Page 152 - Electromagnetics

2
the charge is distributed uniformly with surface density ρ s = Q/4πa , producing a field
2
E = ˆ rQ/4π r external to the sphere. Hence a force density
1 Q 2
t = ˆ r
2 2
2 (4πa )
acts at each point on the surface. This would cause the sphere to expand outward if the
structural integrity of the material were to fail. Integration over the entire sphere yields

1 Q 2
F = ˆ r dS = 0.
2 2
2 (4πa ) S
However, integration of t over the upper hemisphere yields

1 Q 2 2π π/2 2
F = ˆ ra sin θ dθ dφ.
2 2
2 (4πa ) 0 0
Substitution of ˆ r = ˆ x sin θ cos φ + ˆ y sin θ sin φ + ˆ z cos θ leads immediately to F x = F y = 0,
but the z-component is

1 Q 2 2π π/2 2 Q 2
F z = a cos θ sin θ dθ dφ = .
2 2
2 (4πa ) 0 0 32 πa 2
¯
This result can also be obtained by integrating −T e · ˆ n over the entire xy-plane with
¯

ˆ n =−ˆ z. Since −T e · (−ˆ z) = ˆ z E · E we have
2
1 Q 2 2π ∞ rdrdφ Q 2
F = ˆ z = ˆ z .
2 (4π ) 2 0 a r 4 32 πa 2
As a more challenging example, consider two identical line charges parallel to the z-
axis and located at x =±d/2, y = 0 in free space. We can find the force on one line
charge due to the other by integrating Maxwell’s stress tensor over the yz-plane. From
(3.64)we find that the total electric field on the yz-plane is
y ρ l
E(y, z) = ˆ y
2 2
y + (d/2) π 0
¯
where ρ l is the line charge density. The force density for either line charge is −T e · ˆ n,
where we use ˆ n =±ˆ x to obtain the force on the charge at x =∓d/2. The force density
for the charge at x =−d/2 is
2
1 0 y ρ l

¯ ¯
T e · ˆ n = (D · E)I · ˆ x − DE · ˆ x = 2 2 ˆ x
2 2 y + (d/2) π 0
and the total force is
ρ l y
∞ ∞
2 2
F − =− 2 ˆ x dy dz.
2
2
−∞ −∞ 2π 0 y + (d/2) 2
On a per unit length basis the force is
ρ 2 ∞ y 2 ρ 2
F − l l
=−ˆ x dy =−ˆ x .
2 2
2
2
l 2π 0 −∞ [y + (d/2) ] 2πd 0
Note that the force is repulsive as expected.
© 2001 by CRC Press LLC