Solution
The chain of conducting balls connected by wires acts as a single long conductor. When placed in a uniform electric field $E$, the free charges redistribute until the net internal electric field becomes zero. This effectively means the potential difference created by the external field along the length of the chain is cancelled by the potential difference created by the induced charges at the ends.
- External Potential Drop:
The potential drop due to the external field $E$ over the total length $L = (n-1)l$ is: $$\Delta V_{ext} = E \times (n-1)l$$ - Induced Potential Difference:
We can approximate the system as two point charges $+q$ and $-q$ at the ends of the chain (since $l \gg r$). The potential difference between the surface of the positive sphere and the negative sphere is dominated by their self-potentials (interaction terms are negligible due to large distance). $$V_+ \approx \frac{kq}{r}, \quad V_- \approx \frac{k(-q)}{r}$$ $$\Delta V_{ind} = V_+ – V_- = \frac{kq}{r} – \left(-\frac{kq}{r}\right) = \frac{2kq}{r}$$ - Equilibrium Condition:
For the conductor to be an equipotential, the magnitudes of these potential differences must be equal: $$\frac{2kq}{r} = E(n-1)l$$ Substituting $k = \frac{1}{4\pi\epsilon_0}$: $$\frac{2q}{4\pi\epsilon_0 r} = E(n-1)l$$ $$\frac{q}{2\pi\epsilon_0 r} = E(n-1)l$$ $$q = 2\pi\epsilon_0 r (n-1) l E$$
Answer: (b) $2\pi\epsilon_0 r(n-1)lE$