Consider a charged spherical shell of radius $R$ and surface charge density $\sigma$. Choose a point on the surface of the shell and cut a spherical cap of radius $a \ll R$. What is the electric field intensity in the middle of the hole?
This was made up intended as a relatively easy problem to solve, it turns out however it is probably not, as every edition of the textbook shows a different answer and professors at our physics department can't agree upon one. What do you think?
My attempt: First I calculated the intensity on the axis of a disc and integrated over the disc elements: $\vec E = \int_0^{2R-{a^2 \over 2R -1}} {\sigma \over 2\epsilon} ({ \sqrt{2Rh} – |h| })dh$. This, using the fact that $a \ll R^2$ evaluates to ${\sigma R^2 \over 3\epsilon} $, which is wrong.
Best Answer
$\def\ph{\varphi}\def\sign{\operatorname{sign}}\def\eps{\varepsilon}\def\l{\left}\def\r{\right}$ The radially symmetric charge displacement density field of homogeneous surface charge on a sphere with middle point at the origin has only a r-component \begin{align} D_r(r) &= \begin{cases} \frac{4\pi R^2\sigma}{4\pi r^2} &\text{ for } r>0,\\ 0 &\text{ for } r<0. \end{cases} \end{align} On the positive z-axis this can be written as \begin{align} D_r(z) &= \frac{\sigma R^2}{2 r^2}(1+\sign(z-R)) \end{align}
The field of a uniformly charged disk of radius $a$ at $z=R$ with normal in z-direction and with surface charge $-\sigma$ is \begin{align} D_z(z) &= \frac{-\sigma}2\l(\sign(z-R)-\frac{z-R}{\sqrt{a^2+(z-R)^2}}\r). \end{align} This results from the derivative of the potential for this problem which I already described in another answer. For small $a$ we hope to approximate the hole by such a disk with opposite $\sigma$ to the surface charge of the sphere. This should give a first order approximation in the order of $a/R$.
Superposition of the fields gives \begin{align} D_z(z) = \frac{\sigma}2\l(\frac{R^2}{z^2}-1\r)\sign(z-R)+\frac{\sigma R^2}{2z^2}+\frac{\sigma(z-R)}{2\sqrt{a^2+(z-R)^2}} \end{align} and at $z=R$ this gives $D_z(z) = \frac{\sigma}2$ the corresponding E-field is $E_z=\frac\sigma{2\eps_0}$.
Let us evaluate $D_z\l(\sqrt{R^2-a^2}\r)$ as mirgee suggested. Note, that $\sign(z-R)=-1$ in that case. \begin{align} D_z\l(\sqrt{R^2-a^2}\r)&\approx D_z\l(R-\frac{a^2}{2R}\r)\\ &=\frac{\sigma}2+\frac{\sigma\l(-\frac{a^2}{2R}\r)}{2\sqrt{a^2+\l(\frac{a^2}{2R}\r)^2}}\\ &=\frac{\sigma}2+\frac{\sigma\l(-\frac{a^2}{2R}\r)}{2a\sqrt{1+\l(\frac a{2R}\r)^2}}\\ &\approx\frac{\sigma}2\l(1-\frac{a}{2R}\r) \end{align}
To take 2nd order info for the disk into account one assumes that the charged disk is curved. The z-coordinate of the disk $z_\sigma$ in dependence of the radial position $\rho$ from the axis is $$ z_\sigma \approx R\cos\l(\frac \rho R\r) \approx R\l(1-\frac12\l(\frac\rho R\r)^2\r) $$ The potential on the axis becomes \begin{align} \ph(z) &= \frac{\sigma}{4\pi\eps_0} \int_{\rho=0}^a \frac{2\pi\sqrt{1+\l(\frac{\rho}R\r)^2}\rho d\rho}{\sqrt{(z-z_\sigma)^2 + \rho^2}} \end{align}
Now it is time-out for me. I have to go to work.