This is Griffiths, Introduction to Electrodynamics, 2.43, if you have the book.
The problem states Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express your answer in terms of the radius $R$ and the total charge $Q$. Note: I will say its uniform charge is $\rho$.
My attempt at a solution:
My idea is to find the field generated by the southern hemisphere in the northern hemisphere, and use the field to calculate the force, since the field is force per unit charge.
To do this I start by introducing a Gaussian shell with radius $r < R$ centered at the same spot as our sphere. Then in this sphere,
$$\int E\cdot\mathrm{d}a = \frac{1}{\epsilon_0}Q_{enc}$$
Now what is $Q_{enc}$? I feel like $Q_{enc} = \frac{2}{3}\pi r^3\rho$ , since we're just counting the charge from the lower half of the sphere (the part thats in the southern hemisphere of our original sphere). (Perhaps here is my error, should I count the charge from the entire sphere?, if so why?)
Using this we get $$\left|E\right|4\pi r^2 = \frac{2\pi r^3\rho}{3},$$ so $$E = \frac{r\rho}{6\epsilon_0}.$$ Using these I calculate the force per unit volume as $\rho E$ or $$\frac{\rho^2 r}{6\epsilon_0}$$
Then by symmetry, we know that any net force exerted on the top shell by the bottom must be in the $\hat{z}$ direction, so we get $$ F = \frac{\rho^2}{6\epsilon_0} \int^{2\pi}_0\int^{\frac{\pi}{2}}_0\int^R_0 r^3\sin(\theta)\cos(\theta) \mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$
integrating we get $F = \frac{1}{4}\frac{R^4\rho^2\pi}{6\epsilon_0}$.
Now Griffiths requests us to put this in terms of the total charge, and to do so we write $\rho^2 = \frac{9Q^2}{16\pi^2R^6}$
Plugging this back into $F$ we get $$F = (\frac{1}{8\pi\epsilon_0})(\frac{3Q^2}{16R^2})$$
Now the problem is that this is off by a factor of $2$ …
I tried looking back through and the only place I see where I could somehow gain a factor of $2$ is the spot I mentioned in the solution, where I could include the entire charge, however, I can't see why I should include the entire charge, so if that's the reason I would be very grateful if someone could explain to me why I need to include the entire charge.
If that is not the reason, and perhaps this attempt at a solution is just complete hogwash, I would appreciate if you could tell me how I should go about solving this problem instead. (but you don't need to completely solve it out for me.)
Best Answer
The factor of two is coming from the place you identified.
Think about throwing out that factor of two, so you're considering only the bottom hemisphere. When you make your Gaussian shell and have it enclose charge in the bottom hemisphere only, the charge is no longer uniformly distributed inside your Gaussian shell. Thus, the electric field created by the charge you're considering is not the same at all parts of the shell, so you can't find the magnitude of the electric field in the way you described. That only works when the charge distribution has some sort of symmetry you're exploiting. You'd have to do a difficult integral instead.
However, if you don't throw out that factor of two, you're simply finding the electric field inside the shell. Suppose you carry out the rest of your calculation. Then you've found the net force in the z-direction in the north half of the sphere. However, the north half cannot exert any net force on itself, so this entire net force must be the same as the net force from the southern hemisphere.
So you're including all the charge when you make your Gaussian surface because you need to find the true electric field in the shell. The true electric field, when integrated, gives you the net force, which by basic mechanics arguments must be due to the southern hemisphere.