A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
No work is done
Work is done at the cost of the energy already stored in the capacitor before the slab is inserted
Work is done at the cost of the battery
Work is done at the cost of both the capacitor and the battery
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 $V$
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