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 :

The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be

A parallel plate capacitor is connected to a battery and a dielectric slab is inserted between the plates, then which quantity increase

Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

A parallel plate capacitor of capacitance $200 \,\mu {F}$ is connected to a battery of $200 \, {V} .$ A dielectric slab of dielectric constant $2$ is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ……$ J.$