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 potential difference between the plates ? Justify your answer

Explanation

Let the capacitor has a capacitance C and charge on it , when charged by a battery , be Q . Hence potential difference between the plates of capacitor `V = (Q)/(C)` . When the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates , the charge Q remains conserved , Now :
New potential difference between the plates `V = (Q.)/(C.) = (Q)/(KC) = (V)/(K)` .