# On the $E$-field near the edge of a semi-infinite conducting plane

This document (page $$5$$ and $$6$$, it acutally starts from example $$3$$) shows that for a constant potential semi-infinite plane given by $$\,y=0\,$$ and $$\,x>0~$$the $$E$$-field may be written as:

$$\textbf{E}=\frac{\sqrt{2}}{4\sqrt{x^2+y^2}}\Bigg (\sqrt{\sqrt{x^2+y^2}-x},\frac{y}{ \sqrt{\sqrt{x^2+y^2}-x}}\Bigg )$$

Then the author went on to say the $$E$$-field just above the sheet and just to the right of its edge is $$\textbf{E}(\epsilon , \delta)\approx\Big (0, \frac{1}{2\sqrt{\epsilon}}\Big )$$

This result seems to condratic the analytical expression of $$\,\textbf{E}\,$$ above, for example, if we plug $$\epsilon$$ and $$\delta$$ in the first component, we 're essentially going to find the same expression but with $$\epsilon$$ and $$\delta$$ instead of $$x$$ and $$y$$.

Is it just an oversight from them or am I missing something?

It looks to me like this is a case where you can't freely interchange the order of the limits, and the claim holds only when $$y \ll x$$. Let $$r=\sqrt(x^2+y^2)$$ and $$\alpha=r-x$$. Then for $$y \ll x$$, $$\sqrt{\alpha}\sim yx^{-1/2}$$, which is small.
The author has taken the limit as $$\delta\to 0+$$, while keeping $$\epsilon$$ merely small.