Three point charges $3 \text{ nC}$,$6 \text{ nC}$,and $9 \text{ nC}$ are placed at the corners of an equilateral triangle of side $0.1 \text{ m}$. The potential energy of the system is:

Two point charges $q_1 = 6 \mu\text{C}$ and $q_2 = 4 \mu\text{C}$ are kept at points $A$ and $B$ in air where distance $AB = 10 \text{cm}$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

Write an equation of electric potential at a point on the axis of an electric dipole.

An electric dipole with dipole moment $\vec{p}_i = (3\hat{i} + 4\hat{j}) \times 10^{-30} \, \text{C-m}$ is placed in an electric field $\vec{E} = 4000 \hat{i} \, \text{N/C}$. An external agent turns the dipole slowly until its electric dipole moment becomes $\vec{p}_f = (-4\hat{i} + 3\hat{j}) \times 10^{-30} \, \text{C-m}$. The work done by the external agent is equal to:

When the distance between two point charges is increased,the electrostatic potential energy of the system:

Consider the configuration of a system of four charges each of value $+q$. The work done by an external agent in changing the configuration of the system from Figure $(1)$ to Figure $(2)$ is: