91Ó°ÊÓ

As per the law of energy conversations, within a closed system energy remains same. Energy cannot be destroyed or created it only transform from one form to another.

The expression in this electric potential is;

$V=\frac{kq}{r}$

Here in in equation K is column's constant and q is the charge where r denote as distance.

Hence;

$q=\frac{Vr}{k}$

Here;

Sphere potential is 200,000 V

d =12 cm, hence r= 6 cm

and k= 9 x 10⁹Nm²/C²

$\begin{array}{rcl}q& =& \frac{200,000V×{\displaystyle \frac{6}{100}}m}{9×{10}^{9}N{m}^2/C^2}\\ & =& 1.33×{10}^{-6}C\end{array}$

As per the given Condition distance is;

d=2 cm+ 6cm =8 cm

hence the potential difference is;

$v^{\text{'}}=\frac{Kq}{d}$

Substituting all the given data along with the distance of condition;

$\begin{array}{rcl}{V}^{\text{'}}& =& \frac{\left(9×{10}^{9}N{m}^2/C^2\right)×\left(1.33×{10}^{-6}C\right)}{0.08m}\\ V^{\text{'}}& =& 1.5×{10}^{5}V\end{array}$

As law of energy conversations;

kinetic energy is equal to potential energy

$$\begin{gathered} q{V}^{\text{'}}=\frac{1}{2}m{v}^2 \\ or;q=\frac{m{v}^2}{2V^{\text{'}}} \end{gathered}$$

As per given condition m= 5g; v= 3 m/s

Hence;

$$\begin{gathered} q=\frac{0.005kg×{\left(3m/s\right)}^2}{2(1.5×{10}^{5}V)} \\ q=150nC \end{gathered}$$

Hence the required charge can be considered as 150 nC