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 indentical plate `(gamma)`, free to move is located on the other side of the plate with charge q at a distance d, fig. The third plate is released and collidies with the plate `beta`. Assume the collsion is electric 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 charge 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

The given set up is shown in Fig.
(a) The electric field at plate `gamma` due to plate `alpha` is
`E_(1) = - (Q)/(S (2in_(0))` , to the left
The electric field at plate `gamma` to plate `beta` is
`E_(2) = (q)/(S(2 in_(0))` , to the right.
Hence, the net electric field at plate `lamma` before collision is
`E = E_(1) + E_(2) = (q-Q)/(S (2in_(0))` , to the left if `Q gt q`
(b) During collision, plates `beta and gamma` are together. Their potentials become same. Suppose charge on plane `beta is q_(1)` and charge on plate `gamma is q_(2)`. At any point 0, inbetween the two plates, Fig, the electric field must be zero.
Electric field at 0 due to plate `alpha = ((-) Q)/(S (2in_(0)))` , to the left
Electric field at 0 due to plate `beta = (q_(1))/(S(2in_(0)))` , to the right
Electric field at 0 due to plate `gamma = (q_(2))/(S(2in_(0)))` , to the left
As the electric field at 0 is zero, therefore , `(Q+q_(2))/(S(2in_(0))) = (q_(1))/(S(2 in_(0)))`
`:. Q + q_(2) = q_(1) or Q = q_(1) - q_(2)` ...(i)
As there is no loss of charge on collision, `Q + q = q_(1) + q_(2)` ...(ii)
On solving (i) and (ii),we get `q_(1) = (Q + 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)/(2 in_(0) S) + ((Q + q//2))/(2 in_(0) S) = (q//2)/(2 in_(0) S)` to the right.
Just before collision, electric field at plate `gamma` is `E_(1) = (Q-q)/(2 in_(0) S)`
If `F_(1)` is force on plate `gamma` before collision, then `F_(1) = E_(1) Q = ((Q-q)Q)/(2 in_(0) S)`
Similarly, force `F_(2)` on plate `gamma` after collision, `F_(2) = E_(2) (q)/(2) = ((q//2)^(2))/(2in_(0) S)`
Total work done by the electric field in round trip movement of plate `gamma`
`W = (F_(1) + F_(2))d = ([(Q-q) Q+ (q//2)^(2)] d)/(2in_(0) S) = ((Q-q//2)^(2) d)/(2in_(0) S)`
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)/(2in_(0) S)`
`v = (Q - q//2) ((d)/(m in_(0) s))^(1//2)`