Question 86 of 180

A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\dfrac{3}{4}d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given ${C}_{0}=$ capacitance of capacitor with air as medium between plates.)

A$\dfrac{4K{C}_{0}}{3+K}$B$\dfrac{3K{C}_{0}}{3+K}$C$\dfrac{3+K}{4K{C}_{0}}$D$\dfrac{K}{4+K}$

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Answer: (A) $\dfrac{4K{C}_{0}}{3+K}$

From the above diagram, we can write,
$x+y+\dfrac{3d}{4}=d$
$x+y=\dfrac{d}{4}$
Initially, when there is no dielectric inserted between the plates the capacitance will be, ${C}_{0}=\dfrac{{ϵ}_{0}A}{d}$.
When the dielectric slab is inserted, we can assume there are $3$ capacitors that are connected in series.
${C}_{1}=\dfrac{{ϵ}_{0}A}{x}$, ${C}_{2}=\dfrac{4K{ϵ}_{0}A}{3d}$ and ${C}_{3}=\dfrac{{ϵ}_{0}A}{y}$
The equivalent capacitance,
$\dfrac{1}{C}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}+\dfrac{1}{{C}_{3}} \Rightarrow \dfrac{1}{C}=\dfrac{1}{{ϵ}_{0}A}\left(x+\dfrac{3d}{4K}+y\right) \Rightarrow \dfrac{1}{C}=\dfrac{1}{{ϵ}_{0}A}\left(\dfrac{d}{4}+\dfrac{3d}{4K}\right) \Rightarrow C=\dfrac{4{ϵ}_{0}AK}{d\left(3+K\right)} \Rightarrow C=\dfrac{4K{C}_{0}}{3+K}$