Hint: Electric field due to a charged thin plate, $\mathrm{E}=\frac{\mathrm{Q}}{2{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}$.

(a) Step 1: Find the net electric field on the plate.

The net electric field at plate γ before collision is equal to the sum of electric field at plate γ due to plate α and β.

The electric field at plate γ due to plate α is ${\mathrm{E}}_1=\frac{-\mathrm{Q}}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)},$ to the left.

The electric field at plate γ due to plate β is ${\mathrm{E}}_2=\frac{\mathrm{q}}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}$, to the right.

Hence, the net electric field at plate γ before collision.

$\mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2=\frac{\mathrm{q}-\mathrm{Q}}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}$, to the left, if Q>q.

(b) Step 2: Find the net charge on the plates after the collision.

During the collision, the plates β and γ are together. Their potentials become same. Suppose charge on plate β is q₁ and charge on plate γ is q₂. At any point O, in between the two plates, the electric field must be zero.

Electric field at O due to plate α$=\frac{-\mathrm{Q}}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)},$ to the left.

Electric field at O due to plate β$=\frac{{\mathrm{q}}_1}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}$, to the right.

Electric field at O due to plate $\mathrm{\gamma }=\frac{{\mathrm{q}}_2}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}$ to the left

As the electric field at O is zero, therefore

$\frac{\mathrm{Q}+{\mathrm{q}}_2}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}=\frac{{\mathrm{q}}_1}{\mathrm{S}\left(2{\mathrm{\epsilon }}_0\right)}$
$\therefore \mathrm{Q}+{\mathrm{q}}_2={\mathrm{q}}_1$
$\mathrm{Q}={\mathrm{q}}_1-{\mathrm{q}}_2 ............\left(\mathrm{i}\right)$

As there is no loss of charge on collision,

$\mathrm{Q}+\mathrm{q}={\mathrm{q}}_1+{\mathrm{q}}_2 ..................\left(\mathrm{ii}\right)$

On solving equations (i) and (ii), we get,

${\mathrm{q}}_1=(\mathrm{Q}+\mathrm{q}/2)=\mathrm{charge} \mathrm{on} \mathrm{plate} \mathrm{\beta }$
${\mathrm{q}}_2=(\mathrm{q}/2)=\mathrm{charge} \mathrm{on} \mathrm{plate} \mathrm{\gamma }$

(c) Step 3: Find the velocity of the pate after the collision.

After the collision, at a distance d from plate $\mathrm{\beta }$,

Let the velocity of plate γ be v. After the collision, electric field at plate γ is, ${\mathrm{E}}_2=\frac{-\mathrm{Q}}{2{\mathrm{\epsilon }}_0\mathrm{S}}+\frac{(\mathrm{Q}+\mathrm{q}/2)}{2{\mathrm{\epsilon }}_0\mathrm{S}}=\frac{\mathrm{q}/2}{2{\mathrm{\epsilon }}_0\mathrm{S}}$ to the right.

Just before collision, electric field at plate γ is ${\mathrm{E}}_1=\frac{\mathrm{Q}-\mathrm{q}}{2{\mathrm{\epsilon }}_0\mathrm{S}}$.

If ${\mathrm{F}}_1$ is force on plate γ before collision, then ${\mathrm{F}}_1={\mathrm{QE}}_1=\frac{(\mathrm{Q}-\mathrm{q})\mathrm{Q}}{2{\mathrm{\epsilon }}_0\mathrm{S}}$

Total work done by the electric field is round trip movement of plate γ,

$\mathrm{W}=({\mathrm{F}}_1+{\mathrm{F}}_2)\mathrm{d}$
$=\frac{\left[\right(\mathrm{Q}-\mathrm{q})\mathrm{Q}+{(\mathrm{q}/2)}^2]\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{S}}=\frac{{(\mathrm{Q}-\mathrm{q}/2)}^2\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{S}}$

IF m is mass of plate γ, the KE gained by plate $\mathrm{\gamma }=\frac{1}{2}{\mathrm{mv}}^2$

According to work-energy principle, $\frac{1}{2}{\mathrm{mv}}^2=\mathrm{W}=\frac{{(\mathrm{Q}-\mathrm{q}/2)}^2\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{S}}$

$\mathrm{\gamma }=(\mathrm{Q}-\mathrm{q}/2){\left[\frac{\mathrm{d}}{{\mathrm{m\epsilon }}_0\mathrm{S}}\right]}^{1/2}$