Question

Define electric dipole moment. Derive an expression for the electric field intensity at any point along the equatorial line of a dipole.

Solution

Electric dipole moment is defined as the product of either charge and the length of the electric dipole.

Electric field on equitorial line of an electric dipole:

Consider an electric dipole consisting of charges -q and +q seperated by a distance 2a. Let P be a point on equitorial line of the dipole at a distance 'r' from the centre of the dipole as shown in fig.

Let, $\overset{⇀}{E_{A}} and {\overset{⇀}{E}}_{B} be the Electric field at point P due to charge - q at point A and + q at point B . Then, resultant Electric field at point P is given by$

To find the resultant electric field intensity due to the dipole at point P, we will represent ${\overset{⇀}{E}}_{A} a n d {\overset{⇀}{E}}_{B}$ by the two adjacent sides PL and PM of a parallelogram. Then, diagonal PN represents the resultant Electric field due to the dipole acting along Px'. The resultant electric field can also be found using the triangle law of addition of vectors.
In $∆ P A B, \overset{⇀}{P A}, \overset{⇀}{B P} a n d \overset{⇀}{B A} r e p r e s e n t {\overset{⇀}{E}}_{A}, {\overset{⇀}{E}}_{B} a n d \overset{⇀}{E} r e s p e c t i v e l y . ∴ B y t r i a n g l e l a w o f a d d i t i o n o f v e c t o r s, \frac{\overset{⇀}{E}}{B A} = \frac{{\overset{⇀}{E}}_{A}}{P A} = \frac{|{\overset{⇀}{E}}_{B}|}{B P} \Rightarrow \overset{⇀}{E} = {\overset{⇀}{E}}_{A} \times \frac{B A}{P A} = \frac{1}{4 π ε_{\circ}} . \frac{q}{(r^{2} + a^{2})} \times \frac{2 a}{\sqrt{r^{2}} + a^{2}} = \frac{1}{4 π ε_{\circ}} . \frac{q (2 a)}{(r^{2} + a^{2})^{\frac{3}{2}}} N o w, q (2 a) = p i s t h e m a g n i t u d e o f t h e e l e c t r i c d i p o l e m o m e n t o f t h e d i p o l e . ∴ \overset{⇀}{E} = \frac{1}{4 π ε_{\circ}} \cdot \frac{\overset{⇀}{p}}{(r^{2} + a^{2})^{\frac{3}{2}}} T h e d i r e c t i o n o f e l e c t r i c f i e l d a t a p o i n t o n t h e e q u i t o r i a l l i n e o f t h e d i p o l e i s f r o m + q t o - q (i n a d i r e c t i o n o p p o s i t e t o t h e d i r e c t i o n o f e l e c t r i c d i p o l e m o m e n t o f d i p o l e) ∴ \overset{⇀}{E} = - \frac{1}{4 π ε_{\circ}} . \frac{\overset{⇀}{p}}{(r^{2} + a^{2})^{\frac{3}{2}}}$

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Electric Charges And Fields

Define electric dipole moment. Derive an expression for the electric field intensity at any point along the equatorial line of a dipole.

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

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For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

Define electric dipole moment. Derive an expression for the electric field intensity at any point along the equatorial line of a dipole.

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?b) Explain why two field lines never cross each other at any point?

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?