Monday, November 26, 2012

Potential Due To An Electric Dipole

Take the origin at the centre of the dipole. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q
where r1 and r2 are the distances of the point P from q and –q, respectively.

Now, by geometry,
Take r much greater than a ( r >> a ) and retain terms only up to the first order in a/r
 Using the Binomial theorem and retaining terms up to the first order in a/r; obtain,

where rˆ is the unit vector along the position vector OP. The electric potential of a dipole is then given by
Equation is approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin,

From Eq. potential on the dipole axis (θ= 0, π) is given by

(Positive sign for θ= 0, negative sign for θ= π.) The potential in the equatorial plane (θ= π/2) is zero.

(i)                 The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p.

(ii)               The electric dipole potential falls off, at large distance, as 1/r2, not as 1/r, characteristic of the potential due to a single charge.


  1. Useless explanation without a picture.

  2. I have a question, i don't understand the part where we get those two equations from geometry for r1(square) and r2(square). Please explain that using appropriate figure.

  3. whats the use of it as its not helpful as same is given in Pradeep's Physics.
    i came here for an easy explanation as its confusing there but no help was given by this article.

  4. I understood...good explanation

  5. U can make figure by yourselves from equations r1 and r2