Question

$n$ number of identical spherical droplets, each having surface charge density $\sigma$ coalesce to form a bigger drop of surface charge density

• A. ${\left(n\right)}^{1/3}\sigma$
• B. $n\sigma$
• C. ${\left(n\right)}^{2/3}\sigma$
• D. ${\left(n\right)}^{5/3}\sigma$

Answer 1

The correct option is C ${\left(n\right)}^{2/3}\sigma$

Let the potential of $1$ drop $=\frac{ha}{r}=V$

Volume of $n$ drops $=$ Volume of bigger drop

$n\times \frac{4}{3}\frac{\pi r}{3}=\frac{4}{3}\frac{\pi R}{3}R=\left(\right(n{)}^{1/3}{)}^r$

Charges remained conserved

Charge on $1$ drop$=q$

Charge on big drop containing $n$ drops$=nq$

Therefore Potential on big drop$=\frac{4nq}{(n^{1/3}{)}^r}=\left(n^{2/3}\right)\frac{hq}{r}=\left(n^{2/3}\right)\sigma$