A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected ? Justify your answer.

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case.

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case.

A parallel plate capacitor is charged by a battery after some time, the battery is disconnected and a dielectric slab with its thicnkess equal to the plate so reparation is insected between the plates. How will (i) the capacitance of the capacitor (ii) potential difference between the plates & (iii) the energy sotred in the capacitor the affected ? Q_(0) -charge V_(0) - potential difference, C_(0) - capacitance, E_(0) electric field. U_(0) - energy spred, before the dielectric slab is inserted. Q_(0)=C_(0) V_(0), (V_(0))/(d), U_(0)=(1)/(2)C_(0)V_(0)^(2)?