I have been looking at the electric field of a charged disk and have a question about the use of l'Hopital's rule for the limiting case of electric field at points along the axis $z\gg$ disc radius $R$.

$$E = \frac {q}{2\pi\epsilon R^2} \left(1 - \frac {z}{\sqrt{z^2+R^2}}\right)$$

I have applied l'Hôpital's rule in the limit of $R$ approaching zero, and see that the electric field approaches that of a point charge, as intuition suggests. HOWEVER, when I use l'Hôpital's rule in the limit that $z$ approaches infinity, I get a repeating loop of indeterminate forms that doesn't arrive at the point charge expression.

My question is does this difference in results using l'Hopital's rule have any physical or mathematical significance?

  • $\begingroup$ But if you want $z\to\infty $ and $R\to 0$ simultaneously you can not use L'Hôpital's rule $\endgroup$ – Miguel Oct 8 '17 at 21:30
  • $\begingroup$ Thanks, I was solving two different problems, one where I wrote the expression as $\frac{0}{0}$ and the other where I wrote it $\frac{\infty}{\infty}$. I see how the $\frac{0}{0}$ converges to the structure of a point charge. $\endgroup$ – lamplamp Oct 8 '17 at 21:42
  • $\begingroup$ @lamplamp I have written out a solution using L'Hôpital's rule. $\endgroup$ – Farcher Oct 10 '17 at 16:17

To use L'Hôpital you either have to solve a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ kind of limit. I'll rewrite your expression to better show if this is the case: $$E = \frac{q}{2 \pi \epsilon}\frac{\sqrt{z^2+R^2}-z}{R^2 \sqrt{z^2+R^2}}$$ (I just calculated the common denominator and separated the constants from the variables.)

As $R$ approaches 0, we can see that both numerator and denominator go toward 0, so we can use L'Hôpital.

Conversely, as z approaches infinity, we find $\frac{\infty - \infty}{\infty}$, so we cannot use L'Hôpital in this case. We first have to solve the $\infty - \infty$ indeterminate form. The easiest way (or the standard trick, if you prefer) is to multiply numerator and denominator by $\sqrt{z^2+R^2}+z$, so that the formula becomes: $$E = \frac{q}{2 \pi \epsilon}\frac{R^2}{R^2 \sqrt{z^2+R^2}(\sqrt{z^2+R^2}+z)}$$ At this point $z$ disappears from the numerator and thus the indeterminate form is no more indeterminate and we can easily say that the limit goes to 0, as does the field of a point charge.

  • $\begingroup$ Thanks for your reply. I'm seeing two distinct parts to my question: $\endgroup$ – lamplamp Oct 8 '17 at 21:06
  • $\begingroup$ 1) Should L'hopital's rule, when applied successfully, yield only a field diminishing to zero at infinity, OR should it yield an expression diminishing to infinity with the particular mathematical structure of a point charge $\frac{q}{(r^2) 4 \pi \epsilon}$ When I used L'hopital's rule for limit r approaches zero, the expression yielded the form of a point charge, but now I'm wondering if this was just a coincidence, and that the best I could expect was any expression with the limit of zero, but not necessarily a point charge formulation $\endgroup$ – lamplamp Oct 8 '17 at 21:16
  • $\begingroup$ 2) I looked at the expression as a composite, and used l'Hopital's rule only on the portion that yielded $\frac{\infty}{\infty} $: $\frac{z}{((z^2 + R^2))^1/2)}$ and this gave a looping form $\endgroup$ – lamplamp Oct 8 '17 at 21:26
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    $\begingroup$ 1) There is no guarantee that, if you calculate a limit, you will always obtain a formula instead of a single number. I'd say you were lucky with $R \to 0$. $\endgroup$ – GRB Oct 9 '17 at 8:20
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    $\begingroup$ 2) Yes, in this case there is a looping form, but it's sufficient to solve the problem. Since $\lim_{z \to \infty} \frac{z}{\sqrt{z^2+R^2}} = \lim_{z \to \infty} \frac{\sqrt{z^2+R^2}}{z}$, both limits should be equal to 1. $\endgroup$ – GRB Oct 9 '17 at 8:22

Update as the result of a comment from @garyp this time using L'Hôpital's rule. $$E = \frac{q}{2 \pi \epsilon}\frac{(z^2+R^2)^{\frac 12}-z}{R^2 (z^2+R^2)^{\frac 12}}$$

Now differentiate twice with respect to $R$ the numerator and the denominator individually to get something like

$$ \frac {(z^2+R^2)^{-\frac 12} + R(.........)}{2(z^2+R^2)^{\frac 12} +R(.........)} $$

which has the limit $\dfrac{1}{2z^2}$ as $R$ tends to zero and gives the desired equation for the electric field due to a point charge.

$\left(1 - \dfrac {z}{\sqrt{z^2+R^2}}\right)$

Divide top and bottom of the fraction by $z$ and expand using the binomial theorem as far as the second term and for the electric field you will find that the $R^2$ cancels leaving the point charge field in terms of $z$.

  • $\begingroup$ True, and a good way to find the limit. But not an answer to the question. $\endgroup$ – garyp Oct 8 '17 at 20:18
  • $\begingroup$ @garyp Thanks for your comment which has resulted in an update to my answer. $\endgroup$ – Farcher Oct 10 '17 at 10:25

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