Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

$A$ capacitor has a capacitance of $5 \mu F$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material with a dielectric constant of $1.5$,having an area equal to that of the plates but a thickness of $\frac{d}{2}$,is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be $..........\mu F$.

$A$ parallel plate capacitor has a separation between plates of $0.885$ mm. It has a capacitance of $1$ $\mu$$F$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}$ $\Omega$m and resistance $17.7 \times 10^{14}$ $\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is . . . . . . . (Take permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ $F$/m)

$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \text{ pF}$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4 \text{ pF}$. The value of $x$ is

$A$ 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 $\frac{V_{new}}{V_{old}}$.

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field: