There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant $K$ while the other one is charged to potential $V$ having air between its plates. If two capacitors are joined end to end, the common potential will be

A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by

A parallel plate capacitor is made of two plates of length $l$, width $w$ and separated by distance $d$. A dielectric slab ( dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{{\partial U}}{{\partial x}}$ where $U$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$ then the force on the dielectric when it is near the edge is

Between the plates of a parallel plate capacitor a dielectric plate is introduced just to fill the space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference across the plates and the length of the dielectric plate drawn out is

A parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor, then which of the following does not change

A parallel plate capacitor is formed by two plates each of area $30 \pi\, cm ^{2}$ separated by $1\, mm$. A material of dielectric strength $3.6 \times 10^{7} \,Vm ^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\, C$, the value of dielectric constant of the material is

$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$