Comparing the electric field at the centre of a hemispherical shell with an infinite plane

Question

On p. 595 of Purcell and Morin’s Electricity and Magnetism (3rd edition), we obtain the result that at the centre of a hemispherical shell of uniform surface charge density $\sigma$ the electric field has magnitude $\sigma/4\epsilon_0$. Then the authors note the fact that the electric field at any point above an infinite sheet with the same charge density $\sigma$ has magnitude $\sigma/2\epsilon_0$. They write “You should convince yourself why it [the field at the centre of the hemisphere] must be smaller (consider the amount of charge subtended by a given solid angle), although the factor or 2 isn’t obvious.” Our shell is $z>0$, centered at $(0, 0, 0)$ and of radius $R$; they seem to be imagining the infinite plane $z=R$.

It is of course true that a given solid angle subtends more charge on the plane than on the shell, but the shell is nearer than the plane, so why should it be intuitively obvious that the field of the shell is smaller? Have they forgotten about the distance factor?

Answer 1

The magnitude of the electric field created by a patch of surface charge with density $\sigma $ and area $dA$ is $$ dE = \frac{\sigma dA}{4\pi\epsilon r^2}. $$ But the solid angle subtended by this patch is $$ d\Omega = \frac{dA \cos \theta}{r^2} $$ where $\theta$ is the angle between the surface normal and the line between the field point and the patch of charge. And so $$ dE = \frac{\sigma d\Omega }{4\pi\epsilon \cos \theta}. $$

We see from this that the distance of the patch of charge doesn't matter; it's automatically accounted for by writing things in terms of the solid angle, since a patch of charge at a greater distance subtends a smaller solid angle. What does matter is the angle $\theta $, which is zero for all surface patches on the hemisphere but varies over the infinite plane.