(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.

 

Answers (1)

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}}

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