Question 24 Solution
Concept: The conducting thread keeps the two balls at the same potential. An external electric field $E$ induces charges on them.
Step 1: Calculate induced charge ($q$)
Let the separation between balls be $x$.
The external field creates a potential difference $V_{ext} = E \cdot x$.
To maintain zero net potential difference, the induced charges ($q$ and $-q$) must create a counter-potential.
Potential difference due to charges $\approx \frac{kq}{r} – \frac{k(-q)}{r} = \frac{2kq}{r}$ (Assuming $x \gg r$, mutual potential is negligible).
Equating potentials:
Step 2: Calculate Work Done ($W$)
The force pulling the charges apart is $F = qE$ on the positive one and $-qE$ on the negative one. Total work is the integral of force over displacement.
$$W = \int_{l}^{L} F \, dx = \int_{l}^{L} q(x) E \, dx$$ $$W = \int_{l}^{L} (2\pi\epsilon_0 r E x) E \, dx = 2\pi\epsilon_0 r E^2 \int_{l}^{L} x \, dx$$ $$W = 2\pi\epsilon_0 r E^2 \left[ \frac{x^2}{2} \right]_{l}^{L} = \pi\epsilon_0 r E^2 (L^2 – l^2)$$Step 3: Find Velocity
By Work-Energy Theorem ($W = \Delta KE$):
Total KE = $2 \times \frac{1}{2}mv^2 = mv^2$.