Derive the expression for the capacitance of a parallel plate capacitore having plate area A and plate separation d.

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Answer 1

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 $Q$ ,
$E = \frac{σ}{ϵ _{o}}$ , where $σ = \frac{Q}{A}$ (charge / Area)
$\Rightarrow E = \frac{Q}{A ϵ _{o}}$
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]
$\Rightarrow V = \frac{Q d}{A ϵ _{o}} \Rightarrow \frac{Q}{V} = \frac{A ϵ _{o}}{d}$
Now, capacitance is defined by $C = \frac{Q}{V}$ , thus we get :-
$C = \frac{A ϵ _{o}}{d}$

Answer 2

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 = \frac{σ}{ϵ _{o}}

, where

σ = \frac{Q}{A}

(charge / Area)

\Rightarrow E = \frac{Q}{A ϵ _{o}}

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]

\Rightarrow V = \frac{Q d}{A ϵ _{o}} \Rightarrow \frac{Q}{V} = \frac{A ϵ _{o}}{d}

Now, capacitance is defined by

C = \frac{Q}{V}

, thus we get :-

C = \frac{A ϵ _{o}}{d}