Q.Calculate the electric field due to a dipole on its axial line and on its equatorial plane.

Step 1. Setup. A dipole $-q$ at $x=-a$, $+q$ at $x=+a$, midpoint O at the origin, dipole moment $p=q(2a)$ along $+x$.

Step 2. Axial line. For point C at $(r,0)$ with $r\gg a$: field due to $+q$ is $k q/(r-a)^2$ (away from dipole), field due to $-q$ is $kq/(r+a)^2$ (toward $-q$, same direction at C). Adding: $E_{\text{axial}}=kq\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right]=kq\dfrac{4ar}{(r^2-a^2)^2}=\dfrac{2kpr}{(r^2-a^2)^2}$, using $p=2aq$.

Step 3. Axial approximation. For $r\gg a$, $(r^2-a^2)^2\approx r^4$, giving $E_{\text{axial}}\approx 2kp/r^3$, directed parallel to $p$.

Step 4. Equatorial line. For point P at $(0,r)$, both charges are at distance $\sqrt{r^2+a^2}$; the components perpendicular to the dipole axis cancel by symmetry, and the components along the axis (both pointing anti-parallel to $p$) add: $E_{\text{equatorial}}=\dfrac{kp}{(r^2+a^2)^{3/2}}$. …

Superpose the two point-charge fields separately for a point on the axial line and a point on the equatorial …

• Forgetting that the equatorial field points opposite to p while the axial field points parallel to p -- both magnitudes are easy to get right while the directions are reversed. …