Two fixed identical conducting plates a and b each of surface area A are charged to –Q and q respectively (Q > q > 0). A third identical plate c free to move is located on the other side of the plate with charge Q at a distance d. If c is released, it collides with plate b. Assume the collision as elastic and the time of collision is sufficient to redistribute charge among b and c.

(a) Find the electric field acting on the plate c before collision. 

(b) Find the charges on b and c after the collision. 

(c) Find the velocity of plate c after the collision and at a distance d from the plate b if m is mass of the plate.

a) Electric field at C due to plates a and b are $\frac{-Q}{2{\epsilon }_{0}S}\hat{i}\text{ and }\frac{q}{2{\epsilon }_{0}S}\hat{i}$ respectively.

So, before collision net electric field on C is $\frac{(\mathrm{Q}-\mathrm{q})}{2{\epsilon }_0 \mathrm{S}}(-\hat{\mathrm{i}})$

b) When C collides with b, redistribution of charge takes place till the two plates acquire same potential. For a short duration the plates b and c from a single conductor and net electric field at a point p between them is zero. Let q1 and q2 be the changes on b and c after collision. Then net electric field at p will be 

$\overline{\mathrm{E}}=\frac{-\mathrm{Q}}{2{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}}+\frac{{\mathrm{q}}_1}{2{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}}-\frac{{\mathrm{q}}_2}{2{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}}$

$\text{ But }\overline{\mathrm{E}}=0\text{ which means }-\mathrm{Q}+{\mathrm{q}}_1-{\mathrm{q}}_2=0\text{ and }\mathrm{Q}={\mathrm{q}}_1-{\mathrm{q}}_2\text{ from conc. of charge, }$$\mathrm{Q}+\mathrm{q}={\mathrm{q}}_1+{\mathrm{q}}_2\Rightarrow {\mathrm{q}}_1=\mathrm{Q}+{\mathrm{q}}_2\text{ and }{\mathrm{q}}_2=\mathrm{q}/2$

c) Net electric field on plate C after collision is

${\overline{\mathrm{E}}}_{\mathrm{C}}=\frac{\mathrm{Q}}{2{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}}+\frac{{\mathrm{q}}_1}{2{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}}=\frac{\mathrm{q}}{4{\epsilon }_0 \mathrm{S}}\hat{\mathrm{i}} \left(\because -\mathrm{Q}+{\mathrm{q}}_1=\mathrm{q}/2\right)$

Force acting on C till the collision takes place is ${\mathrm{F}}_1=\frac{(\mathrm{Q}-\mathrm{q})\mathrm{Q}}{2{\epsilon }_0 \mathrm{S}}$

Force action on C after the collision is ${\mathrm{F}}_2=\frac{{\mathrm{Q}}^2}{8{\epsilon }_0 \mathrm{S}}$

Total work done = Work done in travelling distance d before collision + work done in travelling distance d after collision 

$\mathrm{W}=\left({\mathrm{F}}_1+{\mathrm{F}}_2\right)\mathrm{d}=\left\{\frac{(\mathrm{Q}-\mathrm{q})\mathrm{Q}}{2{\epsilon }_0 \mathrm{S}}+\frac{{\mathrm{q}}^2}{8{\epsilon }_0 \mathrm{S}}\right\}\mathrm{d}=\frac{(\mathrm{Q}-\mathrm{q}/2{)}^2 \mathrm{d}}{2{\epsilon }_0 \mathrm{S}}$

If V is velocity C after collision and at a distance d from plate b, gain in kinetic energy

$=\frac{1}{2}m{v}^2$

From work energy theorem, $\mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2\Rightarrow \mathrm{v}=(\mathrm{Q}-\mathrm{q}/2)\sqrt{\frac{\mathrm{d}}{{\epsilon }_0\mathrm{mS}}}$