2.1 Electrostatic Phenomena
13
surface S is equal to the total electric charge inside S divided by 0 . Hence, the total
number of electric field lines in Fig. 2.1 is Q// 0 .
Using Gauss’ mathematical theorem, the left side of (2.3) can be rewritten as
V
∇ · EdV =
1
0
V
ρ(r)dV .
(2.5)
In the above ∇ represents a differential operator called nabla and is expressed as
∇ = i x
∂
∂x
+ i y
∂
∂y
+ i z
∂
∂z
(2.6)
in Cartesian coordinates (x, y, z), where i x , i y and i z are unit vectors along the x, y
and z axes, respectively. In (2.5) ∇ · E is also written as divE, where div is called the
divergence. Since (2.5) holds for an arbitrary region V, the integrands on the both
sides are equal to each other:
∇ · E =
ρ
0
.
(2.7)
This is called Gauss’ divergence law. In Cartesian coordinates the left side of (2.7)
is
∇ · E =
∂E x
∂x
+
∂E y
∂y
+
∂E z
∂z
.
(2.8)
Equation (2.7) shows that electric charge causes a divergence of the electric field.
In the case of the point charge shown in Fig. 2.1, for example, the electric field
emerges from the origin at which the point charge is located and expands radially.
This condition is similar to the light emitted from the point source. An electric field
line does not emerge from or terminate at a point where there is no electric charge
(ρ = 0). That is, there is no divergence (∇ · E = 0).
Here we show an example. Suppose that electric charge is uniformly distributed
with density ρ 0 inside a wide slab of thickness 2a parallel to the y-z plane (see
Fig. 2.2a). The electric field is now calculated using Gauss’ law (2.3). In this case
the calculation using Coulomb’s law is not easy. We assume a closed parallelepiped
S, one plane of which stays on the central plane, x = 0, as shown in Fig. 2.2b.
Then, the electric field must be zero on this plane from symmetry with respect to the
x-axis. That is, the same result must be obtained when the right and left sides are
reversed. It can be concluded that the electric field has only the x component (E x )
from symmetry with respect to the y- and z-axes. Hence, the electric field is parallel
to the surface on the four surfaces parallel to the x-axis, and the surface integral of
the electric field on these surfaces is zero. The surface integral only has a nonzero
value on the remaining surface. The position and the electric field on this surface are
denoted by x and E x (x), respectively. Then, the right side of (2.3) is AE x (x), where A
13
surface S is equal to the total electric charge inside S divided by 0 . Hence, the total
number of electric field lines in Fig. 2.1 is Q// 0 .
Using Gauss’ mathematical theorem, the left side of (2.3) can be rewritten as
V
∇ · EdV =
1
0
V
ρ(r)dV .
(2.5)
In the above ∇ represents a differential operator called nabla and is expressed as
∇ = i x
∂
∂x
+ i y
∂
∂y
+ i z
∂
∂z
(2.6)
in Cartesian coordinates (x, y, z), where i x , i y and i z are unit vectors along the x, y
and z axes, respectively. In (2.5) ∇ · E is also written as divE, where div is called the
divergence. Since (2.5) holds for an arbitrary region V, the integrands on the both
sides are equal to each other:
∇ · E =
ρ
0
.
(2.7)
This is called Gauss’ divergence law. In Cartesian coordinates the left side of (2.7)
is
∇ · E =
∂E x
∂x
+
∂E y
∂y
+
∂E z
∂z
.
(2.8)
Equation (2.7) shows that electric charge causes a divergence of the electric field.
In the case of the point charge shown in Fig. 2.1, for example, the electric field
emerges from the origin at which the point charge is located and expands radially.
This condition is similar to the light emitted from the point source. An electric field
line does not emerge from or terminate at a point where there is no electric charge
(ρ = 0). That is, there is no divergence (∇ · E = 0).
Here we show an example. Suppose that electric charge is uniformly distributed
with density ρ 0 inside a wide slab of thickness 2a parallel to the y-z plane (see
Fig. 2.2a). The electric field is now calculated using Gauss’ law (2.3). In this case
the calculation using Coulomb’s law is not easy. We assume a closed parallelepiped
S, one plane of which stays on the central plane, x = 0, as shown in Fig. 2.2b.
Then, the electric field must be zero on this plane from symmetry with respect to the
x-axis. That is, the same result must be obtained when the right and left sides are
reversed. It can be concluded that the electric field has only the x component (E x )
from symmetry with respect to the y- and z-axes. Hence, the electric field is parallel
to the surface on the four surfaces parallel to the x-axis, and the surface integral of
the electric field on these surfaces is zero. The surface integral only has a nonzero
value on the remaining surface. The position and the electric field on this surface are
denoted by x and E x (x), respectively. Then, the right side of (2.3) is AE x (x), where A