Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. $A$ third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now

Two opposite and equal charges each of magnitude $4 \times 10^{-8} \text{ C}$ form a dipole. Their separation is $2 \times 10^{-2} \text{ cm}$. When this dipole is placed in an external electric field of $4 \times 10^8 \text{ NC}^{-1}$,the value of maximum torque and the work done in rotating it through $180^{\circ}$ respectively will be:

Write an equation of electric potential at a point due to an electric dipole.

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?

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

Three point charges of $3 \mu C, 4 \mu C$,and $5 \mu C$ are arranged at the three corners of a right-angled triangle $ABC$ as shown in the figure. The work done in moving the charges at $A$ and $C$,so that the three charges are located at the three corners of an equilateral triangle of side $3 \text{ cm}$ is (in $J$)