# (a) Draw the equipotential surfaces due to an electric dipole. (b) Derive an expression for the electric field due to a dipole of dipole moment $\vec{P}$ at a point on its perpendicular bisector.

a)

(b)

$E=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{r^{2}}$

Hence, $r=\sqrt{a^{2}+x^{2}}$

electric due to positive charge

$\\E_{+}=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{(a^{2}+x^{2})}$ and

electric due to negative charge

$E_{-}=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{a^{2}+x^{2}}$

From the figure, the sine components get mutually cancelled and cosine components can be added.

So,

$E_{net}=E_{+}\cos \theta +E_{-}\cos \theta$

Since due to the same magnitude

$E_{net}=2E_{+}\cos \theta \; \; \; \; \; \; \; (1)$

Now, $\cos \theta =\frac{a}{\sqrt{a^{^{2}}+x^{2}}}$

then,

$E_{net}=2\frac{1}{4\pi \varepsilon _{o}}\frac{q}{a^{2}+x^{2}}\frac{a}{\sqrt{a^{2}+x^{2}}}$

$E_{net}=\frac{1}{4\pi \varepsilon _{o}}\frac{2qa}{(a^{2}+x^{2})^{3/2}}$

The dipole moment, $\vec{P}=2qa$

Now, if $x\gg a$

$E_{net}=\frac{1}{4\pi \varepsilon _{o}}\frac{P}{x^{3}}$

## Related Chapters

### Preparation Products

##### Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-