When a slab of dielectric material is introduced between the parallel plants of a capcitor which remains connected to a battery, then charges on plates relative to earlier charge

When a dielectric material is introduced between the plates of a charged condenser, after disconnected the battery the electric field between the plates

When a dielectric material is introduced between the plates of a charged condenser the electric field between the plates

When a dielectric is introduced between the plates of a charged parallel plate capacitor, whichi one of the following will not change?

While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then

While a capacitor remains connected to a battery, a dielectric slab is slipped between the plates. Then

Two identical capacitors are connected as shown in fignre. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. What will be the capacitance, the charge, potential difference and stored energy for each capacitor?

A parallel plate air capacitor of plate separation d is charged to a potential difference triangle V . A slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected. What happens to the charge on the capacitor?

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is q_(0) , k= dielectric constant). .

If a dielectric slab is introduced between the plates of a parallel plate capacitor after the battery is disconnected, then how do the following quantities change. (i) Charge (ii) Potential (iii) Capacitance (iv) Energy.