Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $.........V.$ (Assuming Dielectric constant $=10$ )

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$

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

The capacity of an air condenser is $2.0\, \,\mu F$. If a medium is placed between its plates. The capacity becomes $ 12\, \,\mu F$. The dielectric constant of the medium will be

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