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I solved the following question(Answer is correct):

Find the force with which two hemisperical parts of a uniformly charged hollow sphere repel each other?(charge density: +$\sigma$)

Answer: enter image description here

Radial force on strip is: $$dF=\left(\frac{\sigma^2}{2\epsilon_0}\right)(2\pi R^2\sin\theta d\theta)$$ Total force on hemisphere is: $$F=\int dF\cos\theta=\int_0^{\frac{\pi}2}\frac{\pi R^2\sigma^2}{\epsilon_0}\sin\theta\cos\theta d\theta$$ $$F=\frac{\pi R^2\sigma^2}{2\epsilon_0}$$


My question is: If the hollow sphere is uniformly charged on one half with a uniform charge density $+\sigma_1$ and its other half is also charged at charge density +$\sigma_2$.Now find the force with which the two halves repel?

Similiar to previous question force can be given as: $$F=\int \frac1{4\pi\epsilon_0}\sigma_1\sigma_2dS_1dS_2$$ But it doesn't give the answer, or I must say it can not be manipulated to an integrable form.Note that this is not a problem of integration.It is simple manipulation, as my teacher says, he suggests using previous problem.

Answer

$$F=\frac{\pi R^2\sigma_1\sigma_2}{2\epsilon_0}$$

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  • $\begingroup$ Which halves are charged? Hemispheres? Inner/outer half (by volume? by radius?)? Some other more complicated geometry? $\endgroup$ – Kyle Oman Aug 12 '14 at 17:31
  • $\begingroup$ @Kyle Since it is a hollow sphere, charge cannot reside on inner surface, only outer.One hemisphere $\sigma_1$, other $\sigma_2$. It cannot be charged by volume, since it is a hollow hemisphere, only on surface $\endgroup$ – RE60K Aug 13 '14 at 5:42
  • $\begingroup$ Nowhere does it specify that your sphere conducts, but I guess you were assuming a conducting material. You should say so. $\endgroup$ – Kyle Oman Aug 13 '14 at 15:10
  • $\begingroup$ aha i took so granted all electrodynamics questions that i forgot to mention it is a conductiong sphere $\endgroup$ – RE60K Aug 13 '14 at 16:15
  • $\begingroup$ @Kyle, It's not a conductor. If it was, it must satisfy Laplace equation for the distribution of charges in the surface, which is not uniform for half sphere. But question imposes: "hollow sphere is uniformly charged". Then cannot be a conductor. $\endgroup$ – Physicist137 Aug 13 '14 at 16:20
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Hm, I would give superposition a go.

Start with two equally charged half spheres with surface charge density $\sigma_{2}$. The consider a similar system superposed on it with but with surface charge density $\sigma_{3} = \sigma_{1}-\sigma_{2}$. We can solve this system as an instance of the first case using $\sigma_{1}$. Let's called this known answer $F_{11}$.

Now, the force on the superposed hemisphere is, by superposition, the force on the hemisphere with charge $\sigma_{2}$ (which is what we want to know!), plus the force on the one with charge $\sigma_{3}$. Call these $F_{12}$ and $F_{13}$, respectively.

The crucial insight now is that the ratio of these forces must be the same as the ratio of the charges of the right hemispheres because (also a consequence of the superposition principle) the force on a system, all other things being equal, scales with the charge on it! This means that:

$$\frac{F_{12}}{F_{13}} = \frac{\sigma_{2}}{\sigma_{3}}$$

and

$$F_{12}+\frac{\sigma_{3}}{\sigma_{2}} F_{12} = F_{11}$$

Therefore,

$$F_{12} = \frac{F_{11}}{1+\frac{\sigma_{3}}{\sigma_{2}}} = \frac{\pi R^{2} \sigma_{1} \sigma_{2}}{2 \epsilon_{0}}$$

Which looks like a reasonable result to me (for one thing, it checks with the case of $\sigma_{1} = \sigma_{2}$).

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  • $\begingroup$ It is a great answer $\endgroup$ – RE60K Aug 13 '14 at 17:38
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If we say we have two objects, with in general charge $Q_1$ and $Q_2$, which experience a certain electrostatic force $F$ between them, then we know that if we double the charge on one, we will double the force - that's just how electrostatics works. Force is proportional to the charge on each of the two objects.

With that insight, we can say that the charge on each hemisphere scales with $\sigma$, and so in general that the force must scale with $\sigma_1\sigma_2$ since the charge is just the integral of surface charge times area.

Finally, we look at the case where $\sigma_1 = \sigma_2$, for which we have the result that you proved:

$$F = \frac{\pi R^2\sigma^2}{2\epsilon_0}$$

Now since we are looking for an expression of the form

$$F \propto \sigma_1 \sigma_2$$

we conclude that the expression (for two different charge densities) must be

$$F = \frac{\pi R^2\sigma_1\sigma_2}{2\epsilon_0}$$

which is the result you were looking for.

SECOND APPROACH

Take two sets of hemispheres. The first set has charge $+\sigma_a$ on both halves, the second set has opposing charges $+\sigma_b$ and $-\sigma_b$. Now the force for the first pair scales with $\alpha\sigma_a^2$, and the second with $-\alpha\sigma_b^2$ where $\alpha$ is the constant of proportionality (which you computed as $\frac{\pi R^2}{2\epsilon_0}$)

The total charge on one side $$\sigma_1=\sigma_a+\sigma_b$$ On the other side it is $$\sigma_2=\sigma_a-\sigma_b$$

Now since we know the force due to each component, we can compute the total net force. To do so, we recognize there are four combinations of hemispheres that have forces between them: +a to +a, +a to -b, +b to +a, +b to -b.

We don't know how to compute +a to -b and +b to +a - but since their signs are opposite, we know the forces, whatever they are must cancel out (by symmetry). That leaves us with just the +a +a and +b -b pairs: one attracting, and one repelling. Thus we write the force between them as

$$\begin{align} F&=\alpha(\sigma_a^2-\sigma_b^2 )\\ &=\alpha(\sigma_a+\sigma_b)(\sigma_a-\sigma_b)\\ &=\alpha \sigma_1 \sigma_2\\ &=\frac{\pi R^2 \sigma_1 \sigma_2}{2 \epsilon_0}\end{align}$$

Giving the same result as before.

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  • $\begingroup$ You didn't show where the $\sigma_1\sigma_2$ term came, that's just working backwards from the answer $\endgroup$ – RE60K Aug 13 '14 at 17:32
  • $\begingroup$ No - I am saying that for two objects, the force scales with the charge. Double the charge, double the force. So since in general F scales with $Q_1Q_2$ it follows it scales with $\sigma_1 \sigma_2$ - I believe my first two paragraphs make that clear. I edited to make more explicit. $\endgroup$ – Floris Aug 13 '14 at 17:34
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I would like to introduce an approach by complementing the hemisphere.

Generally I want to make the left hemisphere bigger. Let the radius of the left hemisphere is $R_1$ and the density $\sigma_1$. The right hemisphere has $R_2$ and $\sigma_2$. ($R_1>R_2$) The force 2 hemisphere acting on each other is $F$,

  • First we complement the left half to make it become a sphere of $\sigma_1$. Then we could see that the inner hemisphere bear no net force. Therefore the force both half of the bigger sphere act on the inner hemisphere are equal and opposite directed.

  • Then we complement the right half (without complementing the left half of course) to have the inner sphere of $\sigma_2$ and the outer hemisphere of $\sigma_1$. By the first conclusion we know that the force the inner sphere acting on the outer hemisphere is $2F$, which is now easily to computed (because in the outside a sphere has electric field just as a poinlike charge)

  • Then we take $F=\frac{2F}{2}$

Of course it also comes from superposition, but it's another way to view the problem.

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