Two charges $7 \ \mu C$ and $-4 \ \mu C$ are placed at $(-7 \ cm, 0, 0)$ and $(7 \ cm, 0, 0)$ respectively. Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$,the electrostatic potential energy of the charge configuration is: (in $J$)

Nine point charges are placed on a cube as shown in the figure. The charge $q$ is placed at the body centre whereas all other charges are at the vertices. The electrostatic potential energy of the system will be

Electric potential at an equatorial point of a small dipole with dipole moment $P$ ($r$,distance from the dipole) is

An electric dipole is formed by two charges $+q$ and $-q$ located in the $xy$-plane at $(0, 2) \text{ mm}$ and $(0, -2) \text{ mm}$,respectively,as shown in the figure. The electric potential at point $P(100, 100) \text{ mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2) \text{ mm}$ and $(1, -2) \text{ mm}$,respectively. What is the value of the electric potential at $P$ due to the new dipole?

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

Derive the expression for the electric potential due to an electric dipole at a general point $P$ at a distance $r$ from its center.