# 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 (Fig 1.13). The third plate is released and collides with the plate β. 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 the collision. (b) Find the charges on β and $\gamma$ after the collision. (c) Find the velocity of the plate $\gamma$ after the collision and at a distance d from the plate β.

(a) Net electric field 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 $\vec{E}_1= \frac{Q}{S(2\varepsilon _0)}(-\hat{i}),$

The electric field at plate $\gamma$  due to plate  $\beta$  is $\vec{E}_2= \frac{q}{S(2\varepsilon _0)}(\hat{i}),$

Hence the net electric field at plate $\gamma$ before collision is
$\\ \vec{E}=\vec{E}_1+\vec{E}_2= \frac{q-Q}{S(2\varepsilon _0)}(\hat{i}),\\ \\ \vec{E}=\vec{E}_1+\vec{E}_2= \frac{Q-q}{S(2\varepsilon _0)}(\hat{i}),$

Or             $\frac{Q-q}{S(2\varepsilon _0)}$ to the left, if $Q>q$

(b) During collision, plates $\alpha\: and\: \beta$  are in contact with each other, hence their potential becomes the same.

Suppose charge on plate   $\beta \: is\: q_1\:$ and charge on plate $\gamma \: is\: q_2\:$. at any point O , in between the two plates, the electric field must be zero.

Electric field at O due to $\alpha$,

$\vec{E}_\alpha = \frac{Q}{S(2\varepsilon _0)}\left (- \hat{i} \right )$

Electric field at O due to $\beta$,

$\vec{E}_2 = \frac{q_1}{S(2\varepsilon _0)}\left ( \hat{i} \right )$

Electric field at O due to $\gamma$,

$\vec{E}_\gamma = \frac{q_2}{S(2\varepsilon _0)}\left (- \hat{i} \right )$

As the electric field at O is zero, therefore

$\frac{Q+q_2}{S(2\varepsilon _0)}=\frac{q_1}{S(2\varepsilon _0)}\\ \\ \because Q+q_2=q_1$

As there is no loss of charge on collision,

$Q+q=q_1+q_2$

On solving Equ. (1) and (2) We get

$q_1=(Q+q/2)=$Charge on plate $\beta$
$q_1=(q/2)=$Charge on plate $\gamma$

(c) Let the velocity be $\nu$ at the distance d from plate $\beta$  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, electric field at plate $\gamma$ is

$\vec{E}_2= \frac{Q}{2\varepsilon _0S}(-\hat{i}),+\frac{\left ( Q+q/2 \right )}{2\varepsilon _0S}\hat{i}=\frac{\left ( q/2 \right )}{2\varepsilon _0S}\hat{i}$
Just before collision, electric field at plate $\gamma$  is

$\vec{E}_1= \frac{Q-q}{2\varepsilon _0S}\: \hat{i}.$

If F1 is force on plate  before collision., then

$F_1=\vec{E}_1Q= \frac{\left ( Q-q \right )Q}{2\varepsilon _0S}\: \hat{i}\: and\: F_2=\vec{E}_2\frac{q}{2}= \frac{\left ( \frac{q}{2} \right )^{2}}{2\varepsilon _0S}\: \hat{i}$

Total work done by the electric field is the round trip movement of plate $\gamma$,

$W=(F_1+F_2)d$

$=\frac{\left [ \left ( Q-q \right )Q+\left ( q/2 \right )^{2} \right ]d}{2\varepsilon _0S}=\frac{\left ( Q-q/2 \right )^{2}d}{2\varepsilon _0S}$

If m is the mass of plate $\gamma$, the KE gained by the plate =

$\frac{1}{2}m\nu ^2$

According to work energy principle,

$\frac{1}{2}m\nu ^2=W \Rightarrow \frac{1}{2}m\nu ^2=\frac{\left ( Q-q/2 \right )^{2}d}{2\varepsilon _0S}$

$\Rightarrow \nu =\left ( Q-q/2 \right )\left ( \frac{d}{m\varepsilon _0S} \right )^{1/2}$

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