What happens to the stored electric potential energy and the electric field between the plates of a charged isolated parallel plate capacitor when the plate separation is increased?

Explanation

When the plate separation is increased, the charge stored in the capacitor remains same and the capacitance decreases. Therefore, potential difference across the plates increases.
`therefore` The stored electric potential energy will increase by `(1)/(2)QDeltaV`.
Hence, electric field between the plates is: `E=(Q)/(in_(0)A)` charge (Q) remains constant. The electric field between the plates remains constant in strength but fills up more volume on increasing the separation.