$A$ given charge is situated at a certain distance from an electric dipole in the end-on position experiences a force $F$. If the distance of the charge is doubled,the force acting on the charge will be:

Charges are placed at the vertices of an equilateral triangle as shown in the figure. What is the resultant dipole moment?

An electric dipole having each charge of magnitude $2 \mu C$ is placed in an electric field of intensity $8 \times 10^{4} \ N/C$. If the maximum torque acting on the dipole is $4 \times 10^{-3} \ N \cdot m$,the length of the dipole is: (in $mm$)

An electric dipole has the magnitude of its charge as $q$ and its dipole moment is $p$. It is placed in a uniform electric field $E$. If its dipole moment is along the direction of the field,the force on it and its potential energy are respectively

Three point charges $q, -2q$ and $q$ are placed along the $x$-axis at $x = -a, 0$ and $a$ respectively. As $a \rightarrow 0$ and $q \rightarrow \infty$ while $qa^2 = Q$ remains finite,the electric field at a point $P$,at a distance $x$ $(x \gg a)$ from $x = 0$ is $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta} \hat{i}$. Then:

$A$ dipole comprises two charged particles of identical magnitude $q$ and opposite in nature. The mass $m$ of the positive charged particle is half of the mass of the negative charged particle $(2m)$. The two charges are separated by a distance $l$. If the dipole is placed in a uniform electric field $E$ such that the dipole axis makes a very small angle $\theta$ with the electric field $E$,the angular frequency of the oscillations of the dipole when released is given by: