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.

Answers (1)

(a) Net electric field field at plate \gamma before collision is vector sum of electric field at plate  \gammadue 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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