A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are

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 has a dielectric slab of dielectric constant $K$ between its plates that covers $1 / 3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C _1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options, igonoring edge effects.

$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{3}{K}$ $(D)$ $\frac{ C }{ C _1}=\frac{2+ K }{ K }$

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation