ab

Is EQ1 equal to EQ2 ? or not? According to Jackson, they are the same,
however I am finding they are not equal. What do you say?

EQ1 = ((v*a*(x^2 - a^2))/(4*Pi))*Integrate[1, {\[Phi]p, 0, 2*Pi}]*
Integrate[(-Sin[\[Theta]p])*((a^2 + x^2 - 2*a*x*Cos[\[Theta]p])^(-3/2) -
(a^2 + x^2 + 2*a*x*Cos[\[Theta]p])^(-3/2)), {\[Theta]p, Pi/2, 0}]

EQ2 = v*(1 - (z^2 - a^2)/(z*Sqrt[z^2 + a^2]))

symbio

That is Jackson's "Classical Electrodynamics", 3rd edition, page 66,
Equation 2.21 and 2.22.

dave

they should only be equal along the positive z axis according to the
simplifying assumption to get from 2.21 to 2.22... (and assuming that my 2nd
edition is the same as the 3rd in that area)