Question

$1)$ Two fixed, identical conducting plates $\left(\alpha \&\beta \right)$ , each of surface area S are charged to $-Q$ and $q$, respectively, where $Q>q>0.$ A third identical plate $\left(\gamma \right)$, free to move is located on the other side of the plate with charge $Q$ at a distance d in 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 $\beta \&\gamma$ .

$A)$ Find the electric field acting on the plate $\gamma$ before collision.

$2)$ Two fixed, identical conducting plates $\left(\alpha \&\beta \right)$ , each of surface area S are charged to $-Q$ and $q$, respectively, where $Q>q>0.$ A third identical plate $\left(\gamma \right)$, free to move is located on the other side of the plate with charge $Q$ at a distance d in 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 $\beta \&\gamma$ .



$B)$ Find the charges on $\beta$ and $\gamma$ after the collision.

$3)$ Two fixed, identical conducting plates $\left(\alpha \&\beta \right)$ , each of surface area S are charged to $-Q$ and $q$, respectively, where $Q>q>0.$ A third identical plate $\left(\gamma \right)$, free to move is located on the other side of the plate with charge $Q$ at a distance d in 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 $\beta \&\gamma$ .


$C)$ Find the velocity of the plate $\gamma$ after the collision and at a distance $d$ from the plate $\beta .$

Answer 1

$A)$

The electric field at γ due to plate $\propto$ is

$=-\frac{\sigma }{2{\epsilon }_o}\hat{x}=-\frac{Q}{2S{\epsilon }_o}\hat{x}$

The electric field at γ due to plate $\beta$ is

$=\frac{\sigma }{2{\epsilon }_o}\hat{x}=\frac{q}{2S{\epsilon }_o}\hat{x}$

Hence, the net electric field is

$\overrightarrow{E_1}=\frac{(Q-q)}{2{\epsilon }_{o}S}(-\hat{x})$

Final Answer:
$\overrightarrow{E_1}=\frac{(Q-q)}{2{\epsilon }_{o}S}(-\hat{x})$

$B)$ Assumption: The collision is elastic, and the time of collision is sufficient to redistribute charge amongst $\beta \&\gamma .$

During the collision plates $\beta \&\gamma$ are together and hence must be at one potential.

Suppose the charge on $\beta$ is $q_1$ and on $\gamma$ is $q_2$. Consider a point $O$. The electric field here must be zero.
Electric field at $O$ due to $\alpha =-\frac{Q}{2{\epsilon }_{o}S)}\hat{x}$

Electric field at $O$ due to $\beta =\frac{q_1}{2{\epsilon }_{o}S}\hat{x}$

Electric Field at $O$ due to $\gamma =-\frac{q_2}{2{\epsilon }_{o}S}\hat{x}$

$\frac{-(Q+q_2)}{2{\epsilon }_{o}S}+\frac{q_1}{2{\epsilon }_{o}S}=0$

$\Rightarrow q_1-q_2=Q\dots \left(i\right)$

Charge is conserved for an isolated conductor:

$Q_f=Q_i$

$q_1+q_2=Q+q\dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow q_1=Q+\frac{q}{2}$

$q_2=\frac{q}{2}$

Thus, the charge on $\beta$ and $\gamma$ are $(Q+\frac{q}{2})$ and $\frac{q}{2},$ respectively.

Final Answer: Charge on $\beta$ and $\gamma$ are $(Q+\frac{q}{2})$ and $\frac{q}{2},$ respectively.

$C)$

Electric field before collision at plate $\left(\gamma \right)$

$\overrightarrow{E_1}=\frac{(Q-q)}{2{\epsilon }_{o}S}(-\hat{x})$

Let the velocity be $v$ at the distance $d$ after the collision. If m is the mass of the plate $\gamma$, then the gain in $K.E.$ over the round trip must be equal to the work done by the electric field.

After the collision, the electric field at $\gamma$ is

$\overrightarrow{E_2}=-\frac{Q}{2{\epsilon }_{o}S}\hat{x}+\frac{(Q+\frac{q}{2})}{2{\epsilon }_{o}S}\hat{x}=\frac{\frac{q}{2}}{2{\epsilon }_{o}S}\hat{x}$

The work done when the plate $\gamma$ is released till the collision is $F_{1}d,$ where $F_1$ is the force on plate $\gamma$.

The work done after the collision till it reaches d is $F_{2}d$ where $F_2$ is the force on plate $\gamma$

$F_1=E_{1}Q=\frac{(Q-q)Q}{2{\epsilon }_{o}S}$

$F_2=\frac{q{E}_2}{2}=\frac{(\frac{q}{2}{)}^2}{2{\epsilon }_{o}S}$

Total work done:

$\frac{1}{2{\epsilon }_{o}S}\left[\right(Q-q)Q+(\frac{q}{2}{)}^2]d$

$=\frac{1}{2{\epsilon }_{o}S}(Q-\frac{q}{2}{)}^{2}d$

$\Rightarrow \left(\frac{1}{2}\right)m{v}^2=\frac{d}{2{\epsilon }_{o}S}(Q-\frac{q}{2}{)}^2$

$v=(Q-\frac{q}{2}){\left(\frac{d}{m{\epsilon }_{o}S}\right)}^{\frac{1}{2}}$

Final Answer:
$v=(Q-\frac{q}{2}){\left(\frac{d}{m{\epsilon }_{o}S}\right)}^{\frac{1}{2}}$