Question

Eight Mercury drops of equal radius and equal charge combine to form a big drop. The capacitance of big drop in comparison the each small drop will be:
(a) 2 times
(b) 8 times
(c) 4 times
(d) 16 times

Answer 1

(a) 2 times
Volume of big drop = 8 × Volume of small drop
\(\frac{4}{3}\)πR³ = 8 × \(\frac{4}{3}\)πr³
R = 2r
Capacitance of small drop C₁ = 4πε₀r
Capacitance of bigger drop C₂ = 4πε₀R
\(\frac{C_{1}}{C_{2}}=\frac{r}{R}=\frac{r}{2 r}\)
C₂ = 2C₂