Question

512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ______ V.

Hint:

For $n$ drops coalescing: - Radius $R = n^{1/3}r$ - Capacitance $C = n^{1/3}c$ - Potential $V = n^{2/3}v$ - Energy $E = n^{5/3}e$

Answer 1

Correct Answer: 128

Step 1: Volume conservation: $\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \Rightarrow R = n^{1/3}r$.
Step 2: Charge conservation: $Q = nq$.
Step 3: Potential of small drop $v = \frac{kq}{r} = 2$ V.
Step 4: Potential of big drop $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \frac{kq}{r} = n^{2/3} v$.
Step 5: $V = (512)^{2/3} \times 2 = (8^3)^{2/3} \times 2 = 8^2 \times 2 = 64 \times 2 = 128$ V.