JEE Main & Advanced Physics Electrostatics & Capacitance Combination of Charged Drops

Suppose we have \[n\] identical drops each having Radius \[-r,\] Capacitance \[-c,\] Charge \[-\,q,\] Potential \[-\,v\] and Energy \[-\,u\].

If these drops are combined to form a big drop of Radius \[-\,R,\] Capacitance \[-\,C,\] Charge \[-\,Q,\] Potential \[-\,V\] and Energy \[-\,U\] then

(1) Charge on big drop :  \[Q=nq\]

(2) Radius of big drop : Volume of big drop \[=n\times \] volume of a single drop i.e., \[\frac{4}{3}\pi {{R}^{3}}=n\times \frac{4}{3}\pi {{r}^{3}}\], \[R={{n}^{1/3}}r\]

(3) Capacitance of big drop : \[C={{n}^{1/3}}c\]

(4) Potential of big drop : \[V=\frac{Q}{C}=\frac{nq}{{{n}^{1/3}}c}\]  \[V={{n}^{2/3}}v\]

(5) Energy of big drop :  \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}({{n}^{1/3}}c)\,{{({{n}^{2/3}}v)}^{2}}\]        \[U={{n}^{5/3}}u\]

(6) Energy difference : Total energy of big drop is greater than the total energy all smaller drop. Hence energy difference

\[\Delta U=U-nu==U-n\times \frac{U}{{{n}^{5/3}}}=U\,\left( 1-\frac{1}{{{n}^{2/3}}} \right)\]