Potential Due to a Electric Dipole

The electrostatic potential due to an electric dipole at an equatorial point is

An electric dipole is formed by two charges $+q$ and $-q$ located in $xy$-plane at $(0, 2) \mathrm{mm}$ and $(0, -2) \mathrm{mm}$, respectively, as shown in the figure. The electric potential at point $P(100, 100) \mathrm{mm}$ due to the dipole is $V_0$. The charges $+q$ and $–q$ are then moved to the points $(-1, 2) \mathrm{mm}$ and $(1, -2) \mathrm{mm}$, respectively. What is the value of electric potential at $P$ due to the new dipole?

Two equal charges $q$of opposite sign separated by a distance $2a$ constitute an electric dipole of dipole moment $p$. If $P$ is a point at a distance$r$from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle $\theta$ with the axis of the dipole, then the potential at $P$ is given by: ($r>>2a$) (Where $p=2qa$)

The electric field and the potential of an electric dipole vary with distance $r$ as

A charge $q$ moves from points $A\left(0,1\right)$ to $B(1,0)$ as shown in figure. If $\left|p_1\right|>\left|p_2\right|$, then work done by electric force in moving charge $q$ from $A$ to $B$, in the vicinity of two short dipoles placed at the origin with their dipole moments as shown, will be

A small dipole of dipole moment $p$ is placed at a distance $2R$ from the centre of a neutral conducting sphere of radius $R.$ The direction of dipole is towards the centre of the sphere. A tangent is drawn from the centre of dipole to the sphere which meets the sphere at point $A.$

Two identical dipoles of dipole moment $P=p_0\hat{i}\left(p_0\right.$ is a positive constant) are placed on $x-$axis at points $A(a, 0, 0)$ and $B(-a, 0, 0)$ as shown. Then pick up the correct statements:

A small charged bead can slide on a circular frictionless, insulating wire frame. A point like dipole is fixed at the centre of circle, dipole moment is $\overrightarrow{p}$, initially the bead is on the plane of symmetric of the dipole. Bead is released from rest. Ignore the effect of gravity. Select the correct options.

The electric potential at the equatorial position due to the dipole is zero.

There is a short dipole along $x-$ axis its centre coincides at origin the ratio of electric potentials at $(3, 4)$ and $(2, 2)$:

The potential at a point $P$ due to a point electric dipole is $1.8\times {10}^5 V$. If $P$ is at a distance of $50 \mathrm{cm}$ from the centre $O$ of the dipole and if vector $\overrightarrow{\mathit{O}\mathit{P}}$ makes an angle ${60}^o$ with the positive side of the axial line of the dipole, the dipole moment of the dipole is ${10}^{-n} \mathrm{C}-\mathrm{m}$ Find the value of $n$.

The distance between charges of a dipole ($\pm e$) is $1.38\mathrm{ }\mathrm{\AA }.$ The potential due to this dipole at a distance of $10\mathrm{ }\mathrm{\AA }$ on the axis of dipole is (Take $e=1.6 \times {10}^{-19} \mathrm{C}$)

Consider a small electric dipole. Find the relation of electric field and the potential with respect to distance $\mathrm{r}$ of a point from it:

Correct statements about an electric dipole of moment $\overrightarrow{P}$ placed in a uniform electric field $\overrightarrow{E}$ are:
$\left(\mathrm{i}\right)$ $\overrightarrow{P}\times \overrightarrow{E}$ is torque on the dipole.
$\left(\mathrm{ii}\right)$ $\overrightarrow{P}\bullet \overrightarrow{E}$ is potential energy of the system.
$\left(\mathrm{iii}\right)$ The resultant force on the dipole is zero.

A charge array is shown in figure and is known as an electric quadrupole. For a point on the axis of the quadrupole, choose the correct dependence of potential on $r$ when $a/r<<1$.

Electric potential at the equatorial point is 

At which of the following points is the electric potential due to a dipole minimum? 

The electrostatic potential due to an electric dipole at a distance $r$ varies as :

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 $-\alpha \left[\frac{ql}{r^2}\right]\times {10}^9 \mathrm{V}$, where the value of $\alpha$ is ______. (Use $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}$)

The electrostatic potential due to a short electric dipole at a distance $r$ varies as: