The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same , the distance between the plates is

Question

Dr. Abubacker · Accepted Answer

(c) Original capacity, with air
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ C = \frac{\varepsilon_{0}A}{d}$
When dielectric plate (medium) of thickness $t$is introduced between the plates, then capacity becomes
$\ \ \ \ \ \ \ \ \ \ \ \ \ C^{'} = \frac{\varepsilon_{0}A}{d^{'} - t\left( 1 - \frac{1}{K} ight)}$
but as given, $C^{'} = \ C$
$	herefore\ \ \ \ \ \ \ \frac{\varepsilon_{0}A}{d} = \ \frac{\varepsilon_{0}A}{d^{'} - t\left( 1 - \frac{1}{K} ight)}$
or $\ d = d^{'} - t + \frac{t}{K}$
or $\ \ \ \ \ \ \ \ 8 = d^{'} - 4 + \frac{4}{2}$
or $\ \ \ \ \ \ 8 = d^{'} - 2$
or $d^{'} = 10\ mm$

Q. The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same , the distance between the plates is

Explanation

Reviewed by Dr. Abubacker

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