Two capacitors of capacities $2 {C}$ and ${C}$ are joined in parallel and charged up to potential ${V}$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant ${K}$. The potential difference across the capacitors will now be

A capacitor of capacity $'C'$ is connected to a cell of $'V'\, volt$. Now a dielectric slab of dielectric constant ${ \in _r}$ is inserted in it keeping cell connected then

Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is $2 \times {10^5}\,V/m$. When the space is filled with dielectric, the electric field becomes $1 \times {10^5}\,V/m$. The dielectric constant of the dielectric material

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be

Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.

Reason : The surface density of charge onthe plate remains constant or unchanged.

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