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 $q_2$.

Find the value of $q$ so that the net electrostatic force on the $10 \mu C$ charge placed at a vertex of the square is zero. The charges are arranged as shown in the figure.

Two charges,each of $1\;C$,are at a distance of $1\;km$ apart. The force between them is:

Two identical balls having like charges $Q_1$ and $Q_2$ are placed at a distance $r$ apart and repel each other with a force $F$. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls, $\frac{Q_1}{Q_2}$, is:

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

For the regular pentagon system shown in the figure,find the net electrostatic force on the charge $q_0$ placed at the center.