Question

A spherical drop of capacitance $1\mu F$ is broken into 8 drops of equal radius. What is the capacitance of each small drop?

Answer 1

Step-1: Given data

The capacitance of the spherical drop is $C=1\mu F.$

The drop is broken into 8 drops of equal radius.

Step-2: Capacitance and potential

Capacitance is the amount of charge in a conductor of unit potential.

Capacitance is defined by the form, $C=\frac{Q}{V}$, where, Q is the total amount of charge in a conductor and V is the potential of the conductor.

Potential V is defined by the form, $V=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{Q}{R}$, where, Q is the charge in the conductor and R is the radius of the conductor.

Step-3: Finding the capacitance of the sphere

As we know, capacitance, $C=\frac{Q}{V}$.

So,

$$\begin{gathered} C=\frac{Q}{V}=\frac{Q}{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{Q}{R}}}=\frac{R}{K}\left(\mathrm{Since},\mathrm{K}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\right) \\ orC=\frac{R}{K}...........\left(1\right) \end{gathered}$$

Step-4: Capacitance of broken parts

We know that the drop is broken into 8 drops of equal radius. Let r be the radius of the small sphere

So, according to the question;

$$\begin{gathered} \frac{4}{3}{\mathrm{\pi R}}^3=8\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)\left(Since,\frac{4}{3}{\mathrm{\pi R}}^3=volumeofsphere\right) \\ \mathrm{or}{\mathrm{R}}^3=8{\mathrm{r}}^3 \\ \mathrm{or}\mathrm{R}=2\mathrm{r} \\ orr=\frac{R}{2}.....................\left(2\right) \end{gathered}$$

Now, the capacitance of the small drop is

$$\begin{gathered} C=\frac{r}{K}=\frac{R}{{\displaystyle \frac{2}{K}}}=\frac{1}{2}\frac{K}{R} \\ orC=\frac{1}{2}\frac{K}{R}=\frac{1}{2}\mu F\left(Since,\frac{K}{R}=1\right) \\ orC=\frac{1}{2}\mu F. \end{gathered}$$

Therefore, the capacitance of each small drop is $\frac{1}{2}\mu F$.