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JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 9)

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :
$$C' = {{3 + K} \over {4K}}{C_0}$$
$$C' = {{4 + K} \over {3}}{C_0}$$
$$C' = {{4K} \over {K + 3}}{C_0}$$
$$C' = {{4} \over {3 + K}}{C_0}$$

Explanation

JEE Main 2021 (Online) 16th March Morning Shift Physics - Capacitor Question 89 English Explanation
$${C_0} = {{{ \in _0}A} \over d}$$

$$ \therefore $$ $${1 \over {C'}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$$

$${1 \over {C'}} = {{(3d/4)} \over {{ \in _0}KA}} + {{(d/4)} \over {{ \in _0}A}}$$

$${1 \over {C'}} = {d \over {4{ \in _0}A}}\left( {{{3 + K} \over K}} \right)$$

$$ \therefore $$ $$C' = {{4K} \over {(K + 3)}}{C_0}$$

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