Question

Eight drops of mercury of equal radii and possessing equal charges combine to form a big drop. The capacitance of bigger drop as compared to capacitance of each individual

• A. 8 times
• B. 32 times
• C. 2 times
• D. 16 times

Answer 1

Answer: (C)

2 times

Answer 2

The problem involves the combination of multiple small mercury drops to form a larger drop. We need to determine how the capacitance changes.

1. Relation between radii:

When $N$ small drops of radius $r$ combine to form one big drop of radius $R$, the total volume remains conserved. Given $N = 8$ small drops.

Volume of $N$ small drops = Volume of 1 big drop $N \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ $8 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ $8r^3 = R^3$

Taking the cube root of both sides: $R = (8)^{\frac{1}{3}} \cdot r$ $R = 2r$

So, the radius of the big drop is twice the radius of a small drop.

2. Capacitance of a spherical conductor:

The capacitance $C$ of an isolated spherical conductor of radius $x$ in a medium of permittivity $\varepsilon$ (or $\varepsilon_0$ for vacuum/air) is given by: $C = 4\pi\varepsilon_0 x$

For a small drop: $C_{\text{small}} = 4\pi\varepsilon_0 r$

For the big drop: $C_{\text{big}} = 4\pi\varepsilon_0 R$

3. Comparison of capacitances:

Substitute $R = 2r$ into the equation for $C_{\text{big}}$: $C_{\text{big}} = 4\pi\varepsilon_0 (2r)$ $C_{\text{big}} = 2 \cdot (4\pi\varepsilon_0 r)$

Since $C_{\text{small}} = 4\pi\varepsilon_0 r$, we can write: $C_{\text{big}} = 2 \cdot C_{\text{small}}$

Therefore, the capacitance of the bigger drop is 2 times the capacitance of each individual small drop.