I am trying to do the following problem;
A uniform surface charge lies in the region $z=0$ for $x^2+y^2 \gt a^2$ and $z=\sqrt{a^2-x^2-y^2}$ for $x^2+y^2 \le a^2$, Find the force on a unit charge that is located at $(0,0,b)$
I am a bit confused because of the changing of the regions.
What I know;
For a point P,
$$dF=\frac{C(1)\sigma dS}{||PB||^{2}} u_{pb}$$
Where C is a constant, sigma the uniform charge, and u the unit vector.
ie
= $$dF=\frac{C(1)\sigma dS PB}{||PB||^{3}}$$
Where $PB=-a\sin(\psi)\cos(\theta)i-a\sin(\psi)\sin(\theta)j+(b-a)\cos(\psi)k$
But I am just really confused on how to approach this specfic problem. I am looking for a solution/help. Thanks
Without loss of generality, lets assume that the point charge $Q$ and the surface charge density $\sigma$ are of the same sign, so that the charge is repelled by each point on the surface. We will represent the net force acting on the charge as $F_{net} = F_0 + F_{sph}$, where $F_0$ is the force created by the ring at $z=0$ with the inner radius $a$ and the outer radius $\rightarrow \infty$, and $F_{sph}$ is the force acting on the point charge due to the presence of hemispherical surface.
First, let us calculate $F_0$. Obviously, this force will be along the positive direction of the $z-$axis. Elementary force reads
$dF_0(r, \phi) = k\frac{\sigma rd\phi dr}{(r^2 + b^2)^{3/2}}b$,
where $k$ is the electrostatic constant, $r$ and $\phi$ are polar coordinates associated with the ring, and extra factor $b/\sqrt{r^2+b^2}$ is due to projection of this elementary force on the $z-$axis. Then, $F_0$ becomes
$F_0 = \int_0^{2\pi}d\phi\int_a^\infty dr k\frac{\sigma r}{(r^2 + b^2)^{3/2}}b = \frac{2\pi k\sigma b}{\sqrt{a^2+b^2}}$.
Now we proceed with $F_{sph}$. Elementary force $dF_{sph}$ reads
$dF_{sph}(\phi,\theta) = k\frac{\sigma a^2 \sin\theta d\theta d\phi}{(a^2+b^2 - 2ab\cos\theta)^{3/2}}(b-a\cos\theta)$,
where $\theta$ is the angle between the $z-$axis and the radius-vector of the point on the hemisphere with respect to the origin. Then, $F_{sph}$ becomes
$F_{sph} = k\sigma a^2 \int_0^{2\pi}d\phi\int_0^{\pi/2}d\theta \frac{(b-a\cos\theta) \sin\theta }{(a^2+b^2 - 2ab\cos\theta)^{3/2}} = \frac{2\pi k \sigma a^2}{b^2}\bigg(\frac{a}{\sqrt{a^2+b^2}} + \frac{b-a}{|b-a|}\bigg)$.
Thus, the net force acting on the unit charge is
$F_{net} = \frac{2\pi k\sigma }{\sqrt{a^2+b^2}}\bigg(b + \frac{a^2}{b^2}\bigg(a + \frac{b-a}{|b-a|}\sqrt{a^2+b^2}\bigg)\bigg)$.
In this solution I have omitted some minimal steps, but they should be easily recovered from the highlights above.
Hope this helps.