Capacitor

At $t \rightarrow \infty$, capacitor

Works as a open circuit as shown in the Figure.

Therefore, current flow in the circuit,

$\begin{aligned} I_1 & =\frac{9}{(12+15) 10^3}=\frac{1}{3} \times 10^{-3} \mathrm{~A} \\ \therefore \quad V_0 & =I_1 \times 15 \times 10^3 \\ & =\frac{1}{3} \times 10^{-3} \times 15 \times 10^3=5 \mathrm{~V} \end{aligned}$

$\therefore$ Charge on capacitor of $5 \mu \mathrm{F}$,

$\begin{aligned} q_0 & =C V_0=5 \times 10^{-6} \times 5 \\ & =25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C} \end{aligned}$

When switch is open, then capacitor starts discharging across $(5+15=20 \mathrm{k} \Omega)$ and instantaneous charge on capacitor is given by

$\begin{aligned} q & =q_0 e^{-t / R C} \\ & =25 e^{-\frac{1}{20 \times 10^3 \times 5 \times 10^{-6}}} \quad[\because t=1 \mathrm{~s}] \\ & =25 e^{-10} \mu \mathrm{C} \end{aligned}$

Given, capacitance of a capacitor

$\begin{aligned} & C=15 \mathrm{~nF}=15 \times 10^{-9} \mathrm{~F} \\ & \varepsilon_r=2.5 \end{aligned}$

Electric field, $E=30 \times 10^6 \mathrm{~V} / \mathrm{m}$

Potential difference, $V=30 \mathrm{~V}$

Area of plate $=$ ?

If $d$ be the distance between the plates, then

$d=\frac{V}{E}=\frac{30}{30 \times 10^6}=10^{-6} \mathrm{~m}$

Capacitance of capacitor,

$\begin{aligned} C & =\frac{\varepsilon_0 \varepsilon_r A}{d} \\ 15 \times 10^{-9} & =\frac{8.85 \times 10^{-12} \times 2.5 \times A}{10^{-6}} \\ \Rightarrow \quad A & =6.7 \times 10^{-4} \mathrm{~m}^2 \end{aligned}$

This combination forms a GP,

$s=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$

Sum of infinite GP,

$\begin{aligned} & s=\frac{a}{1-r} \\ & s=\frac{1}{1-\frac{1}{2}} \\ & s=\frac{1}{\frac{1}{2}} \\ \therefore \quad & s=2 \end{aligned}$

Hence, capacitance of the combination

$C_{\infty}=2 \times 1 \mu \mathrm{F}=2 \mu \mathrm{F}$

Charge of plates remains same as there is no transfer of charge.

The two capacitors are in parallel.

The equivalent capacitance $C^{\prime}=C+\frac{C}{2}=\frac{3 C}{2}$

Work done in charging the equivalent capacitor is stored in the form of potential energy.

$\begin{aligned} W & =U=\frac{1}{2} C^{\prime} V^2 \\ & =\frac{1}{2} \times \frac{3 C}{2} \times V^2 \\ & =\frac{3}{4} C V^2 \end{aligned}$

Net charge on the parallel plate capacitor is $1 \mu \mathrm{C}$

Electric field near one of the Plates due to the other plate is $E=\frac{10^5}{2} \mathrm{~Vm}^{-1}$

Now, force is given by

$\begin{aligned} \qquad F & =q E \\ \text { Substituting } q & =1 \mu \mathrm{C}=1 \times 10^{-6} \mathrm{C} \\ \text { We get } \quad F & =0 \cdot 05 \mathrm{~N} \end{aligned}$