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  4. Between the plates of parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to 1/5th of the distance between the plates of the origin

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Q. Between the plates of parallel plate capacitor of capacity $C$, two parallel plates of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to $1/5th$ of the distance between the plates of the origin

EAMCETEAMCET 2003Electrostatic Potential and Capacitance

Solution:

Thickness of each plates $= \frac{1}{5}$ distance between plates
$= \frac{1}{5} \times d$
The capacitance of capacitor is given by
$C_1 = \frac{\varepsilon_0 A}{d-t} ... (i)$
Thickness of two conductor is given by
$t= 2 (\frac{d}{5}) = \frac{2d}{5}$
Putting the values oft in Eq. (i)
$C_1= \frac{\varepsilon_0 A}{d - \frac{2d}{5}} = \frac{5}{3} \frac{\varepsilon_0 A}{d}$
$= \frac{5}{3}C$ (where $C = \frac{\varepsilon_0 A}{d}$