Question

A charged particle having a positive charge $q$ approaches a grounded metallic sphere of radius $R$ with a constant small speed $v$ as shown in the figure. In this situation,

• A. As the charge draws nearer to the surface of the sphere, a current flows into the ground.
• B. As the charge draws nearer to the surface of the sphere, a current flows out of the ground into the sphere.
• C. As the charged particle draws nearer, the magnitude of current flowing in the connector joining the sphere to the ground increases.
• D. As the charged particle draws nearer, the magnitude of current flowing in the connector joining the sphere to the ground decreases.

Answer 1

Answer: (A), (C)

The correct options are
A As the charge draws nearer to the surface of the sphere, a current flows into the ground.
C As the charged particle draws nearer, the magnitude of current flowing in the connector joining the sphere to the ground increases.

Let the charge induced on the sphere be $Q$. Since $q$ is positive, induced charge $Q$ also has to be positive. Hence, a current flows into the ground. (A)

Then, total electric potential on the surface of the sphere due to the charged particle and the induced charge should be equal to zero.

Let the particle be at a distance $R_0$ from the surface of the sphere initially. Then at time $t$,

$\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}+\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{R_0-vt}\right)=0$

$Q=\frac{Rq}{R_0-vt}$
$\Rightarrow \frac{dQ}{dt}=i=\frac{Rqv}{(R_0-vt{)}^2}$

Hence, as the particle draws nearer (i.e $t$ increases), the current flowing increases. (C)