When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates

A parallel plate capacitor has plate area $100\, m ^{2}$ and plate separation of $10\, m$. The space between the plates is filled up to a thickness $5\, m$ with a material of dielectric constant of $10 .$ The resultant capacitance of the system is $'x'$ $pF$. The value of $\varepsilon_{0}=8.85 \times 10^{-12} F \cdot m ^{-1}$ The value of $'x'$ to the nearest integer is............

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, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then

Following operations can be performed on a capacitor : $X$ - connect the capacitor to a battery of $emf$ $E.$ $Y$ - disconnect the battery $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab in the capacitor

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively ${C_o}$ and ${W_o}$. If the air is replaced by glass (dielectric constant $= 5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be