Vidyamandir Classes
Capacitors
(b)
Parallel Combinations :
When capacitors are connected in parallel, the potential difference
V
across each is same and the charge on
C
1
,
C
2
is different i.e.,
Q
1
and
Q
2
.
E
quivalent capacitance between
a
and
b
is :
C
=
C
1
+
C
2
+
C
4
+ . . . . .
The charges on capacitors is given as :
1
1
1
2
C
Q
Q
C
C
and
2
2
1
2
C
Q
Q
C
C
4.
A parallel plate capacitor contains two dielectric slabs of thickness
d
1
,
d
2
and dielectric constants
k
1
and
k
2
respectively. The area of the capacitor
plates and slabs is equal to
A
.
Considering the capacitor as a combination of two capacitors in series, the
equivalent capacitance
C
is given by :
0
1
2
1
2
A
C
d
d
k
k
In general for more than one dielectric slab :
C
=
0
i
i
A
d
k
If
V
is the potential difference across the plates, the electric fields in the dielectrics are given as :
1
1
2
1
1
2
1
V
E
d
d
k
k
k
2
1
2
2
1
2
1
V
E
d
d
k
k
k
5.(a)
If there exits a dielectric slab of thickness
t
inside a capacitor whose plates
are separated by distance
d
, the equivalent capacitance is given as :
0
A
C
t
d
t
k
The equivalent capacitance is not affected by changing the distance of slab from the parallel plates.
If the slab is of metal, the equivalent capacitance is :
0
A
C
d
t
(for a metal,
k
=
)
+
d
1
d
2
K
1
K
2
+
+
+
+
+
+
–
–
–
–
–
–
–
Things to Remember
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25

Vidyamandir Classes
Capacitors
(b)
Consider a capacitor with two dielectric slabs of same thickness
d
placed
inside it as shown. The slabs have dielectric constants
k
1
and
k
2
and
areas
A
1
and
A
2
respectively. Treating the combination as two capacitors
in parallel,
0
1 1
2
2
C
k A
k A
d
6.
Consider a circuit where an uncharged capacitor
C
is connected to
a cell of emf
E
through a resistance
R
and
a
switch
S
as shown. At
t
= 0, the switch
S
is closed.
Let
q
,
V
c
be the charge and voltage on the capacitor at time
t
and
i
be
the current. This equation gives the expression for charge on capacitor
as a function of time. The charge grows on the plate exponentially as
shown on the graph. Note the following points.
1.
In steady state :
t
and
q
C E
2.
The voltage across capacitor also grows exponentially
towards
E
.
(1
)
t / RC
q
V
E
e
C
3.
The time constant (
) of the circuit is defined as the time after
which the charge has grown upto (1 – 1/
e
) = 0.63
63 % of
its steady-state value.
=
R C
7.
If we connect a charged capacitor
C
across a resistance
R
, the capacitor
begins to discharge through
R
. The excess positive charge on high potential
plate flows through
R
to the negative plate and in steady state, the capacitor
plates become uncharged.

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