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2. Electric Potential and Capacitance
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$N$ identical spherical drops charged to the same potential $V$ are combined to form a big drop. The potential of the new drop will be
A
$V$
B
$V/N$
C
$V \times N$
D
$V \times {N^{2/3}}$
Solution
(d) If the drops are conducting, then
$\frac{4}{3}\pi {R^3} = N\,\left( {\frac{4}{3}\pi {r^3}} \right)$ $==>$ $R = {N^{1/3}}r$. Final charge $Q = Nq$
So final potential $V = \frac{Q}{R}$ $ = \frac{{Nq}}{{{N^{1/3}}r}} = V \times {N^{2/3}}$
Standard 12
Physics
