A parallel plate capacitor is charged to a certain potential difference and the battery used to charged it is disconnected. Now a dielectric slab of dimensions equal to spacing between plates is introduced between the plates. What will be the changes, if any, in the charge, potential difference, capacitance, electric field strength and the energy stored in the capacitor?

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in capacitance?

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in charge on plates?

A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab if thickness d and dielectric constant K is now placed between the plates. what change, if any will take place in electric field intensity between the plates.

After charging a capacitor, the battery is disconnected. Now a dielectric material is introduced between its two plates. What would be the nature of change of its charge, capacitance, potential difference, electric field, energy stored?

If a parallel plate capacitor of capacitance C is kept connected to a supply voltage V and then to just fill the space, dielectric slab is inserted between the plates. What will be the change in the capacitance the charge, the potential difference, the electric field and the energy stored?

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 each with plate area A and separation ‘d’ is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in (i) charge on the plates (ii) electric field intensity between the plates, (iii) capacitance of the capacitor Justify your answer in each case