CBSE Class 12 · 2019 · ElectrostaticsShort Answer

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(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

p

at a point on its perpendicular bisector.

Solution

(a) The equipotentials are symmetric about the axis; the equatorial plane is at zero potential.
(b) Let P be on the perpendicular bisector at distance

r

from the centre O.
Each charge is at distance

r^{2} + a^{2}

from P, so each gives

E_{+} = E_{-} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2} + a ^{2}}

Resolve each field along and perpendicular to the axis.
The perpendicular components cancel; the axial components

E_{+} cos θ

add, with

cos θ = \frac{a}{r ^{2} + a ^{2}}

E = 2 E_{+} cos θ = \frac{1}{4 π ε _{0}} \frac{q ( 2 a )}{( r ^{2} + a ^{2} ) ^{3/2}} = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + a ^{2} ) ^{3/2}}

directed anti-parallel to

p

.
For

r ≫ a

E = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}

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CBSE Class 12 · 2019 · ElectrostaticsShort Answer

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