Question

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is?

• A. $8$ times
• B. $4$ times
• C. $2$ times
• D. $32$ times

Answer 1

The correct option is C $2$ times

Initial capacitance $C_o=4\pi {ϵ}_o{R}_o$

Now, when they are combined to form a bigger drop of radius $R$.

Total volume of mercury is same.

$⟹$ final volume$=8\times$ initial volume of each drop

$⟹\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi R_o^3$

$⟹R^3=8R_o^3$

$⟹R=2R_o$

Final capacitance $C=4\pi {ϵ}_{o}R=2\times 4\pi {ϵ}_o{R}_o$

$⟹C=2C_o$

Answer-(C)