Homework Help: Force between charged and neutral sphere after scaling

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1. Nov 27, 2015

gobbles

1. The problem statement, all variables and given/known data
Consider two solid dielectric spheres of radius $a$ separated by a dis-
tance $R$ ($R\gg a$).
One of the spheres has a charge $q$ and the other is
neutral. We scale up the linear dimensions of the
system by a factor of two. How much charge should reside on the first
sphere now so that the force between the spheres remains the same?

2. Relevant equations
See below.

3. The attempt at a solution
I know that the neutral sphere becomes a dipole and then it's a problem of the force on a dipole in an electric field and that the correct answer is $4\sqrt{2} q$, but I tried a different approach:
We know that a surface charge $\sigma$ is formed on the surface of the neutral sphere, affected by the charged sphere. This surface charge depends on the distance $R$ between the spheres, the radius $a$ of the spheres and the charge $q$ on the charged sphere, and also has some angular dependence (assuming that the origin of axes is in the center of the neutral sphere and the angle ($\theta$) is taken from the horizontal axis).
The dependence of $\sigma$ on the said parameters must be of the form $qa^{-\alpha} R^{\alpha-2}\Theta(\theta)$, because the units must be charge over area and $\Theta(\theta)$ is some angular dependence.
Additionally, the force acting on a surface element $da$ of the neutral sphere is $\frac{q\sigma da}{\left(R(\theta)\right)^2}$, where $R(\theta)$ is the distance from the center of the charged sphere to the surface element of the neutral sphere
Now we can write an expression for the force acting on the neutral sphere ($A$ is the surface of the neutral sphere):

$F=\int_{\mbox{A}} \frac{q\sigma da}{\left(R(\theta)\right)^2}=\int \frac{q a^2 d\Omega}{\left(R(\theta)\right)^2}qa^{-\alpha} R^{\alpha-2}\Theta(\theta)\\ =\int \frac{q^2 a^2 d\Omega}{R^2 + a^2 + 2Ra\cos\theta}a^{-\alpha} R^{\alpha-2}\Theta(\theta)=\int \frac{q^2 R^{\alpha-2}a^{2-\alpha} d\Omega}{R^2 + a^2 + 2Ra\cos\theta} \Theta(\theta)\\ \approx q^2\left(\frac{a}{R}\right)^{2-\alpha}\frac{1}{R^2}\int d\Omega\Theta(\theta)(1-2\cos\theta \frac{a}{r}).$
If we take the first term, we get
$F\propto q^2\frac{a^{2-\alpha}}{R^{4-\alpha}}$.
If we increase the distances $a$ and $R$ by 2, we get:
$q^2\frac{a^{2-\alpha}}{R^{4-\alpha}}=Q^2\frac{a^{2-\alpha}}{R^{4-\alpha}}\frac{2^{2-\alpha}}{2^{4-\alpha}}=Q^2\frac{a^{2-\alpha}}{R^{4-\alpha}}\frac{1}{4},$
So we get $Q=2q,$, unlike the true answer $Q=4\sqrt{2} q$.
What am I missing?

2. Nov 28, 2015

Staff: Mentor

How did you get this factor?

An approach similar to yours, but without an α, leads to a third answer:
The field inside the neutral sphere due to the charged sphere is proportional to q/R2, polarization follows the same factor. Polarized volume scales with a3.
Force then just scales as the product of polarization, volume and another factor q/R2, for a total force
$F\propto q^2 \frac{a^3}{R^4}$
That just gives a factor of $\sqrt{2}$ for q.

I think I found the problem in your approach:
A larger sphere leads to a larger charge separation, the part of the integral you ignored has a factor of "a" hidden inside. If you would approximate a<<R at first order, the integral would be zero.

3. Nov 28, 2015

TSny

Shouldn't the first term in the integral lead to zero when you integrate since the total charge of the second sphere is zero. However, I don't think this will change your result since the second term is proportional to a/R and is constant when scaling the distances. So, I don't see anything wrong with your final result. I don't see how the answer could be 4√2. But I could be overlooking something. [EDIT: See post #5, below]

Last edited: Nov 28, 2015
4. Nov 28, 2015

TSny

Shouldn't the "another factor" be q/R3 rather than q/R2 since the second sphere acts like a dipole and the force on a dipole is proportional to the spatial derivative of the field E rather than the field itself.

5. Nov 28, 2015

TSny

What if we assume that the scaling of all distances does not change the number of molecules in the spheres? This is a bit of a stretch (pun intended), but I think it leads to the given answer of 4√2.

6. Nov 28, 2015

Staff: Mentor

Ah right. I though the factor of a took that into account, but i need another factor of R.
So a3/R5 in agreement with the second approach of gobbles (α=-1).

If we don't change the number of molecules and don't change their properties either, where is the point in scaling up the sphere?

7. Nov 28, 2015

TSny

I am thinking that the polarization per unit volume, P, of a dielectric is proportional to the number density of molecules. (I believe this is true for dielectrics that obey the Clausius-Mossotti equation). Thus, if the number of molecules is kept constant, then P will be inversely proportional to the cube of the scaling factor as all distances are scaled. If you include this effect in your argument in post #2, then I believe you get the answer of 4√2.

8. Nov 28, 2015

Staff: Mentor

The number density and the size of the sphere cancel if we keep the number of molecules constant. Changing "a" would lose its effect, we get 1/R5 and therefore the $4\sqrt 2$ scaling of q, sure, but that doesn't look like the intended way to solve the problem.

9. Nov 28, 2015

TSny

Yes.

I'm not sure.

10. Nov 29, 2015

gobbles

The charged sphere polarizes the neutral one. The polarization density (=dipole density) is proportional to the electric field at the neutral sphere, caused by the charged sphere, so the second sphere behaves as a dipole $\propto \frac{q}{R^2}$. The force on a dipole in an electric field is $\propto \frac{pq}{R^3}$, so the force between the spheres is $\propto \frac{q^2}{R^5}$, giving a factor of $4\sqrt{2}$ when scaling.
Thanks for noting that the first term of the integral is zero. I haven't noticed that bit.
Still, I can't see why my approach gives a different answer.

11. Nov 29, 2015

Staff: Mentor

You ignored the a^3 scaling of the volume of the dipole.

12. Nov 29, 2015

gobbles

Hmm.. the answer I consider to be correct (the dipole approach) is not my answer, I took it from a collection of problems and answers given here:
http://www.princeton.edu/~xxiang/prelim/cahn/Cahn_part1.pdf
This is question 2 from the Electrodynamics section.
I see how the answer given there could've ignored the scaling of the volume of the dipole. You mean that the polarization density should be integrated over the neutral sphere to get the dipole?

13. Nov 29, 2015

TSny

The solution for the net dipole moment of a dielectric sphere of radius $a$ in a uniform external field $E_0$ is worked out in many texts. For example, see equation (4.67) here:
http://physics.oregonstate.edu/~minote/COURSES/ph631/lib/exe/fetch.php?media=emch4.pdf

You can see that the dipole moment $\vec{p}$ is proportional to the volume of the sphere as well as proportional to $E_0$. $$\vec{p} = 4\pi\varepsilon_0 \left(\frac{\varepsilon_r-1}{\varepsilon_r+2}\right)E_0a^3$$
$\epsilon_r$ is the dielectric constant which depends on the polarizability of the individual molecules and the number density of molecules.

According to the Clausius-Mossatti equation (see equation (4.78) in the previous link) the quantity $\left(\frac{\varepsilon_r-1}{\varepsilon_r+2}\right)$ is proportional to the number density of molecules $N$. In this case, you see that $$\vec{p} \propto NE_0a^3$$

The force on either sphere then goes like $$F \propto \frac{q^2Na^3}{R^5}$$

If scaling distances by a factor of 2 is interpreted as doubling $R$ and $a$ while keeping the number density $N$ constant, then $F \propto \frac{q^2a^3}{R^5}$ and you get your result that Q = 2q. This is a natural way to interpret it as it keeps the dielectric properties of the material fixed.

If the scaling is interpreted as keeping the total number of molecules in the sphere fixed so that the distance between molecules is also scaled, then the number density $N$ is inversely proportional to $a^3$. Now $F\propto \frac{q^2}{R^5}$. Then you get the answer Q = 4√2 q.

I think I agree with mfb that this second interpretation is not as natural as the first. The second interpretation changes the intrinsic properties of the dielectric material.

The solution in the book simply ignores the volume factor $a^3$ without any explanation.

14. Nov 30, 2015

gobbles

Thank you TSny for the great explanation! I perfectly understand now.