A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

  • A

    Reduction of charge on the plates and increase of potential difference across the plates

  • B

    Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

  • C

    Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

  • D

    None of the above

Similar Questions

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