Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $$t_1, t_2$$ and $$t_3$$ are thickness of dielectric slabs, $$\mathrm{k}_1, \mathrm{k}_2$$ and $$\mathrm{k}_3$$ are dielectric constants.

In parallel plate capacitor, electric field between the plates is '$$E$$'. If the charge on the plates is '$$Q$$' then the force on each plate is

If the charge on the capacitor is increased by 2 C the energy stored in it increased by 21%. Total original charge on the capacitor is

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates then the energy stored