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Combination of Capacitors
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Derive the expression for the capacitanc
Question
Derive the expression for the capacitance of a parallel plate capacitore having plate area A and plate separation d.
Medium
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Solution
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Let a parallel plate capacitor be applied a voltage
V
across its terminals due to which a charge
Q
develops across its ends.
Separation between plates is
d
and area of plates is
A
.
From gauss's theorem one can show electric field inside the capaciton plates
Q
,
E
=
ϵ
o
σ
, where
σ
=
A
Q
(charge / Area)
⇒
E
=
A
ϵ
o
Q
Now, voltage across the plates is related by :
V
=
E
d
[In general ,
d
V
=
−
E
.
d
r
but we take as
E
is uniform]
⇒
V
=
A
ϵ
o
Q
d
⇒
V
Q
=
d
A
ϵ
o
Now, capacitance is defined by
C
=
V
Q
, thus we get :-
C
=
d
A
ϵ
o
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