Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectrics ($k_1$ and $k_2$) in the following fashion. Which of the following is true? $(k_1 > k_2)$

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

$A$ parallel-plate capacitor is connected to a resistanceless circuit with a battery until the capacitor is fully charged. The battery is then disconnected from the circuit and the plates of the capacitor are moved to half of their original separation using insulated gloves. Let $V_{new}$ be the potential difference across the capacitor plates when the plates have moved. Let $V_{old}$ be the potential difference across the capacitor plates when they were connected to the battery. Find the ratio $\frac{V_{new}}{V_{old}}$.

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then:

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ capacitor has capacitance $C_0$ when there is no dielectric between its plates. Two slabs of dielectric constant $K_1$ and $K_2$ respectively,with area equal to the area of the plates but thickness half of the distance between the plates,are placed in between the plates. Then the new capacitance is