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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.)

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Your choice for Qenc would be correct if you consider only half the shell (Eda -> 2*pir^2) –  user12642 Sep 30 '12 at 17:46

3 Answers 3

up vote 3 down vote accepted

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.

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This is essentially what I was thinking after proposing the question, and I'm glad to have the thoughts in my head confirmed by someone who knows what they're doing! Thank you very much –  Deven Ware Mar 31 '12 at 4:42

If you are off by a factor of two, it's probably because the volume of a sphere is

$\frac{4}{3}\pi r^3$ and not $\frac{2}{3}\pi r^3$

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Perhaps I did not understand the question correctly, but it seems to me that you cannot use a Gaussian shell in this case, because the intensity of the field $E$ would be different at different points of the shell. If you want the following expression to hold, $$ \int E\cdot da = E \int da $$ then $E$ must be equal to the same value all over the Gaussian shell. I believe this might be the source of your mistake.

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Hi Marc, welcome. I don't precisely know what the question is about, but while reviewing your answer, I think I can see that is is perhaps already stated by the currently accepted answer. Is that correct? (If that is the case, perhaps you should consider deleting your answer. If it is not correct, then ignore this comment.) –  Glen The Udderboat May 13 '13 at 14:16

protected by Qmechanic May 13 '13 at 11:53


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