Electric Field on the surface of charged conducting spherical shell

[vela]

Perhaps you need to abandon the notion that a charge has to either be enclosed or not enclosed. It can be neither, just like 0 is neither positive nor negative.

 

[Delta2]

Perhaps you need to abandon the notion that a charge has to either be enclosed or not enclosed

Well this notion works well for all charge densities except the ones that involve Dirac delta functions.

 

[vela]

What bothers me now is that, if we accept the results of case III or the Griffiths argument presented by @vela in post #33, the implication is that one can do a delta function integral when the upper limit is the value where the argument of the delta function vanishes: $$E 4\pi R^2 =\frac{1}{\epsilon_0}\int_0^R \frac{Q}{4\pi R^2}\delta (r-R)4\pi r^2~dr=\frac{Q}{2}.$$Can one assert that this is true for the specific case of a charged conductor but not in general? Although Griffiths's argument is based on physical grounds, the Coulomb integral in case III is a mathematical argument.

The delta function can be defined more rigorously by
$$\int_{-\infty}^\infty f(x)\delta(x-a)\,dx = \lim_{n\to\infty} \int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx = f(a)$$ for a suitably chosen sequence of functions $\delta_n(x)$. (See for example https://dlmf.nist.gov/1.17 and https://mathworld.wolfram.com/DeltaSequence.html.) Most of the examples on the MathWorld page are symmetric about $x=0$, but symmetry isn't a requirement. For example,
$\delta_n(x) = n[u(x)-u(x-1/n)]$ is a perfectly valid delta sequence.

If we were to use the first example on the MathWorld page, we could say
\begin{align*}
\int_0^\infty f(x)\delta(x)\,dx &= \lim_{n\to\infty} \int_0^\infty f(x)\delta_n(x)\,dx \\
&= \lim_{n\to\infty} n \int_0^{1/2n} f(x)\,dx \\
&= \frac 12 f(0)
\end{align*} But if we used $\delta_n(x) = n[u(x)-u(x-1/n)]$ to evaluate the same integral, we'd get
\begin{align*}
\int_0^\infty f(x)\delta(x)\,dx &= \lim_{n\to\infty} \int_0^\infty f(x)\delta_n(x)\,dx \\
&= \lim_{n\to\infty} n \int_0^{1/n} f(x)\,dx \\
&= f(0)
\end{align*} When the argument of the delta function vanishes at one of the endpoints of integration, the value of the integral depends on how you define the delta function. In this problem, the factor of 1/2 makes physical sense.

 

[alan123hk]

Dose it mean that we can understand this way, even if the electron is so small, the thickness of the layer accumulated by the electrons on the surface of the spherical conductor will not be zero, so the position of the boundary of the conductor should be defined according to the distribution of charge. If we assume the gaussian surface contains half of the total charge, then the electric field will become $\frac{kQ}{2R^2}$ ?

 

[kuruman]

When the argument of the delta function vanishes at one of the endpoints of integration, the value of the integral depends on how you define the delta function. In this problem, the factor of 1/2 makes physical sense.

So if I understand you correctly, by choosing an appropriate sequence of functions $\delta_n(x)$, one may get an arbitrary factor $k$, not just $\frac{1}{2}$, multiplying $f(0)$ after $k$ has been determined independently by some other method. This I didn't know. Thank you for pointing it out to me.