Two fixed, identical conducting plates $(\alpha $ and $\beta )$, each of surface area $S$ are charged to $-\mathrm{Q}$ and $\mathrm{q}$, respectively, where $Q{\rm{ }}\, > \,{\rm{ }}q{\rm{ }}\, > \,{\rm{ }}0.$ 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 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 $.
Two fixed, identical conducting plates $(\alpha $ and $\beta )$, each of surface area $S$ are charged to $-\mathrm{Q}$ and $\mathrm{q}$, respectively, where $Q{\rm{ }}\, > \,{\rm{ }}q{\rm{ }}\, > \,{\rm{ }}0.$ 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 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 $.
$(a)$ Net electric field at plate $\gamma$ before collision is vector sum of electric field at plate $\gamma$ due to plate $\alpha$ and $\beta$.
The electric field at plate $\gamma$ due to plate $\alpha$ is,
$\overrightarrow{\mathrm{E}}_{1}=\frac{-\mathrm{Q}}{\mathrm{S}\left(2 \in_{0}\right)}(-\hat{i})$
The electric field at plate $\gamma$ due to plate $\beta$ is,
$\overrightarrow{\mathrm{E}}_{2}=\frac{q}{\mathrm{~S}\left(2 \epsilon_{0}\right)}(\hat{i}$
$\therefore$ Hence, the net electric field at plate $\gamma$ before collision is,
$\overrightarrow{\mathrm{E}} =\overrightarrow{\mathrm{E}_{1}}+\overrightarrow{\mathrm{E}_{2}}=\frac{q-\mathrm{Q}}{\mathrm{S}\left(2 \epsilon_{0}\right)}(\hat{i})$
$\therefore \overrightarrow{\mathrm{E}} =\overrightarrow{\mathrm{E}_{1}}+\overrightarrow{\mathrm{E}_{2}}=\frac{\mathrm{Q}-q}{\mathrm{~S}\left(2 \epsilon_{0}\right)}(-\hat{i})$
$\therefore \frac{Q-q}{S\left(2 \epsilon_{0}\right)} \text { to the left, if } Q>q$
$(b)$ During collision plates $\beta$ and $\gamma$ are in contact with each other, hence their potentials become same.
Suppose charge on plate $\beta$ is $q_{1}$ and charge on plate $\gamma$ is $q_{2}$. At any point $\mathrm{O}$ in between the two plates, the electric field must be zero.
Electric field at $\mathrm{O}$ due to plate $\alpha$,
$\overrightarrow{\mathrm{E}}_{a}=\frac{\mathrm{Q}}{\mathrm{S}\left(2 \epsilon_{0}\right)}(-\hat{i})$
Electric field at $O$ due to plate $\beta$,
$\overrightarrow{\mathrm{E}}_{2}=\frac{q_{1}}{\mathrm{~S}\left(2 \epsilon_{0}\right)}(\hat{i})$
Electric field at $\mathrm{O}$ due to plate $\gamma$,
$\overrightarrow{\mathrm{E}}_{\gamma}=\frac{q_{2}}{\mathrm{~S}\left(2 \epsilon_{0}\right)}(-\hat{i})$
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