1. Potential energy of a conducting spherical drop

For an isolated conducting sphere, $U = \frac{1}{2}QV$ Also, $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ So, $Q = 4\pi\varepsilon_0 R V$

For one small drop, let:

• radius $= r$
• charge $= q$
• potential $= 12\,\text{V}$
• energy $= E$

Thus, $E = \frac{1}{2}qV$

2. When 8 identical drops combine

If 8 similar spherical drops combine, volume is conserved: $\frac{4}{3}\pi R^3 = 8\left(\frac{4}{3}\pi r^3\right)$ Hence, $R^3 = 8r^3 \Rightarrow R = 2r$

Total charge is also conserved: $Q = 8q$

3. Potential of the big drop

Potential of the big drop is $V' = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ Substitute $Q=8q$ and $R=2r$: $V' = \frac{1}{4\pi\varepsilon_0}\frac{8q}{2r} = 4\left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\right) = 4V$ So, $V' = 4 \times 12 = 48\,\text{V}$

4. Energy of the big drop

Potential energy of the big drop: $U' = \frac{1}{2}QV' = \frac{1}{2}(8q)(4V)$ $U' = 16qV$

But for one small drop, $E = \frac{1}{2}qV$ So, $qV = 2E$ Therefore, $U' = 16(2E) = 32E$

5. Final answer

The potential energy of the bigger drop is $\boxed{32E}$ So the required integer is: $\boxed{32}$