A parallel plate capacitor is connecyed to a battery of emf V volts as shown. Now a slab of dielectric constant `k=2` is inserted between the plates of capacitor without disconnecting the battery. The electric field between the plates of capacitor after inserting the slab is `E= (PV)/(2d)`. Find the value of P.

A parallel plate capacitor is connected to a battery of emf V volts. Now a slab of dielectric constant k=2 is inserted between the plates of capacitor without disconnecting the battery. The electric field between the plates of capacitor after inserting the slab is E= (PV)/(2d) . Find the value of P.

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

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 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.