In worked example 4.1 of Intermolecular and Surface Forces by Jacob Israelachvili, he is calculating the electric field on a test charge due to the dipole shown in the picture.enter image description here

He assumes $r\gg l$ and: $$AB\approx r-\frac{1}{2} l\cos{\theta}$$ $$AC\approx r+\frac{1}{2} l\cos{\theta}$$ Using this, he writes that the magnitude of the electric field on A due to the negative charge at B is: $$E_{-}=q / 4 \pi \varepsilon_{0} \cdot A B^{2} \approx\left(q / 4 \pi \varepsilon_{0} r^{2}\right)\left(1+\frac{l}{r} \cos \theta\right)$$ But I don't see where that comes from. Why is the squared distance in the numerator? and what approximation is used on the right-hand side?


1 Answer 1


This is the binomial approximation, where you can approximate $$(1 + x)^\alpha \approx 1 + \alpha x, \quad \text{(if $|\alpha x|\ll 1$)}.$$

The squared distance is in the denominator, as you'd expect: $$E_- = \frac{q}{4\pi\epsilon_0 \cdot AB^2}.$$

If you substitute for $AB$, you can easily show that it's just $$E_- = \frac{q}{4\pi\epsilon_0 r^2} \frac{1}{(1 - \frac{l}{2r}\cos\theta)^2}.$$

We can now use the binomial approximation (with $\alpha = -2$ and $x = -l\cos\theta/2r$) on $$ \frac{1}{(1 - \frac{l}{2r}\cos\theta)^2} \approx \left(1 + (-2)\times \left(-\frac{l}{2r}\right)\cos\theta \right),$$

so that

$$\frac{q}{4\pi\epsilon_0 \cdot AB^2} \approx \frac{q}{4\pi\epsilon_0 r^2} \left(1 + \frac{l}{r}\cos\theta \right).$$

  • $\begingroup$ Thank you very much. $\endgroup$
    – Logi
    Jan 16, 2021 at 18:31

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