Name: Electric Charges and Electric Fields: A Review
Uploaded: 2019-07-19
Duration: 6 min 41 s
Description: In this video, we are going to review the electric charges, the law of charge, conservation of charges and electric field in detail. Watch this to learn more.

Question

Electric dipole moment is vector or scalar, if it is vector? Show its direction and obtain an expression for an electric field due to an electric dipole at a point on the axis of the electric dipole.

Accepted Answer

Electric dipole moment is a vector physical quantity expressed as $\vec{P} = q \times \vec{2 l}$

Its direction is from $- q$ Charge to $+ q$ charge

For points on axial line

The axial line of a dipole is the line passing through the positive and negative charges of the electric dipole.

Consider a system of charges (-q and +q) separated by a distance 2a. Let ‘P’ be any point on an axis where the field intensity is to be determined.
Electric field at P (EB) due to +q

$E_{B} = \frac{1}{4 {πε}_{0}} \frac{q}{(BP)^{2}} along BP = \frac{1}{4 {πε}_{0}} \frac{q}{(r - a)^{2}}$

Electric field at P due to -q (EA)

$E_{A} = \frac{1}{4 {πε}_{0}} \frac{q}{(AP)^{2}} along PA = \frac{1}{4 {πε}_{0}} \frac{q}{(r + a)^{2}}$

Net field at P is given by

$E_{p} = E_{B} - E_{A} = \frac{1}{4 {πε}_{0}} [\frac{q}{(r - a)^{2}} - \frac{q}{(r + a)^{2}}]$

Simplifying, we get

$E_{p} = \frac{q}{4 {πε}_{0}} \frac{4 ar}{(r^{2} - a^{2})^{2}} E_{p} = \frac{2 qa}{4 {πε}_{0}} \frac{2 r}{(r^{2} - a^{2})^{2}} (2 aq = p, K = \frac{1}{4 {πε}_{0}}) or E_{p} = \frac{2 kpr}{(r^{2} - a^{2})^{2}} along BP$
As a special case :

If $2 a ≪ r, E_{p} = \frac{2 kp}{r^{3}} along BP$

Electric dipole moment is vector or scalar, if it is vector? Show its direction and obtain an expression for an electric field due to an electric dipole at a point on the axis of the electric dipole.

Important Questions on Electric Charges and Fields

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