Answer to Question #193092 in Physics for Abhi

Question

64 drops of radius 0.02m and each carrying a charge 5 microcoloumb are combined to form a bigger drop. Find how surface density electrification will change if no charge lost

Answer 1

Find the total volume of N individual drops:

$$V=N·\frac43\pi r^3.$$

And the bigger drop will have the same volume but different radius:

$$V=\frac43 \pi R^3,\\\space\\ Nr^3=R^3,\\ R=r\sqrt[3]{N}.$$

The initial surface density of charge was

$$\sigma_i=\frac{Nq}{NA_i}=\frac{q}{4\pi r^2.}$$

The final:

$$\sigma_f=\frac{Nq}{A_f}=\frac{Nq}{4\pi R^2}.$$

The change:

$$\Delta \sigma=\sigma_f-\sigma_i=\frac{q}{4\pi}\bigg(\frac{1}{r^2}-\frac{1}{R^2}\bigg),\\\space\\ \Delta\sigma=\frac{q}{4\pi r^2}\bigg(1-\frac{1}{\sqrt[3]{N^2}}\bigg)=21.2\space\mu C.$$