Sixty four conducting drops each of radius \(0.02 m\) and each carrying a charge of \(5 \,\mu C\) are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be ............

\(r=\) radius of smaller drop

Volume will remain same

\(\frac{4}{3} \pi R ^{3}=64 \times \frac{4}{3} \pi r ^{3}\)

\(R =4 r\)

\(Q =64 q ;\)

\(q\) : charge of smaller drop

\(Q\) : Charge of combined drop

\(\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{\frac{ Q }{4 \pi R ^{2}}}{\frac{ q }{4 \pi r ^{2}}}=\frac{ Q }{ q } \cdot \frac{ r ^{2}}{ R ^{2}}\)

\(=64 \frac{ r ^{2}}{16 r ^{2}}=4\)

\(\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{4}{1}\)