If $OP = 1 \, cm$ and $OS = 2 \, cm$,calculate the work done by the electric field in shifting a point charge $q = \frac{4\sqrt{2}}{27} \, \mu C$ from point $P$ to $S$ in the given figure. The dipole moment is $p = 2 \times 10^{-6} \, C \cdot m$.

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:

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 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?

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic potential energy of the system is zero,the value of $Q$ is