As I was reading the solution of this problem the author gave the electric field in the point P as follows:

$$ \vec{E} = \sigma /(2 \epsilon ) [1-x/(x^2+R^2)^{1/2}]\ \hat \imath$$


  • $\sigma$ is the surface charge density on the disk
  • $x$ is the distance from the center of the disk to the point P
  • $R$ is the radius of the disk

Here the question comes: for $R \rightarrow 0$ with keeping $Q$ constant (the total charge) why is it that $E$ should go to a point charge?

According to my knowledge $E$ goes to zero as we calculate the limit of $E$ while $R$ tends to zero. Even if the absolute value of $x$ gives $-x$ we still get another constant.

  • 1
    $\begingroup$ Did you mean $R \rightarrow \infty$? $\endgroup$
    – Jeff
    Apr 10, 2016 at 22:00
  • 1
    $\begingroup$ Is this the source for the problem?: web.mit.edu/8.02t/www/materials/InClass/IC_Sol_W02D1_4.pdf $\endgroup$ Apr 11, 2016 at 1:19
  • $\begingroup$ @sagardipak has a complete solution, but his comments can be expanded. Any finite object will "look like" a point source if you get far enough away, assuming that you have real measuring equipment, with real limitations on precision. $\endgroup$
    – garyp
    Apr 11, 2016 at 1:57
  • $\begingroup$ @Jeff i mean R->0 $\endgroup$ Apr 11, 2016 at 5:56
  • $\begingroup$ exactly that's the source @Adpiavoc $\endgroup$ Apr 11, 2016 at 5:59

2 Answers 2


First, I think a little intuition could help. If you imagine a disk with a charge $Q$ get smaller, but all the while keeping the charge Q intact, shouldn't it geometrically approach a point with charge Q? So, in theory, we should expect the field to approach that due to a point charge.

Now, intuition aside, let's go to the mathematics. While the factor $\left[ 1-x/(x^2 + R^2)^{1/2} \right]$ does go to zero as $R\to0$, the charge density $\sigma = Q/(\pi R^2)$ goes to infinity. This is the source of your problem, because you end up with an indeterminate form $\infty \times 0$ while calculating the limit, and not 0.

To evaluate the limit, you could use l'Hôpital's rule, after rewriting your formula as an appropriate fraction (and substituting $\sigma$ for its expression in terms of $R$).

As a bonus, a quick way to do this would be to use this handy approximation, which works for small $y$ values: $$ (1+y)^n \approx 1+ny. $$

You can use this if you factor $x^2$ from the square root: $$ \frac{x}{ \left(x^2 + R^2\right)^{1/2} } = \frac{x}{ \lvert x \rvert \left(1 + (R/x)^2\right)^{1/2} } = \frac{x}{\lvert x \rvert}\left(1+\left(R/x\right)^2 \right)^{-1/2}\approx\frac{x}{\lvert x \rvert} \left( 1 - \frac{R^2}{2x^2}\right). $$

Here I used $y=R/x$ and $n=-1/2$. If $R$ goes to 0, then $y$ goes to 0, and this approximation gets better. Replacing in the expression you have given, we end up with $$ \vec E = \frac{Q}{4 \pi \epsilon x^2} \frac{x \hat \imath}{\lvert x \rvert} = \frac{Q}{4 \pi \epsilon x^2} \frac{\vec x}{\lvert x \rvert}, $$ which is the expression for a field due to a point charge. (Notice that the term $\vec x / \lvert x \rvert$ only gives you the direction of the field, but doesn't change its magnitude.)

Edit: if you try to do the calculations for $x<0$ you'll end up in trouble. The actual formula for the electric field should be $$ \vec E = \frac{\sigma}{2\epsilon}\left[ \frac{x}{\lvert x \rvert} - \frac{x}{\left(x^2 + R^2 \right)^{1/2}} \right] \hat \imath, $$ which you can see if you follow the derivation of the equation.

  • $\begingroup$ till now it's good. but you said if i try to do calculations for x<0 i'll end up in trouble. that's right i did so. i found E =(4Qx^2-QR^2)/4 \pi \epsilon x^2R^2 so the limit here (R->0) doesn't exist right? what does this mean physically? is it for Q<0 we can't find E at the interior of disk?! $\endgroup$ Apr 14, 2016 at 21:51
  • $\begingroup$ Yes, in that case the limit will be infinity. There isn't any interpretation -- the formula that you were given just isn't right for those cases. You can check this if you consider the following. For x>0, the field is positive, so it "runs away" from the disk. But everything is symmetric about the plane containing the disk. So, for the other side of the plane, you should get the same value for the field but with a minus sign, so the field would be running away in the opposite direction. If you change x with -x in the formula that I gave in the end, that's exactly what you obtain. $\endgroup$
    – sagardipak
    Apr 14, 2016 at 23:23

I think the mystery comes from the fact that as $R \rightarrow 0$ you will find that $\sigma$, the area charge density, will become infinitely great. The two effects "cancel out" such that the resulting electric field is that of a single point charge.

Note that $dq=\sigma dA$, and because we are preserving $Q$, as $R$ shrinks, the charge on the disk will become more and more compactly assorted. Namely, $dq$ will grow larger for the given strip of area. Consequently $\sigma$ will grow.

In other words, (roughly) think of $\sigma=\frac{dq}{dA}$, where we are decreasing $dA$ while keeping the charge dq constant. Consequently $\sigma$ increases.

  • $\begingroup$ When the surface becomes a point (on tending the radius to zero), the surface charge density reduces to a point charge. But you may misunderstand that as radius tends to zero, the surface charge density blows up. But that's misleading. We have no surface here. So there is only a point and there is no need to think about a surface charge. there is only a point charge now. It's similar to the case of observing the surface charge from a far off distance. In both cases, the electric field reduces to that of a point charge. $\endgroup$
    – UKH
    Apr 11, 2016 at 15:23
  • $\begingroup$ @Unnikrishnan.K.H Right, but the surface disappears in a limiting manner, hence the blowing up of the density. We don't immediately become a point charge but it becomes one as the surface area disappears. I'll revisit the language in my answer as soon as I get to a computer... $\endgroup$ Apr 11, 2016 at 19:02
  • $\begingroup$ But is it necessary to speak a point charge as something with infinite charge density.....Point charge is a structure less entity $\endgroup$
    – UKH
    Apr 12, 2016 at 11:21
  • $\begingroup$ Mathematically it would make sense for a point charge to have an "infinite" charge density-- just in the same way that division by zero may analytically seem to go to infinity. I agree though completely, that it is pointless to talk about the surface charge density in a physically meaningful way; as you say point charges are somewhat structure less. It is just a manifestation of the formula; the resulting electric field is that of a point charge. Perhaps I should clarify that the changing surface density is a purely mathematical object. $\endgroup$ Apr 12, 2016 at 11:30

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