Charge Problem

A simple question if you look at it carefully. Since the sphere is earthed when switch is closed, potential at the centre of it is $0$

Also let the charged particle of charge $q$ be at distance $x$

Now if we say that at any time charge on it be ${q}_{1}$

Hence we write :

$V=\frac{kq}{x}+\frac{k{q}_{1}}{R}=0$

$\Rightarrow {q}_{1}=\frac{-Rq}{x}$

To write current we would differentiate both sides with respect to $x$

${i}_{ammeter}= \frac{Rq}{{x}^{2}} \frac{dx}{dt}=\frac{Rq}{{x}^{2}} {v}_{particle}$ (i)

Given $v=-(2t+4)$ . Integrating both sides with respect to $t$

$x=-{t}^{2}-4t+C$

Since at $t=0$ we have $x=10 m$ hence $C=10$

$x=10-{t}^{2}-4t$

Put $t=1$ to get :

$x=5 m$

Also we get ${v}_{particle}=-2(1)-4=-6 m/s$

Put all this in (i) to get :

$\boxed{6q=25 C}$