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 r

_{1}and r_{2}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

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/r

_{2}, not as 1/r, characteristic of the potential due to a single charge.
Useless explanation without a picture.

ReplyDeletesoo confusing

ReplyDeletesoo confusing

ReplyDeleteI 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.

ReplyDeletewhats the use of it as its not helpful as same is given in Pradeep's Physics.

ReplyDeletei came here for an easy explanation as its confusing there but no help was given by this article.

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