# Electric field of a full disk - when $R \to 0$ - it's not equal to coulomb law

An MIT document states that the electric field of a full disk, when $R \to 0$, is similar to Coulomb's law

$$\mathbf E_{disk}=2\pi k_e\sigma\left[1-\frac{x}{\left(x^2+R^2\right)^{1/2}}\right]\hat{i}=\frac{\sigma}{2\varepsilon_0}\left[1-\frac{x}{\left(x^2+R^2\right)^{1/2}}\right]\hat{i}$$ Either version is fine, its just a different way of writing the constant. You should also check the limits: for $R\to0$ (but keep $Q$ constant!) it should go to a point charge. For $R\to\infty$ (infinite plane) it should be a constant.

Though, I don't think that it works that way, it is easily seen that when $R \to 0$, then $\mathbf E_{disk} = 0$.

Can somebody help me figure out how to arrive at the stated result - what am I missing to get the field of a point charge when the disk size goes to zero?

• But when $R \to 0$ you have no more charge. The charge on a surface equal to zero is $\sigma \cdot 0 = 0$. Commented Mar 12, 2015 at 18:32
• @Sofia: why the "but" ? I think similar to you :) I don't understand why did they write that. What are we missing here?
– Dor
Commented Mar 12, 2015 at 18:34
• Aaaa! I apologize, I didn't notice that you keep $Q$ constant when shrinking the radius. But if so, the formula is different, $\vec E = \frac {Q}{\pi R^2} \left[1 - \frac{x}{\sqrt {x^2 + R^2}}\right]$. So, when $R \to 0$ you get $\frac {0}{0}$. Commented Mar 12, 2015 at 18:40
• When you take the limit $R\rightarrow 0$ you cannot simply plug in $R = 0$. Commented Mar 12, 2015 at 18:45
• This is the setup for a good question but it doesn't actually ask anything, so I'm closing it. Dor, can you edit to make clear what you're asking? Once you do it'll be reopened. Commented Mar 12, 2015 at 18:46

Remember that we keep leading order terms. So for the second part of the expression in parentheses, as $R \rightarrow 0$, we don't just get 1. Using the taylor expansion, we get $$\frac{1}{\sqrt{1+\frac{R^2}{x^2}}}\Rightarrow 1 - \frac{1}{2}\frac{R^2}{x^2}+....$$ Plug this into the original equation while remembering $\sigma= \frac{Q}{\pi R^2}$ gives $$\vec{E}_{disc}= \frac{\sigma}{2 \epsilon_0}\left[\frac{R^2}{2x^2}\right] = \frac{Q}{4 \pi \epsilon_0 x^2}$$ which is exactly the field of a point charge that we want.
• The next taylor term is proportional to $R^4/x^4$. You could keep this term around and then you would add a term looking like $R^2/x^4$ in the final expression. But remember that we're taking the limit as $R$ goes to zero. If $R$ is small, say $10^{-10}$, then a term that scales like $1/x^2$ dominates the electric field compared to a term like $R^2/x^4$. So we just ignore the extra terms when we find 1 nonzero term, because we can show any extra terms should be small compared to the first term. Commented Mar 12, 2015 at 20:20