The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

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

Write the formula for the capacitance of a capacitor having a dielectric constant $K = 2$.

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)