Two point charges $q_1 = 6 \mu C$ and $q_2 = 4 \mu C$ are kept at points $A$ and $B$ in air where $AB = 10 \ cm$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \ cm$ (in $J$)?
$\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right)$

Three charges $Q$,$(-2q)$ and $(-2q)$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. The net electrostatic potential energy is zero if $Q$ is equal to

$A$ system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased,the potential energy of the system,

In the first configuration $(1)$ as shown in the figure,four identical charges $(q_0)$ are kept at the corners $A, B, C$ and $D$ of a square of side length $a$. In the second configuration $(2)$,the same charges are shifted to the midpoints $G, E, H$ and $F$ of the sides of the square. If $K = \frac{1}{4 \pi \varepsilon_0}$,the difference between the potential energies of configuration $(2)$ and $(1)$ is given by:

The work required to be done to assemble four particles,each having a charge $+q$,at the four corners of a square of side $a$ is .......

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: