Question

A point charge q is moving towards conducting spherical shell at constant velocity v. Shell is connected to ground.

• A. As charge moves towards shell, current flows from shell to earth.
• B. As charge approaches shell, current increases.
• C. Current increases linearly.
• D. Current decreases as charge approaches shell.

Answer 1

Answer: (A), (B)

As charge approaches shell, current increases.

Potential at O is zero
$\therefore \frac{kq}{x}+\frac{k{q}^{\prime }}{R}=0$
$q^{\prime }=\frac{-qR}{x}$

$i=\frac{d{q}^{\prime }}{dt}=\frac{d}{dt}(-\frac{qR}{x})=\frac{qR}{x^2}\left(\frac{dx}{dt}\right)=\frac{-qRV}{x^2}$
By analysing this equation, we can conclude the correct options.