CBSE Class 12 · 2007 · ElectrostaticsShort Answer

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Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric potential of a dipole at a point as compared to that due to a single charge.

Solution

Let P be a point on the axis at distance

r

from the centre O.
Then P is at

(r - a)

from

+ q

and

(r + a)

from

- q

.
Potential is a scalar, so the net potential is the algebraic sum:

V = \frac{1}{4 π ε _{0}} \frac{q}{( r - a )} + \frac{1}{4 π ε _{0}} \frac{( - q )}{( r + a )}

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

Taking the common denominator,

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

For

r ≫ a

r^{2} - a^{2} \approx r^{2}

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

Contrast: the dipole potential falls as

1/ r^{2}

(faster than the

1/ r

of a single charge) and depends on direction.

CBSE Class 12 Board · 2007 · Subjective

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CBSE Class 12 · 2007 · ElectrostaticsShort Answer

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