The circular plates $A$ and $B$ of a parallel plate air capacitor have a diameter of $0.1 m$ and are $2 m m$ apart. the plates $C$ and $D$ , of a similar capacitor have a diameter $0.12 m$ and are $3 m m$ apart. Plate $A$ is earthed and the plates $B$ and $D$ are connected together. Plate $C$ is connected to the positive pole of a $120 V$ battery whose negative terminal is earthed. Calculate
The combined capacitance of the arrangement.

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$14 p F$

$15 p F$

$16 p F$

$17 p F$

answer is D.

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Detailed Solution

The capacitance of capacitor $(A, B)$ is

$C_{1} = \frac{ε_{0} A_{1}}{d_{1}} = \frac{ε_{0} {πR}_{1}^{2}}{d_{1}} \Rightarrow C_{1} = \frac{(\frac{1}{36 π \times 10^{9}}) π {(\frac{0.1}{2})}^{2}}{2 \times 10^{- 3}} = \frac{1}{300 \times 10^{8}} F$

The capacitors $C_{1}$ and $C_{2}$ are in series. Therefore, equivalent capacitance $C$ is given by

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = 288 \times 10^{8} + 300 \times 10^{8} \Rightarrow \frac{1}{C} = 588 \times 10^{8} \Rightarrow \frac{1}{C} = \frac{1}{588 \times 10^{8}} = 17 \times 10^{- 12} F = 17 p F$