CBSE Class 12 · 2017 · Delhi · ElectrostaticsShort Answer

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(i) Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
(ii) Depict the equipotential surfaces due to an electric dipole.

Solution

(i) Let P be an axial point at distance

r

from the centre O, so it is at

(r - a)

from

+ q

and

(r + a)

from

- q

.
Potential is a scalar:

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

V = \frac{1}{4 π ε _{0}} \frac{p}{r ^{2} - a ^{2}}

For

r ≫ a

V = \frac{1}{4 π ε _{0}} \frac{p}{r ^{2}}

(ii) The dipole's equipotential surfaces are symmetric about its axis — positive near

+ q

, negative near

- q

— and the plane bisecting the axis perpendicularly (the equatorial plane) is the zero-potential surface.

CBSE Class 12 Board · 2017 · Delhi · Subjective

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