Question

an electric dipole of dipole moment p is placed at origin along the x-axis.the angle made by

Answer 1

Field of an electric dipole We first calculate the potential and then the field: V = q 4π0 1 r+ − 1 r− ! (1) where r± are the distances from the +ve(-ve) charge to the point r. Now r 2 ∓ = |r ± a/2| 2 = (r 2 ± a · r + a 2 /4) = r 2 1 ± a r cos θ + a 2 4r 2 ! Now consider the “far field limit” r a 1 r± = 1 r 1 ∓ a r cos θ + a 2 4r 2 !−1/2 ' 1 r 1 ± a 2r cos θ + O((a/r) 2 ) where O((a/r) 2 ) indicates terms proportional to (a/r) 2 or higher powers Thus we obtain in the far-field limit V = qa cos θ 4π0r 2 = p · ˆr 4π0r 2 (2) One can check that away from the charges this is a solution of Laplace’s equation (see tut) 1 The components of the electric field E= −∇V are simplest in spherical polar coordinates: Er = − ∂V ∂r = 2p cos θ 4π0r 3 Eθ = − 1 r ∂V ∂θ = p sin θ 4π0r 3 (3) To get a co-ordinate free form of the electric field we can use (see tutorial sheet 1) E = −∇V = − 1 4π0 ∇ p · ˆr r 2 ! = 1 4π0 3(p · r)r r 5 − p r 3 ! (4) The important point to note is that a dipole field is 1/r3 , whereas a point charge field is 1/r2 N.B. The above ‘far-field’ limit r a can also be presented as an ideal dipole which is the limit a → 0, q → ∞ but p finite. The ideal dipole is a useful approximation to the ‘physical dipole’ for which the potential is given by (1) and E = −∇V = q 4π0 " r − a/2 |r − a/2| 3 − r + a/2 |r + a/2| 3 # (5)