The value of the electric permittivity of free space $(\varepsilon_0)$ is ........

Two point charges $+10 \mu C$ and $+4 \mu C$ are placed $10 \ cm$ apart in air. The work required to be done to bring them $2 \ cm$ closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right)$. (in $J$)

$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?

State the principle of superposition for electric forces.

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. For a given distance $R$,the force between them is maximum when:

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting,massless strings of same length. At equilibrium,the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $\rho_l = 800 \ kg \ m^{-3}$ and dielectric constant $K = 21$. If the angle between the strings remains the same after the immersion,then
$(A)$ electric force between the spheres remains unchanged
$(B)$ electric force between the spheres reduces
$(C)$ mass density of the spheres is $840 \ kg \ m^{-3}$
$(D)$ the tension in the strings holding the spheres remains unchanged