Question 664: Electrostatics & Capacitance

Question

Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.

Accepted Answer

From law of conservation of charge
$$
\begin{aligned}
&q_{i}=\mathrm{q}_{\mathrm{f}} \Rightarrow 8 	imes\left(4 \pi \mathrm{E}_{0} \mathrm{R}ight) 	imes 12=\left(4 \pi E R^{1}ight) 	imes \mathrm{V}_{f} \\
&\Rightarrow 96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \mathrm{R}^{1} \\
&	ext { And, } 8 	imes \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{1} 3 \\
&8=\left(\frac{R^{1}}{R}ight)^{3}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{R}^{1}=2 \mathrm{R}
\end{aligned}
$$
From (i) & (ii), we get
So, $96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} 	imes 2 \mathrm{R} \Rightarrow \mathrm{V}_{\mathrm{f}}=48$ Volt
$$
\begin{aligned}
&V_{f}=\frac{1}{2} C_{f} V_{f}^{2}=\frac{1}{2} 	imes\left(4 \pi \varepsilon_{0} \mathrm{R}^{1}ight) \mathrm{V}_{f}^{2} \\
&=\frac{1}{2} 	imes\left(4 \pi \varepsilon_{0} 	imes 2 \mathrm{R}ight) 	imes 48^{2} \\
&=\left(\frac{1}{2} 	imes 4 \pi \varepsilon_{0} R 	imes 12^{2}ight) 	imes \frac{48^{2} 	imes 2}{12^{2}}=32 \,E
\end{aligned}
$$

Electrostatics & Capacitance

Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.

Integer / Numeric Answer Type