Question

Two fixed, identical conducting plates ( α β)

, each of surface area S are charged to Q and q, respectively, where Q q 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . 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 βγ . (a) Find the electric field acting on the plate γ before collision. (b) Find the charges on β and γ after the collision. (c) Find the velocity of the plate γ after the collision and at a distance d from the plate β

Answer 1

Here initially charge on both the plates are 0 , when the plate γ moves towards plate β then the collision occurs and both distribute the charge . so final charge on both the plates after collision is
q_f = (0+q)/2= q/2 .
now electric field before collision on the plate γ = 0 ( since both plate has charge 0)
now using work energy theorem
change in kinetic energy = work done by all the forces on the body.
$\frac{1}{2}m{v}^2$ = Fd = q_fEd = $\frac{\sigma }{2{\epsilon }_0}x{q}_{f}xd=\frac{q_f}{sx2{\epsilon }_0}x{q}_{f}xd=\frac{{q_f}^{2}d}{2s{\epsilon }_0}$
v =$\sqrt{\frac{{q_f}^{2}d}{2ms{\epsilon }_0}}=q_f\left\{\sqrt{\frac{d}{2ms{\epsilon }_0}}\right\}$$\sqrt{\frac{{q_f}^{2}d}{2ms{\epsilon }_0}}=q_f\left\{\sqrt{\frac{d}{2ms{\epsilon }_0}}\right\}$ (mass is not mentioned in question)