The battary remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be the effect on its (i) capacity (ii) charge , (iiI) potential difference (iv) electric field, (v) energy stored ? Justify your answer.

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?

If the capacitor is kept connected with the battery and then the dielectric slab is inserted between the plates, then what will be the change in the charge, the capacitance the potential difference, the electric field an the stored energy?

A dielectric slab is introduced between the plates of a capacitor. Then the potential difference and the capacity respectively

The potential enery of a charged parallel plate capacitor is U_(0). If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

The potential enery of a charged parallel plate capacitor is U_(0) . If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

The potential enery of a charged parallel plate capacitor is U_(0) . If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

The potential enery of a charged parallel plate capacitor is U_(0) . If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be