Two fixed, identical conducting plates `(alpha and beta)`, each of surface area S are charged to -Q and q, respectively, where Q `gt` q `gt` 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 (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.`

Explanation

(a). Find electric field at plate `gamma` before collision is equal to the sum of electric field at plate `gamma` due to plate `alpha and beta`.
The electric field at plate `gamma` due to plate `alpha` is `E_(1)=(-Q)/(S(2epsi_(0)))` to the left.
The electric field at plate `gamma` due to plate `beta` is `E_(2)=(q)/(S(2epsi_(0)))`, to the right
Hence, the ent electric field at plate `gamma` before collision,
`E=E_(1)+E_(2)=(q-Q)/(S(2epsi_(0)))`, to the left if `Q gt q`
(b) During collision, plates `beta and gamma` are together. Their potentials become same.
Suppose charge on plate `beta ` is `q_(1)` and cahrge 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 plate `alpha=(-Q)/(S(2epsi_(0)))`, to the left
Electric field at O due to plate `beta=(q_(1))/(S(2epsi_(0)))`, to the right ltBrgt Electric field at O due to plate `gamma=(q_(2))/(S(2epsi_(0)))`, to the left
As the electric field at O is zero, therefore ltBrgt `(Q+q_(2))/(S(2epsi_(0)))=(q_(1))/(S(2epsi_(0)))`
`thereforeQ+q_(2)=q_(1)`
`Q=q_1+q_2` . . .(i)
As there is no loss of charge o collision, ltBrgt `Q+q=q_(1)+q_(2)`
On solving eqs. (i) and (ii), we get
`q_(1)=(Q+q//2)`=charge on plate `beta`
`q_(2)=(q//2)`=charge on plate `gamma`
(c) After collision, at a distance d from plate `beta`.
Let the velocity of plate `gamma` be v. After the collision, electric field at plate `gamma` is
`E_(2)=(-Q)/(2epsi_(0)S)+((Q+q//2))/(2epsi_(0)S)=(q//2)/(2epsi_(0)S)` to the right
just before collision, electric field at plate `gamma` is `E_(1)=(Q-q)/(2epsi_(0)S)`
If `F_(1)` is force on plate `gamma` before collision, then `F_(1)=E_(1)Q=((Q-q)Q)/(2epsi_(0)S)`
Total work done by the electric field is round trip movement of plate `gamma`
`W=(F_(1)+F_(2))d`
`=([(Q-q)Q+(q//2)^(2)]d)/(2epsi_(0)S)=((Q-q//2)^(2)d)/(2epsi_(0)S)` ltBrgt If m is mass of plate `gamma`, the KE gained by plate `gamma=(1)/(2)mv^(2)`
According to work-energy principle, `(1)/(2)mv^(2)=W=((Q-q//2)^(2)d)/(2epsi_(0)S)`
`gamma=(Q-q//2)((d)/(mepsi_(0)S))^(1//2)`.