December 2014
Intermediate to advanced
418 pages
16h 17m
English
Content preview from Electricity and Magnetism
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6–48 Electricity and Magnetism
Application of Gauss’s Law yields
E(R) 4 π R
2
≡
Q
Total
R
3
a
3
ǫ
0
=
Q
Total
ǫ
0
a
3
R
3
=⇒ E(R) =
Q
Total
4 π ǫ
0
a
3
R =
k Q
Total
a
3
R .
Thus, within the insulating material, the electric field grows linearly from zero at the
centre to its value at the surface.
To convince ourselves that the exterior and interior solutions are consistent, let’s consider
Figure 6.2 and the following comments.
0 R
r
INTERIOR EXTERIOR
E
r
∝ r ∝
1
r
2
FIGURE 6.2 Radial Field of a Spherically Symmetric Uniformly Charged Ball
⊙ The electric field vanishes at the centre, as it must, in accord with the isotropy
of the spherically symmetric distribution.
The only vector that points ...
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