
Also charges Q 1 , Q 2, Q 3 on C1 , C2 , C 3
respectively are same
q
=
3.14 µC/m
Q1
=
Q2
=
Q3
=
Q
=
3.14 × 10–6 C
Q
=
C1 × V1
r
=
1m
=
5 × 10–6 × 80
To Find :
=
4 × 10–4 C
E
=
?
–4
Q1
=
Q2
= Q3
= 4 × 10 C Formula :
The charge on capacitor C4 is Q4 and is
q
E
=
given by
4πε0 r 2
Q4
=
C4V
Solution :
=
5 × 10–6 × 240
=
12 × 10–4
q
E
=
Q4
=
12 × 10–4 C
4πε0 r 2
Electrostatics
MAHESH TUTORIALS SCIENCE
∴
E
=
E
=
.. 69
3.14 × 10 –6
4 × 3.14 × 8.85 × 10 –12 × ( 1 )
0.02825 × 106 N/C
2
To Find :
K
=
F
=
Formula :
F
5.
=
A long cylinder of radius 2 cm carries a
charge of 5µC/m kept in a medium of Solution :
dielectric constant 10 ? Find the electric
field intensity at a point situated at a
F
=
distance 1 m from the axis of cylinder.
Given :
R
=
2 cm
λ
=
=
=
=
k
r
To Find :
E
=
Formula :
E
=
Q
=
l
5 × 10–6 C/m
10
1m
5 µC/m
F
σ2
×A
2kε0
2
Q
 
A
×A=
2kε 0
?
=
2λ
2 πkε 0 r
=
=
σ2
×A
2kε0
Q2
2kε 0 A
Q
... ∵ σ = 
A


Solution :
E
=
?
?
2λ
4 πε 0 kr
∴
1 2λ
.
4 πε 0 kr
F
=
=
( 10–4 )
2
2 × 1 × 8.85 × 10–12 × ( 0.01)
10 –8
2 × 8.85 × 10–14
5.6497 × 10–4 N
56500 N
The energy density at a point in a
medium of dielectric constant 8 is
26.55 × 106 J/m2. Calculate electric field
9
–6
9 × 10 × 2 5 × 10
intensity at that point.
=
Given :
10 × 1
k
=
8
∴
E
=
9 × 103 V/m
U
=
26.55 × 106 J/m2
2
6.
A metal plate of area 0.01 m carries a To Find :
charge of 100 µC. Calculate the outward
E
=
?
pull on plate.
Formula :
Given :
=
(
Q
A
=
=
=
=
) (
100 µC
100 × 10–6 C
10–4C
0.01 m2
7.
)
U
=
1
ε kE2
2 0
=
1
ε kE2
2 0
Solution :
U
Electrostatics
MAHESH TUTORIALS SCIENCE
.. 70
E
=
∴
E
ii)
2 × 26.55 × 106
=
=
=
∴
2u
ε0 k
8.85 × 10
–12
∴
iii)
8.66 × 10 N/C
8.66 × 102 µN/C
8.
A parallel plate air capacitor has
rectangular plates each of length 20 cm
and breadth 10 cm. They are separated
by a distance of 2 mm. If the potential
difference between the plates is 500 volt.
Calculate
i)
capacitance
ii)
charge on each plate
iii) electric field intensity between
the two plates.
Given :
=
20 cm =
0.2 m
l
b
=
10 cm =
0.1 m
d
=
2 mm
=
2 × 10–3 m
k
=
1
A
=
l×b
=
0.2 × 0.1 = 0.02 m2
V
=
500 V
To Find :
i)
C
=
?
ii)
Q
=
?
iii) E
=
?
Formula :
C
=
ii)
Q
=
iii)
E
=
Aε 0 k
d
CV
C
Electrostatics
=
=
8.85 × 10–11 F
CV
8.85 × 10–12 × 500
4.425 × 10–9 C
E
=
V
d
E
=
2 × 10–3
∴
E
=
2.5 × 105 V/m
500
9.
A parallel plate air capacitor with air as
dielectric has a capacity of 20 µF. What
will be the new capacity if
i)
the distance between two plates
is doubled ?
ii)
a marble slab of dielectric constant
8 is introduced between the two
plate such that entire space
between plates is filled by marble
slab ?
Given :
C1
=
20 µF
=
2d
d2
k1
=
k2
= k(air)
=
8
k3
To Find :
i)
C2
=
?
ii)
C3
=
?
Formula :
C
=
Aε 0 k
d
C
=
Aε 0 k
d
C1
=
Aε 0 k
d1
and
C2
=
Aε 0 k
d2
=
Solution :
V
d
Solution :
C
=
=
=
=
×8
8
i)
C
Q
Q
Q
Aε 0 k
d
0.02 × 8.85 × 10 –12 × 1
2 × 10–3
Aε 0 k
2d 1
MAHESH TUTORIALS SCIENCE
∴
∴
∴
C2
C1
=
C2
C1
=
C2
=
=
∴
C2
Aε0 k
2d1
Aε0 k
d1
iii) E
Formula :
1
2
C1 ×
1
2
C3
d2 = 2d1
=
?
Aε 0 k
d
CV
i)
C
=
ii)
Q
=
iii)
E
=
C
=
Aε 0 k
d
C1
=
Aε 0 k 1
d1
V
d
Solution :
1
20 ×
2
10 µF
Aε 0 k 3
Aε 0 × 8
=
d
d
[ ∵ Distance between plates remains
same]
=
8 C1
=
8 × 20 = 160 µF
C3
=
160 µF
ii)
∴
=
.. 71
=
A parallel plate capacitor has rectangular
plates each of length 20 cm and breadth
10 cm. The separation between the plates
is 1 mm.
i)
Calculate the potential difference
between the plates if 1 nC is given
to capacitor.
ii)
With the same charge of 1 nC, if
the separation between plates is
doubled what is the new
potential difference ?
iii) Electric field between the plates.
Given :
l
=
20 cm
= 0.2 m
b
=
10 cm
= 0.1 m
d
=
1 mm
=
× 10–3 m
A
=
l×b
=
0.2 × 0.1
= 0.02 m2
To Find :
i)
V1
=
?
when Q = 1 nC = 1 × 10–9 C
ii)
V2
=
?
when Q = 1 nC = 1 × 10–9 C
=
( 0.2 × 0.1) × 8.85 × 10–12 × 1
1 × 10–3
17.70 × 10–11 F
CV
C1
Q
=
=
V1
=
V1
=
V1
=
E
=
V
d
E
=
V1
d1
V1
d1
=
V2
2d 1
Q
C1
10.
∴
1 × 10–9
17.7 × 10 –11
5.65 V
=
V2
d2
∴
∴
V2
V2
=
=
2V1 = 2 × 5.65
11.30 V
∴
E
=
V
, gives,
d
E
=
E
=
=
1
∴
5.65
V2 – V1
=
d
1 × 10–3
5.65 × 103
5650 N/C
Electrostatics
MAHESH TUTORIALS SCIENCE
.. 72
11.
A parallel plate capacitor consists of two
identical metal plates. Two dielectric
slabs having dielectric constants K1 and
K2 are introduced in the space between
two plates as shown below.
P
C2
C1
k1
d
k2
A
2
A
2
K1
K2
d
Q
d
Ceffective = C1 + C2
A
2
A
A
k 1ε0   k 2 ε0  
2
 +
2
=
d
d
A
2
(i)
=
A
d
2
d
2
K1
K2
(ii)
Show that capacity in arrangement (i)
is given by
=
P
2ε0 K1K 2 A
(K1 + K2 ) d
=
Solution :
The arrangement (i) can be regarded as
two capacitors connected in parallel as
shown below
( K1 + K 2 ) ε0A
(proved)
2d
The arrangement (i) can be regarded as
two capacitors connected in series as
shown below,
ε0 ( K 1 + K 2 ) A
C
=
2d
Also show that the capacity of capacitor
in arrangement (ii) is given by
C
ε0A ( K 1 + K 2 )
2d
d
2
k1
d
2
k2
A
Q
1
C effective
P
1
1
+
C2
C1
=
P
A
k1
k2
d
C1
k1
d
2
k2
d
2
A
A
2
Electrostatics
Q
A
2
C2
Q
MAHESH TUTORIALS SCIENCE
=
.. 73
1
1
+
k 1 ε0 A
k 2 ε0A
d
d
2
2
=
d
d
+
2k 1 ε 0 A
2k 2 ε 0 A
=
d  1
1 
 + 
2ε0 A  k 1 k 2 
=
d  k1 + k2 


2ε0 A  k 1 k 2 
∴
Ceffective
=
∴
Ceffective
=
2ε 0 A ( k 1 k 2 )
d (k1 + k2 )
2 ( k 1k 2 ) ε0A
(k1 + k2 ) d
(proved)
Electrostatics