# Electric field of a disc

1. Oct 25, 2015

### whatisreality

1. The problem statement, all variables and given/known data
What is the field of a disc radius a, uniformly charged with surface charge density σ, for x>>a?

2. Relevant equations

3. The attempt at a solution
I've worked out the equation for the field of a disc:
$E =$ $\frac{\sigma x}{2 \epsilon}$ $\frac{\sqrt{a^2+x^2}-x}{x \sqrt{a^2+x^2}}$

I'm also given a hint, which is that the answer shouldn't be zero. As far as I can tell, if x>>a, this definitely becomes zero! I have looked up my answer for E and it is right. To work out E for x>>a, I essentially ignored all a as irrelevant - is that wrong?

$E =$ $\frac{\sigma x}{2 \epsilon}$ $\frac{\sqrt{x^2}-x}{x \sqrt{x^2}}$ is most definitely zero!

2. Oct 25, 2015

### TSny

For a finite value of $x$ that is much greater than $a$, $E$ will be small but it won't be zero. You have "over approximated". In this type of problem, you want to keep the lowest order nonzero approximation.

Instead of approximating $\sqrt{x^2+a^2}$ in the numerator as $\sqrt{x^2}$, find a more accurate approximation in terms of the small quantity $a/x$.
Hint: https://en.wikipedia.org/wiki/Binomial_approximation

You might find it easier to first simplify your expression for E by cancelling some $x$'s and splitting the fraction involving the square roots into the difference of two fractions. You will then have only one square root to worry about.

3. Oct 28, 2015

### whatisreality

Simplifying the expression gives

$E = \frac{\sigma}{2\epsilon}(1 - \frac{1}{\sqrt{\frac{a^2}{x^2}+1}})$
And then use a binomial approximation for $\sqrt{\frac{a^2}{x^2}+1}$.
Thank you, gave the right answer :)