$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

$A$ parallel plate capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be:

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. The capacitance of the capacitor is $17.71 \, pF$. If this capacitor is connected to a $100 \, V$ supply,and a $3 \, mm$ thick mica sheet (of dielectric constant $k = 6$) is inserted between the plates,calculate the new capacitance,charge,and potential difference in the following cases:
$(a)$ While the voltage supply remains connected.
$(b)$ After the supply is disconnected.

Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?