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Browsing an archive of problems of a local physics olympiad, i stumbled upon a problem which seems not a very trivial.

Given two identical metal spheres in vacuum, with mass $m$ and radius $R$. One sphere with a charge $Q$ and other with no charge(neutral). At initial moment, they are very far from each other. After releasing, due to the electrostatic attraction, the spheres come together and collide. Find the velocity of the spheres after the collision if the collision is:

a) perfectly elastic
b) inelastic

Ignore gravity forces and a possible spark discharge.

The part b) seems especially tricky.

Edit:

Yes, there was probably a typo. The part b) must be simply "inelastic collision" that means kinetic energy is not conserved and the spheres do not stick. Sorry for confusion!

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Sorry, (b) is zero, isn't it? An inelastic collision is one in which the spheres merge into one object moving by one velocity, and by momentum conservation, its momentum is zero, isn't it? Am I missing something? –  Luboš Motl Jun 11 '11 at 12:41
    
Without thinking too much I'd say b is zero too. –  Diego Jun 11 '11 at 12:47
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Ben. There will be an attraction, because the charge (net zero) in the second sphere will be unevenly distributed. If say the charged sphere is positively changed, negative charges will accumulate towards the side facing the charged sphere, i.e. the charge redistributes to maintain a constant potential on the surface. I think you would want to expand the potential in terms of spherical harmonics term with m=0. Seems pretty messy to workout without a computer. B is not a giveaway, after the collision, each sphere has half change, and they repel each other, which adds further energy.... –  Omega Centauri Jun 11 '11 at 16:29
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@anna: My interpretation of elastic/inelastic is exactly the same as Carl's. Elastic implies the kinetic energy of approach is totally transferred to the kinetic energy in the opposite direction after collision. Inelastic implies the kinetic energy of approach is lost (presumably converted to heat).It does not imply cohesiveness (i.e. welding). –  Omega Centauri Jun 12 '11 at 22:58
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Just for the record: An inelastic collision is a collision that is not elastic, cf. en.wikipedia.org/wiki/Inelastic_collision In a perfectly inelastic collision, the colliding particles stick together, cf. en.wikipedia.org/wiki/… @Martin Gales: If the collision in (b) is just (unspecified) inelastic (v2), it could mean anything in between elastic and perfect inelastic, and it seems that we have not enough information to determine the speed. Btw.: Speed at what time after the collision? Do you mean just after the collision? –  Qmechanic Jun 13 '11 at 16:37

2 Answers 2

This is an edited answer following the remark by Qmechanic pointing out that the capacitance used in the original answer assumes electrical contact between the spheres. Here I assume that the actual contact time is very small, so no discharge occurs. On, the other hand it is assumed that the motion is slow enough that the laws of electrostatics apply. The capacitance formulas are taken from the article: "Capacitance coefficients of two spheres" by John Lekner. The same notation is used, where the charged sphere is denoted with the subscript $a$ and the other sphere with the subscript $b$. Also CGS units are used (The capacitance has units of length).

Substituting $ Q_a = Q$, $Q_b = 0$, we obtain the following formula for the effective capacitance:

$$ C_s \equiv \frac{V_a}{Q} = \frac{C_{aa}C_{bb}-C_{ab}^2}{C_{bb}},$$

where $s$ is the separation distance. The expressions for the capacitance matrix elements at infinite separations $s=\infty$ are given in the text eq. (8) resulting the following result:

$$ C_{\infty} = R.$$

Of course, this is just the capacitance of the charged sphere.

The capacitance matrix elements become singular when the spheres approach contact $s\to 0$, but using the asymptotic formulas (16)-(18), the effective capacitance has a well defined limit:

$$C_0 = \lim_{u \to 0}\frac{R}{2} \frac{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))^2-(\ln(\frac{2}{u})+\gamma)^2}{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))}= -R (\psi(\frac{1}{2})+\gamma) = 2 R \ln 2,$$

where $\gamma$ is the Euler constant and $\psi$ is the digamma function, and the following identity was used

$$\gamma +\psi(\frac{1}{2}) = -2 \ln 2.$$

By energy conservation

$$2 \frac{m v^2}{2} + \frac{Q^2}{C_0}= \frac{Q^2}{C_{\infty}}.$$

We obtain:

$$v = \sqrt{\frac{Q^2 (2\ln 2-1)}{2m R \ln 2}},$$

which is the speed just after collision in the case of an elastic collision. Of course, the speed is zero in the case (b) of inelastic collision.

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That looks correct, but the hard part is definitely setting up the method of images properly, and performing the sum. Afaik, thats the only possible way to get to the analytic solution. –  Columbia Jun 13 '11 at 13:27
    
@David: Looks good! But why do you think the speed is zero in the case of inelastic collision? –  Martin Gales Jun 14 '11 at 5:35
    
For a completely inelastic collision, the formulas in the wikipedia page: en.wikipedia.org/wiki/Inelastic_collision can be used to compute the velocities after the collision. Actually, the case of inelastic collision can be explained intuitively: In this case the two colliding objects become "glued" together. Since by symmetry, their velocities before the collision are equal in magnitude and opposite in direction, the total momentum is zero, thus by conservation of momentum it must stay zero after the collision, therefore the speed of the combined body is zero. –  David Bar Moshe Jun 14 '11 at 6:22
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Dear @David Bar Moshe: Could you please explain how Maxwell's formula for two identical spheres in (geometrical and electrical) contact (and hence with equal charges) is relevant for the elastic case (a)? See also J.C. Maxwell, "A Treatise on Electricity and Magnetism", page 219-221. –  Qmechanic Jun 14 '11 at 6:43
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@Qmechanic you are correct, I missed the point that this formula is valid for the case of electrical contact, which was not my initial aim. I'll edit the answer with the correct formula of the capacitance (and its actual computation). –  David Bar Moshe Jun 14 '11 at 7:50

There is an electrostatic attraction between the two spheres even though one has a net zero charge. On the side of the uncharged sphere that faces toward the (say) positively charged sphere, there will be felt a strong electromagnetic field arising from the charged sphere. This field will attract conduction band electrons and cause them to rush to the side facing the charged sphere. As a result, the far side of the uncharged sphere will have a shortage of electrons thus will be positively charged.

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If I were ambitious enough to solve it, I would use the mirror technique, partially shown here:en.wikipedia.org/wiki/Method_of_images –  anna v Jun 11 '11 at 16:56

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