Question

Derive expression for eletric field due to short dipole

Answer 1

Answer:

Consider an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q. We know that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between vector r and p. Further, the field falls off, at large distance, not as 1/r2 (typical of field due to a single charge) but as 1/r3. To determine the electric potential due to a dipole, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. The potential due to the dipole is the sum of potentials due to the charges q and –q is given by where r1 and r2 are the distances of the point P from q and –q, respectively.  Now, by geometry,  r12 = r2 + a2 - 2ar cosθ and r22 = r2 + a2 + 2ar cosθ We take r much greater than a ( r >>1) and retain terms only upto the first order in a/r  Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain,  where ˆr is the unit vector along the position vector ܱܲOP. The electric potential of a dipole is then given by when (r >> a) -------- (2)  Equation (1) is, as indicated, 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, Eq. (1) is, however, exact.  From Eq. (2), 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. Thus  (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.