A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
Reduction of charge on the plates and increase of potential difference across the plates
Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates
Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
None of the above
Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be
The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :
A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are
The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$
An insulator plate is passed between the plates of a capacitor. Then the displacement current