An electric dipole is formed by placing a charge of $4 \times 10^{-8} \ C$ at a distance of $2 \times 10^{-2} \ cm$. If it is placed in an electric field of $4 \times 10^8 \ N/C$,what is the maximum torque acting on it and the work done to rotate the dipole by $180^{\circ}$?

As shown in the figure,a particle $A$ of mass $2m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q$. The system is placed in an electric field $\vec E$. The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E$. After the system settles into equilibrium,one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find the maximum tension in the rod.

Three charges are arranged on the vertices of a right-angled triangle as shown in the figure. The magnitude of the dipole moment of the combination in the unit of $C-cm$ is:

An electric dipole of moment $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) \times 10^{-29} \; C \cdot m$ is at the origin $(0, 0, 0)$. The electric field due to this dipole at $\overrightarrow{r} = \hat{i} + 3\hat{j} + 5\hat{k}$ (note that $\overrightarrow{r} \cdot \overrightarrow{p} = 0$) is parallel to

The distance between charges $+q$ and $-q$ is $2l$ and between $+2q$ and $-2q$ is $4l$. The electrostatic potential at point $P$ at a distance $r$ from centre $O$ is $-\alpha \left[ \frac{ql}{r^2} \right] \times 10^9 \text{ V}$,where the value of $\alpha$ is . . . . . . . (Use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$)

An electric dipole will have minimum potential energy when it subtends an angle with the direction of the electric field.