A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$
$\frac{l}{4}$
$\frac{l}{2}$
$\frac{l}{3}$
$\frac{2l}{3}$
Two condensers of capacities $2C$ and $C$ are joined in parallel and charged upto potential $V$. The battery is removed and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The $p.d.$ across the capacitors will now be
A parallel plate capacitor with plate separation $5$ $\mathrm{mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is_____.
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
A capacitor is half filled with a dielectric $(K=2)$ as shown in figure A. If the same capacitor is to be filled with same dielectric as shown, what would be the thickness of dielectric so that capacitor still has same capacity?
A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :