Torque acting on an electric dipole in a uniform electric field is maximum when the angle between $\vec{p}$ and $\vec{E}$ is . . . . . . . (in $^{\circ}$)

$A$ positive charge $Q$ is distributed uniformly over a circular ring of radius $R$. $A$ point particle having a mass $m$ and a negative charge $-q$ is placed on its axis at a distance $x$ from the centre. Assuming $x \ll R$,find the time period of oscillation of the particle,if it is released from there [neglect gravity].

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

An electric dipole of mass $m$,charge $q$,and length $l$ is placed in a uniform electric field $\vec{E} = E_0 \hat{i}$. When the dipole is rotated slightly from its equilibrium position and released,the time period of its oscillations will be:

If a dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$,then the torque acting on it is given by:

An electric dipole is placed at an angle $60^{\circ}$ with an electric field of strength $4 \times 10^5 \,N/C$. It experiences a torque equal to $8 \sqrt{3} \,Nm$. Calculate the charge on the dipole,if the dipole length is $4 \,cm$.