1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $z$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.