$A$ charge $q_1$ exerts some force on a second charge $q_2$. If a third charge $q_3$ is brought near,the force of $q_1$ exerted on $q_2$:

Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

Point charges $+4q, -q$ and $+4q$ are kept on the $x$-axis at points $x = 0, x = a$ and $x = 2a$ respectively. Then:

$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

$A$ certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?