A parallel plate capacitor with air between the plates has capaci… - Vidyadip

MCQ

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 is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1 = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now.......$pF$

A
$20.25$
B
$1.8 $
C
$45 $
✓
$40.5$

✓

Answer

Correct option: D.

$40.5$

d
The given capacitance is equal to two capacitances connected in series where

$C_{1}=\frac{k_{1} \in_{0} A}{d / 3}=\frac{3 k_{1} \in_{0} A}{d}=\frac{3 \times 3 \in_{0} A}{d}=\frac{9 \in_{0} A}{d}$

and

$C_{2}=\frac{k_{2} \in_{0} A}{2 d / 3}=\frac{3 k_{2} \in_{0} A}{2 d}=\frac{3 \times 6 \in_{0} A}{2 d}=\frac{9 \in_{0} A}{d}$

The equivalent capacitance $C_{\mathrm{eq}}$ is

$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{9 \in_{0} A}+\frac{d}{9 \in_{0} A}=\frac{2 d}{9 \in_{0} A}$

$\therefore C_{e q}=\frac{9}{2} \frac{\in_{0} A}{d}=\frac{9}{2} \times 9 \,p F=40.5 \,p F$

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