My TB gives the formulae for electric field and potential due to a dipole as:

$$|E| = \dfrac{k p}{r^3} \sqrt{1+ 3\cos^2θ}$$

$$|V| = \dfrac{k p}{r^2} \cosθ$$

Why are these expressions are not satisfying $E = \dfrac{-dV}{dr}$?

Edit: Is the following derivation correct?

Let a test charge be placed at origin and the dipole be at a distance x on the x-axis.

Equatorial field due to x component of dipole $E_x = \dfrac{2kp}{x^3} \cosθ$

Axial field due to y component of dipole $E_y = \dfrac{kp}{x^3} \sinθ$

If the particle is moved from origin to infinity (to the left) along x-axis, the potential drop will be only due to $E_x$:

$$|V| = \int^\infty_x E_x dx = \dfrac{kp \cosθ}{x^2} $$

And, $$|E| = \sqrt{E_x ^ 2 + E_y ^ 2} = \dfrac{k p}{x^3} \sqrt{1+ 3\cos^2θ}$$

  • $\begingroup$ Remember that $\mathbf E$ is a vector $(-dV/dx, -dV/dy)$ so $|E| = \sqrt{(-dV/dx)^2 + (-dV/dy)^2}$. It is not as simple as $|E| = d/dr(|V|)$. $\endgroup$ May 14, 2022 at 8:02
  • $\begingroup$ Exactly. The formula you are expecting to find is valid for a spherically symmetric field, which the dipole field is not! $\endgroup$
    – kricheli
    May 14, 2022 at 8:39

1 Answer 1


Remember that $\mathbf E$ is a vector to be calculated with the gradient operator $$\mathbf E=-\nabla V$$

If you want to do it in spherical coordinates ($r,\theta,\phi$), then you need to use the gradient operator in spherical coordinates. Doing this calculation with the given dipole potential $V=\frac{kp}{r^2}\cos\theta$, you will find that $\mathbf E$ has components in $r$ and $\theta$ direction (because $V$ depends on $r$ and $\theta$): $$\begin{align} E_r&=-\frac{\partial V}{\partial r}=\frac{2kp}{r^3}\cos\theta \\ E_\theta&=-\frac{1}{r}\frac{\partial V}{\partial \theta}=\frac{kp}{r^3}\sin\theta \end{align}$$

When you then calculate the magnitude of $\mathbf E$, you find $$|\mathbf E|=\sqrt{E_r^2+E_\theta^2} =\frac{kp}{r^3}\sqrt{1+3\cos^2\theta}$$ which is the formula from your textbook.

  • $\begingroup$ @Shub I thought your textbook would derive the dipole potential $V(r,\theta)$. You can find a derivation also here. $\endgroup$ May 14, 2022 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.