Two fixed, identical conducting plates , each of surface area S are charged to –Q and q, respectively, where
. A third identical plate (
), 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
(a) Find the electric field acting on the plate before the collision.
(b) Find the charges on β and after the collision.
(c) Find the velocity of the plate after the collision and at a distance d from the plate β.
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(a) Net electric field field at plate
before collision is vector sum of electric field at plate 
due to plate 
.
The electric field at plate
due to plate 
is 
The electric field at plate
Hence the net electric field at plate
before collision is
Or
to the left, if 
(b) During collision, plates
are in contact with each other, hence their potential becomes the same.
Suppose charge on plate
and charge on plate 
. at any point O , in between the two plates, the electric field must be zero.
Electric field at O due to
,
Electric field at O due to
,
Electric field at O due to
,
As the electric field at O is zero, therefore
As there is no loss of charge on collision,
On solving Equ. (1) and (2) We get
(c) Let the velocity be
at the distance d from plate 
after the collision.
, then the gain in K.E. over the round trip must be equal to the work done by the electric field.
is
If m is the mass of the plate
After the collision, electric field at plate
Just before collision, electric field at plate
If F1 is force on plate before collision., then
Total work done by the electric field is the round trip movement of plate
,
If m is the mass of plate
, the KE gained by the plate =
According to work energy principle,