Given that,
A parallel plate capacitor is charged by a battery.
The battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates.
(i). We need to calculate the charge
The battery is disconnected so the charge on capacitor remains constant.
(ii). We need to calculate the capacitance of the capacitor
The capacitance without dielectric is
When dielectric slab is inserted, then the capacitance
Where, k = constant
(iii) We need to calculate the electric field intensity between the plates
Using formula of electric field
Where, V = potential difference and
d = separation between the plates.
Potential difference decreases and d remains the same and electric field also decreases.
(iv). We need to calculate the potential difference
Using formula of potential difference
As C increases and Q remains the same since the battery is disconnected, the potential difference between the plates decreases.
(v) We need to calculate the electric field intensity between the plates
Using formula of store energy
As Q is constant
So, Capacitance will be increases then stored energy will be decreases.
Hence, This is the required answer.