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I was calculating potential energy due to point charge at origin using relationship

$$U=\dfrac{1}{2}\epsilon_{0}\displaystyle\int \mathbf {\vec E.\vec E\ } \ d\tau$$

and it comes out to be infinity because $\mathbf {\vec E.\vec E\ }=|\vec E|^2 $ falls off as $\propto$ $\dfrac{1}{r^4}$ and volume $d\tau$ grows as $4\pi r^2 dr$ resulting in overall fall off as $\propto$ $\dfrac{dr}{r^2}$ and which in turn lead to blowing up of integral at $r=0$ and U approaching $\infty$

Question:

it is written in my textbook as it is that:

"it takes infinite energy to pack a point charge(like electrons, protons) into zero volume."

i want to know physical justification of this result

(not mathematical derivation like above )

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    $\begingroup$ Imagine that you have already packed a finite amount of charge into a point, say you are half-way through the total. The important thing is that you have a finite amount of charge so the electric field intensity is already proportional to $1/r^2$. Now how much work is needed to finish packing the remaining other half into that point? $\endgroup$
    – hyportnex
    Commented Aug 27, 2018 at 21:32

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In the limit $ r \to 0$, the energy is indeed infinite. But the idea is that, you cannot have $ r \to 0$ in any 'classical' calculation. Note that when we talk of quantum mechanics, no charge can be point-like, because it will violate uncertainty principle. So, in this case, the resolution is that there is a quantity, known as "classical electron radius" given by $ r_{0} = \frac{e^2}{m_{e}c^2}$, it has the numerical value $2.82 \times 10^{−13}$ cm, about one one-hundred-thousandth of the diameter of an atom. So, one still has a finite value of U.

I would highly recommend reading Chapter 28 Volume II of Feynman Lectures to understand this. No one can answer this better than that book in my opinion at this level without invoking any ideas of quantum theory. http://www.feynmanlectures.caltech.edu/II_28.html

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  • $\begingroup$ thanks ...... got it that $r$ isn't practically zero (but theoratically) but in range of femto metres and that is why energy is very large . $\endgroup$
    – user182868
    Commented Aug 27, 2018 at 21:55
  • $\begingroup$ @veeresh pandey The problem with this answer is that the upper limit of the electron radius is believed to be much smaller than this value. I do not have a way out. $\endgroup$
    – my2cts
    Commented Aug 27, 2018 at 22:56