# Field and potential due to dipole

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

• 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 $|E| = d/dr(|V|)$. May 14, 2022 at 8:02
• Exactly. The formula you are expecting to find is valid for a spherically symmetric field, which the dipole field is not! May 14, 2022 at 8:39

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