Physics - Electrostatics Question with Solution | TestHub

64 drops have formed a signal drop of radius$R$. Volume of big drop = Volume of small' drops

$\therefore \frac{4}{3}\pi R^3=64\times \frac{4}{3} \pi r^3$

$\Rightarrow R =4r$

So, the total current is

$Q_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=64q$

As,$C^{\prime }=\frac{Q}{V}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R};C^{\prime }=4\pi {\epsilon }_{0}R$

$C^{\prime }=\left(4\pi {\epsilon }_0\right)4r$

$\Rightarrow C^{\prime }= 4C$