W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 81

S

a r S

r

1

2

σo

Figure 17: A spherical shell of radiusa, carrying a uniform charge per unit areaσ 0. Two spherical
concentricGaussian surfacesS 1 (with radiusr < aandS 2 (with radiusr > a) are shown.

Example 2.3.1: Spherical: A spherical shell of charge

Suppose you are given a spherical shell of charge with a uniform charge per unit areaσ 0

and radiusa. Find the field everywhere in space.

As you can see in figure 17, there are two distinct regions where we must find the field:inside
the shell andoutsidethe shell. Draw asphericalGaussian surfaceS 1 inside the sphere (forr < a).
From the symmetry of the distribution we know that the fieldE~must point in the direction of~r
and (hence) be perpendicular and constant in magnitude at all points on the Gaussian surfaceS 1.
Hence:

φe=

∮

S 1

E~·ˆrdA=Er

∮

S 1

dA=Er(4πr^2 ) (101)

where it is presumed that everybody knows how to integrate to evalute the area of a sphereand
knows the result.

The total chargeQSinside this sphere iszeroby inspection – the fingers and toes thing. That
was easy! Now we write Gauss’s law:

φe=

∮

S 1

E~·rˆdA=Er(4πr^2 ) =QS^1

ǫ 0

= 0 (102)

and solve forEr:

Er(4πr^2 ) = 0

=^0

4 πr^2

Er = 0 forr < a (103)

We’ve just shown thatin generalthe electric field of a spherical shell of charge (like the gravitational
field of a spherical shell of mass last semester)vanishesinside, but using Gauss’s law the derivation
wastrivial!

Outside the shell we draw asecondspherical Gaussian surfaceS 2 atr > a. Again, the field must
be constant and normal to all points on this surface from symmetry. The flux integral isalgebraically
identical:

φe=

∮

S 2

E~·ˆrdA=Er

∮

S 2

dA=Er(4πr^2 ) (104)

and in fact it willalwayshave this algebraic form for a spherical problem, to the point wherewe
will get bored writing this line out umpty times doing homework. Don’t letthat stop you! Do it