W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 91

from Gauss’s Law plus our knowledge that the field vanishes inside. )

• Is theinteriorof the conductor electrically neutral? (Sure, it must be. If it weren’t the charges
  there would create a field (see Gauss’s Law!) and move away from oneanother until they reach
  the surface and become part of the surface charge distribution.)


• Can we tell just from the figure whether or not the conductor is overall electrically neutral
  (has a net charge or not)? (No, not really. The lines of force in the figure above suggest that it
  might be, butwedrew them in response to the question above, right? So there isn’t any real
  reason to rely on them. What wedoknow is that if it isn’t neutral, all of the surplus charge
  will be located on the surface of the conductor, arranged in just the right way that the field
  lines leave the surface at right angles.)

Make sure that you understand the ideas underlying all of these answers.

Example 2.4.2: Two Thick Plates Plus Wires (Capacitor)

+ + + + + + + +

− − − − − − − −

cancel

add!

cancel

+σ

−σ

Figure 24: Opposite charges placed on two facing conducting platesspread out to formsurface
charge layers. This is exactly what is needed tocancel the fields of the two layers in the
plates themselveswhileadding together in the space between the plates.

In the figure above, two conducting plates with facing areaA, with wires attached to them are
schematically illustrated. The plates are deliberately drawn to bethickand the gap between the
plates is similarly exaggerated. We assume that the plates arelargecompared to this gap.

Suppose equal and opposite charges±Qare placed on the plates (and prevented from flowing
together through the conducting wires). We know that the field inside the shaded metal region
must be zero once the plates are in electrostatic equilibrium. We also know that the charges have
to spread out on the surface(s) of the conductors. Finally, we know that the oppposite charges will
attractacross the gap between the plates.

The charge distribution illustrated above, with the charges spreadout uniformly on the facing
surfaces of the plates as±σ=±Q/Asatisfies all of these conditions. As we have seen, the field of

a single plane sheet of charge isE=

σ

2 ǫ 0

= 2πkeσ, directedawayfrom a positive surface charge

density.

The field lines from the upper plate go up above the surface layer +σand down below it. Similarly
the field lines go down above the surface layer−σand up beneath it. The idealized field lines from