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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
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Deduce an expression for the electric field at a point on the equatorial plane of an electric dipole of length $2a$.

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Electric dipole moment is a vector quantity.
Consider an electric dipole AB consists of two charges +q and q separated by a distance 2a. We have to find electric field at point P on Equipotential line separated by a distance r.
Electric field at point P, due to charge + q.
$E_1=\large\frac{1}{4 \pi \in_0 } \frac{q}{\bigg[ \sqrt {(r^2 +a^2)}\bigg]^2}$
$\qquad=\large\frac{1}{4 \pi \in_0 } \frac{q}{(r^2 +a^2)}$ along AP
Due to charge -q
$E_2=\large\frac{1}{4 \pi \in_0 } \frac{q}{(r^2 +a^2)}$ along PB
On resolving $E_A$ and $E_B$ into rectangular components, we get resultant electric field at point P,
$E= E_A \cos \theta+ E_B \cos \theta$
$\quad= \large\frac{1}{4 \pi \in _0} .\frac{q}{(r^2+a^2)} \times \frac{a}{\sqrt {(r^2+l^2)}}$
$\quad= 2 \times \large\frac{1}{4\pi \in_0} . \frac{q}{(r^2+a^2)} \times \frac{a}{\sqrt {(r^2+l^2)}}$
$\qquad = \large\frac{1}{4 \pi \in_0}. \frac{q.2a}{(r^2+a^2)}$
But $q_1 .2a =p$
$E= \large\frac{1}{4 \pi \in_0} . \frac{ p}{(r^2+a^2)^{3/2}}$
if $r > > a $ then
$\large\frac{1}{4\pi \in_0} .\frac{p}{r^3}$
Hence A is the correct answer.
answered Jun 2, 2014 by meena.p
 

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