$A$ proton is moving away from an electron. Find the change in the potential energy of the system.

An electric dipole with dipole moment $p$ is placed in an electric field of intensity $E$. Initially,the dipole is parallel to the field. What is the work done by an external agent to rotate the dipole by $180^{\circ}$ in the electric field?

Charges $Q$,$+q$,and $+q$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. If the total electrostatic potential energy of the system is zero,then $Q = $ . . . . . . .

The value of electric potential at any point due to an electric dipole is:

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:

Three particles,each having a charge of $10 \mu C$,are placed at the corners of an equilateral triangle of side $10 \ cm$. The electrostatic potential energy of the system is (Given $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):