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Chapter 3: Q3.2P (page 119)

In one sentence, justify Earnshaw's Theorem: A charged particle cannot be held in a stable equilibrium by electrostatic forces alone. As an example, consider the cubical arrangement of fixed charges in Fig. 3.4. It looks, off hand, as though a positive charge at the center would be suspended in midair, since it is repelled away from each comer. Where is the leak in this "electrostatic bottle"? [To harness nuclear fusion as a practical energy source it is necessary to heat a plasma (soup of charged particles) to fantastic temperatures-so hot that contact would vaporize any ordinary pot. Earnshaw's theorem says that electrostatic containment is also out of the question. Fortunately, it is possible to confine a hot plasma magnetically.]

Short Answer

Expert verified

The cube leaks from the centre of each surface.

Step by step solution

01

Stable equilibrium of particle.

If the particle is disturbed by the internal or external force, which tends to the displacement of particle and the particle comes again in the equilibrium, it is known as the stable equilibrium. The particle can have instability only in one order.

02

Determine the leaks in the cubic arrangement.

The saddle point in the free space is defined as the stationary point with a non-local extremum. The saddle point in free space has an increasing curve in one direction and a decreasing curve in the other.

For the given arrangement, the potential is contour increasing along the diagonal and decreasing along the x and y-axis. Therefore, the potential due to the positive charge will rise in one direction and drop in another.

Hence the cube leaks from the centre of each surface.

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Most popular questions from this chapter

Buckminsterfullerine is a molecule of 60 carbon atoms arranged

like the stitching on a soccer-ball. It may be approximated as a conducting spherical shell of radius R=3.5A°. A nearby electron would be attracted, according to Prob. 3.9, so it is not surprising that the ion C60-exists. (Imagine that the electron on average-smears itself out uniformly over the surface.) But how about a second electron? At large distances it would be repelled by the ion, obviously, but at a certain distance r (from the center), the net force is zero, and closer than this it would be attracted. So an electron with enough energy to get in that close should bind.

(a) Find r, in A°. [You'll have to do it numerically.]

(b) How much energy (in electron volts) would it take to push an electron in (from

infinity) to the point r? [Incidentally, the C60-ion has been observed.]

Two infinite parallel grounded conducting planes are held a distanceapart. A point chargeqis placed in the region between them, a distance xfromone plate. Find the force on q20Check that your answer is correct for the special

cases a→∞and x=a2.

RFind the average potential over a spherical surface of radius Rdue to

a point charge qlocated inside (same as above, in other words, only with z<R).(In this case, of course, Laplace's equation does not hold within the sphere.) Show that, in general,

role="math" localid="1657706668993" Vave=Vcenter+Qenc4πε0R

where Vcenteris the potential at the center due to all the external charges, andQenc is the total enclosed charge.

(a) Using the law of cosines, show that Eq. 3.17 can be written as follows:

V(r,θ)=14πε0[qr2+a2−2racos⁡θ−qR2+(ra/R)2−2racos⁡θ]

Whererand θare the usual spherical polar coordinates, with the zaxis along the

line through q. In this form, it is obvious thatV=0on the sphere, localid="1657372270600" r=R.

(a) Find the induced surface charge on the sphere, as a function of θ. Integrate this to get the total induced charge . (What should it be?)

(b) Calculate the energy of this configuration.

Show that the average field inside a sphere of radius R, due to all the charge within the sphere, is

Eave=-14πε0ρR3

Where ρis the total dipole moment. There are several ways to prove this delightfully simple result. Here's one method:

(a) Show that the average field due to a single chargeqat point r inside thesphere is the same as the field at r due to a uniformly charged sphere with

ρ=q/(43πR3), namely

14πε0(43πR3)∫qr2rdζ'

Where r is the vector from r to dζ

(b) The latter can be found from Gauss's law (see Prob. 2.12). Express the answerin terms of the dipole moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

(d) While you're at it, show that the average field over the volume of a sphere, dueto all the charges outside, is the same as the field they produce at the center.
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