Question 224: Electrostats & Capacitance

Question

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 :

Accepted Answer

$${C_0} = {{{ \in _0}A} \over d}$$
$$ 	herefore $$ $${1 \over {C'}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$$
$${1 \over {C'}} = {{(3d/4)} \over {{ \in _0}KA}} + {{(d/4)} \over {{ \in _0}A}}$$
$${1 \over {C'}} = {d \over {4{ \in _0}A}}\left( {{{3 + K} \over K}} ight)$$
$$ 	herefore $$ $$C' = {{4K} \over {(K + 3)}}{C_0}$$

Answer

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

Answer

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

Answer

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

Answer

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

Electrostats & Capacitance

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