A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process

If the capacitor is kept connected with the battery and then the dielectric slab is inserted between the plates, then what will be the change in the charge, the capacitance the potential difference, the electric field an the stored energy?

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

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

Two capacitors are connected in parallel to a battery. If a dielectric medium is inserted between the plates of one of the capacitors then the enegry stored in the system will increase. On inserting dielectric medium between the plates of a capacitors, its capacity increases.

A capacitor is charged with a battery and then disconnected. A dielectric slab is then inserted between the plates. How are the The stored energy related to the capacitor affected?

A parallel plate capacitor having capacitance C farad is connected with a battery of emf V volts. Keeping the capacitor cannected with the battery, a dielectric slab of dielectric constant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates. Then,

A capacitor is charged with a battery and then disconnected. A dielectric slab is then inserted between the plates. How are the Charge.