com www.GOALias.blogspot.com

www.GOALias.blogspot.comPhysicsthe field falls off, at large distance, not as1/r 2 (typical of field due to a single charge)but as 1/r 3 . We, now, determine the electricpotential due to a dipole and contrast itwith the potential due to a single charge.As before, we take the origin at thecentre of the dipole. Now we know that theelectric field obeys the superpositionprinciple. Since potential is related to thework done by the field, electrostaticpotential also follows the superpositionprinciple. Thus, the potential due to thedipole is the sum of potentials due to thecharges q and –qFIGURE 2.5 Quantities involved in the calculationof potential due to a dipole.Now, by geometry,r = r + a − ar cosθ2 2 21 21 ⎛qq ⎞V = −4πε⎜⎝rr ⎟⎠0 1 2(2.9)where r 1and r 2are the distances of thepoint P from q and –q, respectively.r = r + a + ar cosθ (2.10)2 2 22 2We take r much greater than a ( r >> a ) and retain terms only uptothe first order in a/r22 2 2acosθa1= r 1 − +2r⎛⎜⎝2 ⎛ 2acosθ⎞≅r⎜1−⎝⎟r ⎠Similarly,2 2 ⎛ 2acosθ⎞r2≅ r ⎜1+⎝⎟r ⎠rr⎞⎟⎠(2.11)(2.12)Using the Binomial theorem and retaining terms upto the first orderin a/r ; we obtain,561− 1/21 1 ⎛ 2acosθ⎞ 1⎛ a ⎞≅ ⎜1− ≅ 1+cosθr r ⎝⎟r ⎠⎜ ⎟r ⎝ r ⎠2− 1/21 1⎛ 2acosθ⎞ 1 ⎛ a ⎞≅ ⎜1+ ≅ 1−cosθr r ⎝⎟r ⎠⎜ ⎟r ⎝ r ⎠Using Eqs. (2.9) and (2.13) and p = 2qa, we getq 2acosθpcosθV = =4πεr 4πεrNow, p cos θ =2 20 0pr gˆ[2.13(a)][2.13(b)](2.14)