Question

Derive an expression for the electric field due to a dipole of dipole moment at a point on its perpendicular bisector

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Point P at a very large perpendicular distance from the axis is

$\vec{E}=\frac{-\bar{p}}{4 \pi \varepsilon_{0} y^{3}} \quad( for y > > a)$

Explanation:

Given: In this case we determine the electric field at a point P which is located on the perpendicular bisector of the line joining the two charges. Let the distance of P from this line be y.

To find: Electric Field Intensity due to an Electric Dipole at a Point M which is on the Equatorial line and at a distance r from the center of a dipole.

Solution:

Magnitude of electric field at $P$ due to $-\mathrm{q}$ :

$\left|\vec{E}_{-q}\right|=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{y^{2}+a^{2}}$

$\bar{E}_{-q}$ is directed from P to A. It has vertical component directed in the -y direction, and a horizontal component directed in the $-x$ direction.

Magnitude of $E$ at $P$ due to $+q$ :

$\left|\vec{E}_{+q}\right|=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{y^{2}+a^{2}}$

$\vec{E}_{+q}$ is directed from B to P. It has vertical component directed in the $+\mathrm{y}$ direction, and a horizontal component directed in the $-\mathrm{x}$ direction.

The net electric field at $\mathrm{P}$ is calculated by adding vectorially $\vec{E}_{-q}$ and$\vec{E}_{+q} .$

As their vertical components, $\bar{E}_{-q} \sin \theta$ and $\vec{E}_{+q} \sin \theta$, are equal in magnitude but opposite in direction. Hence they cancel out.

And, the horizontal components are both in the $-x$ direction. Hence they add up. Thus,

$\begin{aligned}&\vec{E}=\left(\left|\vec{E}_{-\uparrow}\right|+\left|\vec{E}_{+q}\right|\right) \cos \theta(-\hat{i}) \\&\vec{E}=\frac{2 q a}{4 \pi \varepsilon_{0}}\left(\frac{1}{y^{2}+a^{2}}\right) \cos \theta(-\hat{i})\end{aligned}$

From simple trigonometry, where we can substitute $\cos \theta$ in terms of a and $y$, we get

$\begin{gathered}\cos \theta=\frac{A O}{A P}=\frac{a}{\sqrt{y^{2}+a^{2}}} \\\vec{E}=\frac{2 q}{4 \pi \varepsilon_{0}}\left(\frac{1}{y^{2}+a^{2}}\right) \frac{a}{\sqrt{y^{2}+a^{2}}}(-\hat{i}) \\=\frac{2 q a}{4 \pi \varepsilon_{0}\left(y^{2}+a^{2}\right)^{\frac{3}{2}}}(-\hat{i})\end{gathered}$

For $y > > a$, neglect $a^{2}$ in the denominator.

Thus, we get

$\vec{E}=\frac{2qa}{4\pi \varepsilon_{0} y^{3}}(-\hat{i})$

But the dipole moment of an electric dipole is

$\vec{p}=2 q a \hat{i}$

Hence, for a point $P$ at a very large perpendicular distance from the axis,

$\vec{E}=\frac{-\bar{p}}{4 \pi \varepsilon_{0} y^{3}} \quad( for y > > a)$