Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be.....$m$

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be

A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V\ volt$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric material are separated by a conducting foil $F$. The voltage of the conducting foil is $.....V$

In a parallel plate capacitor set up, the plate area of capacitor is $2 \,m ^{2}$ and the plates are separated by $1\, m$. If the space between the plates are filled with a dielectric material of thickness $0.5\, m$ and area $2\, m ^{2}$ (see $fig.$) the capacitance of the set-up will be $.........\, \varepsilon_{0}$

(Dielectric constant of the material $=3.2$ ) and (Round off to the Nearest Integer)

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by