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This is problem 2.40 from Introduction to Electrodynamics by D. J. Griffiths:

A point charge $q$ is inside a cavity (not necessarily spherical or anything similarly regular) in an uncharged conductor. Is the force on $q$ necessarily zero?

What my understanding tells me is that if the charge on the inner surface of the conductor is just enough to cancel the field from the point charge from every direction, then the force from any two opposite pieces on the inner surface will produce equal but opposite forces. Thus, the net force is zero.

Is my reasoning correct?

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No, your reasoning is incorrect, because there's no reason for the forces to cancel in general. Actually, the charge in general will be attracted by the field of the induced opposite charge on the inside surface of the conductor. This is easy to see by use of the fact that $\nabla^2 V=0$ in the region devoid of charges implies that $V$ is a harmonic function, therefore it has extrema on the boundary only. This implies that it is impossible to have stable equilibrium in absence of charges in electrostatic configuration (Earnshaw's theorem). Let's assume that the charge is in equilibrium, i.e. the electrostratic force on it due to the induced charges vanishes. Since the conducting shell has arbitrary shape, we are allowed to deform it slightly, therefore the position of equilibrium will change and the new configuration of induced charges will exert a force that is attractive to the conductor, because the theorem implies that the potential tends to push the charge away from its position. Equivalently, we may move the particle slightly, which has the same effect, as the problem is posed.

This is easy to see with a concrete example. The simplest I could find is the spherical shell conductor with a charge inside. It is easy to find the potential of the induced charges with the method of images and it is always attractive.

Also, if we had a spherical mass shell, then inside it $\vec{g} =\vec{0}$, because of the symmetry of the problem and Gauss's law. In this case, though, a particle inside it will feel no force at all, because the field is fixed.

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  • $\begingroup$ Nice reasoning on stability. If I understand right, (1) put the point charge in (2) let surface charge settle (3) freeze surface charge (4) remove point charge (5) deduce from harmonic potential that potential of surface charge takes on extrema at boundary (6) if you put back the point charge in frozen field any equilibrium cannot be stable. IMO not that simple. The surface charges are DOFs of system. You use that equilibria of subsystems of stable systems are stable. Right for finite DOFs because of positive definite modal stiffness matrix implies that property for subsystems. Dunno infinite $\endgroup$ – Tobias Apr 11 '14 at 7:08
  • $\begingroup$ @Tobias Thanks. If you change the distribution of induced charges, this will affect the charge in the cavity before the induced charges settle down in their new distribution, because the former propagates at the speed of light and the latter at lower speed. You don't need to freeze the system arbitrarily. Also, Earnshaw's theorem implies that the configuration with the charge in the cavity is always unstable, I'm not sure where I implied that this is a stable system. $\endgroup$ – auxsvr Apr 11 '14 at 7:33
  • $\begingroup$ I did not say that you implied that the system is stable. I meant that you deduce instability of the overall system from the instability of a subsystem (with the other DOFs frozen). This would follow from the fact that subsystems of stable systems are stable. I think I should have a closer look at "Earnshaw's theorem" when I have time. Thanks, Tobias $\endgroup$ – Tobias Apr 11 '14 at 7:51
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The force on the charge $q$ is given by the electric field $\mathbf E_m = -\nabla \phi_m$ of charges of the metal shell surrounding it. This field does not vanish inside the metal, because total field $\mathbf E_m + \mathbf E_q$ does. It follows that the potential of metal charges $\phi_m$ is not constant throughout the metal and its inner surface is not necessarily equipotential surface of $\phi_m$. Hence the region inside the inner surface is not equipotential of $\phi_m$ either and there are places inside where the electric field $\mathbf E_m$ is non-zero. The force on charge $q$ is thus non-zero in general and will lead the charge to move towards the metal and be absorbed by it.

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The answer to your question is "no". Put the point charge close to the wall at some spot of the wall. There will a surface charge be collected at this spot of the wall that attracts the point charge.


In the following I give an example for which one could even calculate the attracting force analytically. Nevertheless, I keep a bit informal here since the calculation would be quite involved.

Consider a cubic cavity with a positive point charge at a small distance $\delta$ from the left wall and centered in all other coordinates.

The potential boundary conditions (the potential at the boundary must be constant) are satisfied with the principle of image charges. This is illustrated in the following Figure. There the red dots stand for positive charges $Q$, the blue ones for negative charges $-Q$.

principle of image charges

Note, that this grid must be continued ad infinitum in all three space directions to have reflection symmetry at the walls.

Now, let $\delta$ get smaller and smaller. The force of the $2\delta$ close mirror charge on the actual point charge will grow with $\frac{Q^2}{4\pi\varepsilon_0(2\delta)^2}$ while all other image charges form dipoles where the charge keeps constant and the distance of the pair charges gets smaller. The influence of the dipol fields will shrink to zero. (The product of charge and distance would have to converge towards a nonzero constant to let the potential field approach a nonzero constant. In our case the product of charge and distance goes down to zero with shrinking distance.)

This methodology can also be made more rigorous with convergence proofs and so on. But, that is no fun anymore.

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  • $\begingroup$ The electric field inside a hollow conductor is zero, no? $\endgroup$ – gj255 Apr 8 '14 at 20:05
  • $\begingroup$ No, not if you put a charge in. $\endgroup$ – Tobias Apr 8 '14 at 20:06
  • $\begingroup$ Yes, but then the only E-field is that generated by the charge, which cannot be experienced by it? $\endgroup$ – gj255 Apr 8 '14 at 20:07
  • $\begingroup$ No, you have a surface charge on the conductor which also creates a field. $\endgroup$ – Tobias Apr 8 '14 at 20:08
  • $\begingroup$ According to Griffiths (2.5.2) an outside charge does attract an uncharged conductor. I don't see a reason why this wouldn't be the case for one in a cavity, but I might be wrong about that. $\endgroup$ – Kvothe Apr 8 '14 at 20:09
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Let us assume that the conductor and cavity are asymmetric random shapes with $q$ placed at any general location in the cavity. This charge q, will induce a charge -q on the cavity surface. The leftover charge after -q is drawn is +q and that distributes itself over the conductor exterior surface.

Force on the charge, $F=q/vec(E)$ is to be made zero. So Electric field at the location of the charge must be zero, assuming that it's a general point, we need the field to be zero at all points of the cavity. Now, what is producing this field?, When calculating forces, the field due to the othercharges is only important, so we can assume q to be removed entirely and calculate field due to -q and +q. On page 101 ehich is two pages behind the problem, Griffiths clearly states that the third field due to +q is separately zero than the sum of q and -q in the conductor. So we neglect +q in this calculation, this is understopd by the fact that conductors have infinite permittivity leading to symmetrical (zero) field on a gaussian surface in the conductor and as the integral due to it is zero as the surface does not enclose it, and as such distance in conductors for field is of no consequence as it does not decrease with distance, a symmetrical field will give zero integral that is the field due to +q is always zero. This is a logical and not mathematical proof.

The field is now only due to -q. We have already assumed the cavity as asymmetrical. Taking a gauss integral ( surface integral of electric field and dot product with perpendicular) is equal to the charge enclosed (zero) divided by permittivity. Now for the integral to be zero there are three possibilities, that the $/vec(E)$ is zero at all points of the surface (i) or it is perpendicular to the surface (ii) or it is equally positive as it is negative on the surface (iii). As we are free to assume any shape of the surface for gauss integral irrespective of cavity (assuming no limitation due ease of calculation) we can simply take a spherical shape that would make the curl of E non zero fo case (ii), which rules out (ii). Field can only be zero at all points for all gaussian surfaces for a symmetrical surface, specifically a sphere. Thus (iii) is the only possibilty and as proven by the second uniquemess theorem, if there is a field and charge distribution related to each other, you can be sure that it is a totally one to one relation for all space considered (not for localized areas, as is the case for method of images).

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First of all, as the charge is in the cavity of conductor, it will feel no external electric field due to any charge present outside the cavity. This implies that net external field on the charge is null, i.e., the charge will feel no external electric force. Also it can't apply force on itself, so the net force on it is null.

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  • $\begingroup$ Please check punctuations and grammar. $\endgroup$ – Anubhav Goel Feb 14 '16 at 11:17
  • $\begingroup$ How about the induced charges on the inner surface of the conductor (around the cavity)? This may exert a force on the charge. See the other answers. $\endgroup$ – Ron Mar 17 '16 at 15:00

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