The electric field of radius $R$ and a uniform positive surface charge density $\sigma$ at a distance $x$ from its center is given as $$E = \frac{\sigma}{2 \epsilon_0}\left( 1 - \frac{1}{\left(\frac{R^2}{x^2}\right) + 1}\right).$$

I am asked to show that for $x\gg R$, that $E = \frac{Q}{4\pi\epsilon x^2}$.

This is what I've done (but it's wrong):

So since $\sigma$ is the charge density of the disk, $\sigma = \frac{Q}{\pi R^2}$. Substituting this, we get $$E = \frac{Q}{2\pi R^2 \epsilon_0}\left( 1 - \frac{1}{\left(\frac{R^2}{x^2}\right) + 1}\right).$$ Further, as $x \gg R$, then the $\frac{R^2}{x^2}$ term evaluates to 0, so $E$ is therefore 0 at these conditions.

This is wrong, so can someone please explain why the answer is $E = \frac{Q}{4\pi\epsilon x^2}$?

  • $\begingroup$ $1/(u+1) \sim 1 - u $ for small $u$ $\endgroup$
    – Andrea
    Jun 25, 2016 at 19:51
  • $\begingroup$ Sum the two terms in the parentheses and observe that when $x$ is large the denominator will be $\simeq 1$ $\endgroup$
    – valerio
    Jun 25, 2016 at 19:58
  • $\begingroup$ There is a minor mistake in your question: The electric field should be $$E = \frac{\sigma}{2 \epsilon_0}\left( 1 - \frac{1}{\sqrt{\frac{R^2}{x^2}+1}}\right).$$ This gives the extra factor of $\frac{1}{2}$ noted by A.A $\endgroup$
    – jim
    Jun 26, 2016 at 19:32

2 Answers 2


You are not completely mistaken. What you have to do to get the desired formula is a Taylor expansion of the term which contains $R/x$ and then consider the limit when $x \to \infty$.

Also be careful with notation, $\sigma=Q/ (\pi R^2)$.

As Andrea Di Biagio mentions in his comment, $1/(1-u) \approx 1 - u$ when $u$ is small. In your situation $u=R^2/x^2$. So you get:

$$E= \frac{Q}{2 \pi \varepsilon_0 R^2} \frac{R^2}{x^2}$$

There is a factor 2 missing but that might be an errata or maybe they define $\varepsilon = \varepsilon_0/2$.

  • $\begingroup$ Thanks! So do I use the Taylor series of $1/x$ about 0? $\endgroup$
    – lithium123
    Jun 25, 2016 at 18:10
  • $\begingroup$ @lithium123 Not exactly, take into account that when $x \to \infty$ the fraction tends to 1. So you can consider $z=(R\x +1)$ and then use the series of $1/z$ at 1. $\endgroup$
    – A. A.
    Jun 25, 2016 at 18:14
  • $\begingroup$ Sorry, I still don't quite understand. Would you mind writing it out for me? $\endgroup$
    – lithium123
    Jun 25, 2016 at 18:19
  • $\begingroup$ @lithium123 I have edited my answer, I hope it's clearer now. $\endgroup$
    – A. A.
    Jun 25, 2016 at 21:09

If you are asking for an explanation of the result, it is because the finite disk behaves as a point charge for x>>R. Or are you specifically interested in mathematically proving the result?


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