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 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 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 at 10:42

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