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

  • A

    No work is done

  • B

    Work is done at the cost of the energy already stored in the capacitor before the slab is inserted

  • C

    Work is done at the cost of the battery

  • D

    Work is done at the cost of both the capacitor and the battery

Similar Questions

Match the pairs

Capacitor Capacitance
$(A)$ Cylindrical capacitor $(i)$ ${4\pi { \in _0}R}$
$(B)$ Spherical capacitor $(ii)$ $\frac{{KA{ \in _0}}}{d}$
$(C)$ Parallel plate capacitor having dielectric between its plates $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$
$(D)$ Isolated spherical conductor $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$

The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is

The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$

Two parallel plate capacitors of capacity $C$ and $3\,C$ are connected in parallel combination and charged to a potential difference $18\,V$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be $V$

  • [JEE MAIN 2022]

The distance between the plates of a parallel plate capacitor is $d$. A metal plate of thickness $d/2$ is placed between the plates. The capacitance would then be