Two fixed,identical conducting plates $(\alpha)$ and $(\beta)$,each of surface area $S$,are charged to $-Q$ and $q$,respectively,where $Q > q > 0$. $A$ third identical plate $(\gamma)$,free to move,is located on the other side of the plate with charge $q$ at a distance $d$ as per the figure. The third plate is released and collides with the plate $(\beta)$. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst $(\beta)$ and $(\gamma)$.
$(a)$ Find the electric field acting on the plate $(\gamma)$ before collision.
$(b)$ Find the charges on $(\beta)$ and $(\gamma)$ after the collision.
$(c)$ Find the velocity of the plate $(\gamma)$ after the collision and at a distance $d$ from the plate $(\beta)$.

(N/A) The net electric field at plate $(\gamma)$ before collision is the vector sum of the electric fields due to plates $(\alpha)$ and $(\beta)$.
The electric field at $(\gamma)$ due to $(\alpha)$ is $\vec{E}_1 = \frac{Q}{2S\epsilon_0} \hat{i}$.
The electric field at $(\gamma)$ due to $(\beta)$ is $\vec{E}_2 = \frac{q}{2S\epsilon_0} \hat{i}$.
Thus,$\vec{E}_{net} = \frac{Q+q}{2S\epsilon_0} \hat{i}$.
$(b)$ During collision,plates $(\beta)$ and $(\gamma)$ are in contact,so they share the total charge $q_{total} = q + 0 = q$ (assuming plate $\gamma$ was initially uncharged as per standard interpretation of such problems,or if $\gamma$ had charge $q_{\gamma}$,then $q_{\beta} + q_{\gamma} = q + q_{\gamma}$). Given the plate $\gamma$ is identical and neutral,$q_{\beta} = q_{\gamma} = q/2$.
$(c)$ The force on plate $(\gamma)$ before collision is $F = q_0 E = (0) E = 0$. Wait,if plate $\gamma$ is neutral,it experiences no force. If it has charge $q_{\gamma}$,$F = q_{\gamma} E$. Assuming $\gamma$ is neutral,it must be given an initial impulse or the problem implies it is charged. Re-evaluating: The force on plate $\gamma$ is $F = q_{\gamma} E$. After collision,$q_{\gamma} = q/2$. The field between $\alpha$ and $\beta$ is $E = \frac{Q-q}{2S\epsilon_0}$. The work done $W = Fd = \frac{1}{2}mv^2$. Thus $v = \sqrt{\frac{2Fd}{m}}$.