Question

*A parallel plate capacitor is charged and the charging battery is disconnected. Then a dielectric slab is introduced between the plates. The quantity that remains unchanged is:* 1️⃣ Potential 2️⃣ Capacity 3️⃣ Energy 4️⃣ Charge​

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(i) The capacitance increases as the dielectric constant K>1. 

(ii) Potential difference V=CQ.  As C increases and Q remains the same since the battery is disconnected, the p.d. between the plates decreases.

(iii) Electric field E=dV where V is the p.d. and d the separation between the plates. As V decreases and d remains the same, electric field also decreases.

(iv) Energy stored in a capacitor U=21CQ2. As Q is constant and C increases, U decreases.