Question 394: Electrostats & Capacitance

Question

A capacitor has capacitance 5$$\mu$$F when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $$\frac{d}{2}$$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $$\mu$$F.

Accepted Answer

$C=\frac{\varepsilon_{0} A}{d} K$
When completely air filled
$C=5 \mu \mathrm{F}=\frac{\varepsilon_{0} A}{d} \quad...(1)$
When half filled with $K=1.5$
$$
\begin{gathered}
\frac{1}{C_{\mathrm{eq}}}=\frac{\frac{d}{2}}{\varepsilon_{0} A}+\frac{\frac{d}{2}}{\varepsilon_{0} A K} \\
C_{	ext {eq }}=\left(\frac{2 K}{K+1}ight) \frac{\varepsilon_{0} A}{d} \quad...(2)
\end{gathered}
$$
From (1) & (2)
$C_{\mathrm{eq}}=\left(\frac{2 	imes 1.5}{1.5+1}ight) 5 \mu \mathrm{F}=6 \mu \mathrm{F}$

Electrostats & Capacitance

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