Question 392: Electrostatics & Capacitance

Question

A parallel plate capacitor with air between the plates has capacitance of $9$ $pF.$ The separation between its plates is $'d'.$ The space between the plates has dielectric constant ${k_1}$ $=3$ and thickness ${d \over 3}$ while the other one has dielectric constant ${k_2} = 6$ and thickness ${{2d} \over 3}$. Capacitance of the capacitor is now

Accepted Answer

The given capacitance is equal to two capacitances connected in series where
${C_1} = {{{k_1}{ \in _0}A} \over {d/3}} = {{3{k_1}{ \in _0}A} \over d}$
$ = {{3 	imes 3{ \in _0}A} \over d} = {{9{ \in _0}A} \over d}$
and
${C_2} = {{{k_2}{ \in _0}A} \over {2d/3}} = {{3{k_2}{ \in _0}A} \over {2d}}$
$ = {{3 	imes 6{ \in _0}A} \over {2d}} = {{9{ \in _0}A} \over d}$
The equivalent capacitance ${C_{eq}}$ is
${1 \over {C{}_{eq}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$
$ = {d \over {9{ \in _0}A}} + {d \over {9{ \in _0}A}}$
$ = {{2d} \over {9{ \in _0}A}}$
$	herefore$ ${C_{eq}} = {9 \over 2}{{{\varepsilon _0}A} \over d} = {9 \over 2} 	imes 9pF = 40.5pF$

Answer

$1.8$ $pF$

Answer

$45$ $pF$

Answer

$40.5$ $pF$

Answer

$20.25$ $pF$

Electrostatics & Capacitance

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