Chapter 4

1 q σ 1 q σ
E = = , E = = (4.3.10)
1 1 2 2
1 2
4πε0
r1 ε0
2
2
4πε0 r2
ε0
w here σ 1 and σ 2 are the surface charge densities on spheres 1 and 2, respectively. The
two equations can be combined to yield
E1 σ1
r2
= = (4.3.11)
E σ r
2 2 1
With
the surface charge density
being inversely proportional to the radius, we conclude
that the regions with the smallest radii of curvature have the greatest σ . Thus, the
electric field strength on the surface of a conductor is greatest at the sharpest point. The
design of a lightning rod is based on this principle.
4.4 Force on a Conductor
We have seen that at the boundary surface of a conductor with a uniform charge density
σ, the tangential component of the electric field is zero, and hence, continuous, while the
normal component of the electric field exhibits discontinuity, with ∆ En = σ / ε 0 . Consider
a small patch of charge on a conducting surface, as shown in Figure 4.4.1.
Figure 4.4.1 Force on a conductor
What is the force experienced by this patch? To answer this question, let’s write the total
electric field anywhere outside the surface as
where
E
patch
E=E E
(4.4.1)
′
patch +
is the electric field due to charge on the patch, and ′
E is the electric field due
to all other charges. Since by Newton’s third law, the patch
cannot exert a force on itself,
the force on the patch m ust come solely from E′ . Assuming the patch to be a flat surface,
from Gauss’s law, the electric field due to the patch is
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