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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

Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends only on r. Do the same for cylindrical coordinates, assuming v depends only on s.

A uniform line charge ฮปis placed on an infinite straight wire, a distanced above a grounded conducting plane. (Let's say the wire runs parallel to the x-axis and directly above it, and the conducting plane is the xyplane.)

  1. Find the potential in the region above the plane. [Hint: Refer to Prob. 2.52.]
  2. Find the charge density ฯƒ induced on the conducting plane.

In Ex. 3.2 we assumed that the conducting sphere was grounded ( V=0). But with the addition of a second image charge, the same basic modelwill handle the case of a sphere at any potentialV0 (relative, of course, to infinity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

Prove that the field is uniquely determined when the charge density ฯ

is given and either V or the normal derivative a โˆ‚V/โˆ‚n is specified on each boundary surface. Do not assume the boundaries are conductors, or that V is constant over any given surface.

In Prob. 2.25, you found the potential on the axis of a uniformly charged disk:

V(r,0)=ฯƒ2ฮต0(r2+R2โˆ’r)

(a) Use this, together with the fact that Pi(1)=1to evaluate the first three terms

in the expansion (Eq. 3.72) for the potential of the disk at points off the axis, assuming r>R.

(b) Find the potential for r<Rby the same method, using Eq. 3.66. [Note: You

must break the interior region up into two hemispheres, above and below the

disk. Do not assume the coefficients A1are the same in both hemispheres.]

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