Two point charges $q_1$ and $q_2$ are '$l$' distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

Two point charges $q_2 = 3 \times 10^{-6} \ C$ and $q_1 = 5 \times 10^{-6} \ C$ are located at $B(3, 5, 1) \ m$ and $A(1, 3, 2) \ m$ respectively. Find the magnitude of the force on $q_1$ due to $q_2$.

$A$ proton and an anti-proton come close to each other in vacuum such that the distance between them is $10 \, cm$. Consider the potential energy to be zero at infinity. The velocity at this distance will be ........... $\, m/s$.

Charges $Q$ are placed at the ends of a diagonal of a square and charges $q$ are placed at the other two corners. The condition for the net electric force on $Q$ to be zero is

Two charges $q_1 = +6q$ and $q_2 = -3q$ are placed as shown in the figure. $A$ proton is placed on the $x$-axis away from $q_2$. To keep the proton in equilibrium,the distance between $q_1$ and the proton is:

$A$ total charge $Q$ is broken into two parts $Q_1$ and $Q_2$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when: