Solution to Question 52
Physics Principle: Superposition of Potentials.
Method: We treat the shell with a hole as a complete shell of charge \(Q\) minus a small disc of radius \(r\) and charge density \(\sigma\) placed at the hole. $$ \sigma = \frac{Q}{4\pi R^2} $$
1. Potential at the Center (O):
Potential due to full shell: \( V_{shell,O} = \frac{Q}{4\pi\epsilon_0 R} \).
Potential due to removed disc (charge \(q_d = \sigma \pi r^2\)) located at distance \(R\) from O:
$$ V_{disc,O} = \frac{q_d}{4\pi\epsilon_0 R} $$
Net Potential at O:
$$ V_O = \frac{Q – q_d}{4\pi\epsilon_0 R} $$
2. Potential at the Hole (P):
Potential due to full shell at surface: \( V_{shell,P} = \frac{Q}{4\pi\epsilon_0 R} \).
Potential due to the disc at its own center: \( V_{disc,P} = \frac{\sigma r}{2\epsilon_0} \).
Net Potential at P:
$$ V_P = \frac{Q}{4\pi\epsilon_0 R} – \frac{\sigma r}{2\epsilon_0} $$
3. Kinetic Energy Calculation:
The particle \(q\) is released from O and moves to P.
$$ K = q(V_O – V_P) $$
$$ K = q \left[ \left( \frac{Q}{4\pi\epsilon_0 R} – \frac{q_d}{4\pi\epsilon_0 R} \right) – \left( \frac{Q}{4\pi\epsilon_0 R} – \frac{\sigma r}{2\epsilon_0} \right) \right] $$
Terms involving \(Q\) cancel out:
$$ K = q \left[ \frac{\sigma r}{2\epsilon_0} – \frac{q_d}{4\pi\epsilon_0 R} \right] $$
Substitute \(q_d = \sigma \pi r^2\):
$$ K = q \left[ \frac{\sigma r}{2\epsilon_0} – \frac{\sigma \pi r^2}{4\pi\epsilon_0 R} \right] = \frac{q \sigma r}{2\epsilon_0} \left( 1 – \frac{r}{2R} \right) $$
Since \(r \ll R\), we neglect the second term. Substitute \(\sigma = \frac{Q}{4\pi R^2}\):
$$ K \approx \frac{q r}{2\epsilon_0} \left( \frac{Q}{4\pi R^2} \right) $$