The charges $+q$,$+q$,and $Q$ are placed at the vertices of an equilateral triangle of side $a$. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to:

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 .......

$A$ moving positive charge approaches a negative charge. What will happen to the 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:

Two charges $+3.2 \times 10^{-19} \, C$ and $-3.2 \times 10^{-19} \, C$ kept $2.4 \, \mathring{A}$ apart form a dipole. If it is kept in a uniform electric field of intensity $4 \times 10^5 \, V/m$,what will be its electrical potential energy in stable equilibrium?

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?