n small drops of same size are charged to V volt each. If they coalesce to form a single large drop, then its potential will be

Question

n small drops of same size are charged to V volt each. If they coalesce to form a single large drop, then its potential will be (A) V n (B) Vn -1 (C) V n1 / 3 (D) V n2 / 3

Tardigrade Admin · Accepted Answer

As (4/3) π R3=n × (4/3) π r3 ∴ R=n1 / 3 r. New potential V'=(n q/4 π ε0 R) =(n q/4 π ε0(n1 / 3 r)) =n2 / 3 (q/4 π ε0 r)=n2 / 3 V

Q. $n$ small drops of same size are charged to $V$ volt each. If they coalesce to form a single large drop, then its potential will be

$Vn$

$V n^{- 1}$

$V n^{1/3}$

$V n^{2/3}$

As $\frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$
$∴ R = n^{1/3} r$ .
New potential $V^{'} = \frac{n q}{4 π ε _{0} R}$
$= \frac{n q}{4 π ε _{0} ( n ^{1/3} r )}$
$= n^{2/3} \frac{q}{4 π ε _{0} r} = n^{2/3} V$

Q. n small drops of same size are charged to V volt each. If they coalesce to form a single large drop, then its potential will be

Vn

Vn−1

Vn1/3

Vn2/3