The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

Two charges $+10 \mu C$ and $-10 \mu C$ are separated by $10 \text{ cm}$. The magnitude of the force acting on another charge $5 \mu C$ placed at the midpoint of the line joining the two charges will be: [Use $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$] (in $\text{ N}$)

$A$ charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum,is

In Millikan's oil drop experiment,an oil drop carrying a charge $Q$ is held stationary by a potential difference of $2400\,V$ between the plates. To keep a drop of half the radius stationary,the potential difference had to be made $600\,V$. What is the charge on the second drop?

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

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.