$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.

Two capacitors $C_1$ and $C_2$ are connected to a battery as shown in the figure. The space between the plates of $C_1$ is filled with air,and the space between the plates of $C_2$ is filled with a dielectric material. Which of the following is true regarding the charges $Q_1$ and $Q_2$ on the capacitors?

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

In a parallel plate capacitor setup,the plate area of the capacitor is $2 \, m^{2}$ and the plates are separated by $1 \, m$. If the space between the plates is filled with a dielectric material of thickness $0.5 \, m$ and area $2 \, m^{2}$ (see figure),the capacitance of the setup will be $......... \, \varepsilon_{0}$. (Dielectric constant of the material $= 3.2$) (Round off to the nearest integer).

Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

$A$ parallel plate capacitor with air as the dielectric has capacitance $C$. $A$ slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be