When one electron is moved towards another electron,the electric potential energy of the system:

Derive the formula for the electric potential energy of a system of two point charges.

In the figure,the value of $Q$ so that the electrostatic potential energy of the system becomes zero is

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

Three point charges shown in the figure lie along a straight line. The energy required to exchange the position of the central charge with one of the negative charges is

Two particles of charges $4 \text{ nC}$ and $Q$ are kept in air with a separation of $10 \text{ cm}$ between them. If the electrostatic potential energy of the system is $1.8 \mu \text{ J}$, then $Q$ is: (in $\text{ nC}$)