Gauss’s Law

Consider a positive point charge Q located at the center o a sphere of radius r, as shown in
Figure 4.2.1. The electric field due to the charge Q is , which points in
the radial direction. We enclose the charge by an imaginary sphere of radius r called the
“Gaussian surface.”

Figure 1 A spherical Gaussian surface enclosing a charge Q.

In spherical coordinates, a small surface area element on the sphere is given by (Figure 2)

(1)

Figure 2 A small area element on the surface of a sphere of radius r.

Thus, the net electric flux through the area element is

 (2)
The total flux through the entire surface is

(3)

The same result can also be obtained by noting that a sphere of radius r has a surface area
, and since the magnitude of the electric field at any point on the spherical surface
is
, the electric flux through the surface is

(4)

In the above, we have chosen a sphere to be the Gaussian surface. However, it turns out that
the shape of the closed surface can be arbitrarily chosen. For the surfaces shown in Figure 3,
the same result is obtained. whether the choice is S1,S2,S3

Figure 3 Different Gaussian surfaces with the same outward electric flux.

The statement that the net flux through any closed surface is proportional to the net charge
enclosed is known as Gauss’s law. Mathematically, Gauss’s law is expressed as

(5)

where qenc is the net charge inside the surface. One way to explain why Gauss’s law holds is
due to note that the number of field lines that leave the charge is independent of the shape of
the imaginary Gaussian surface we choose to enclose the charge.
To prove Gauss’s law, we introduce the concept of the solid angle. Let be an
area element on the surface of a sphere S1 of radius r1 , as shown in Figure 4.

Figure 4

The solid angle Δ Ω subtended by at the center of the sphere is defined as

(6)
Solid angles are dimensionless quantities measured in steradians(sr). Since the surface area of
the sphere S1 is 4πrr 12 , the total solid angle subtended by the sphere is

(7)
The concept of solid angle in three dimensions is analogous to the ordinary angle in two
dimensions. As illustrated in Figure 5, an angle Δφ is the ratio of the length of the arc to the
radius of a circle:

Δφ=Δs/r (8)

Figure 5 The arc Δs subtends an angle Δφ

Since the total length of the arc is S=2πr, the total angle subtended by the circle is

(9)

In Figure 4, the area element makes an angle θ with the radial unit vector then the solid
angle

(10)

Where is the area of the radial projection of ΔA2 onto a second sphere s2
0f radius r2, concentric with S1.

(11)

Now suppose a point charge Q is placed at the center of the concentric spheres. The electric
field strengths E1 and E2 at the center of the area ΔA1 and ΔA2 are related by Coulomb’s law

(12)

(13)

(14)
Thus, we see that the electric flux through any area element subtending the same solid angle is
constant, independent of the shape or orientation of the surface. In summary, Gauss’s law
provides a convenient tool for evaluating electric field. However, its application is limited
only to systems that possess certain symmetry, namely, systems with cylindrical, planar and
spherical symmetry.

Infinitely Long Rod of Uniform Charge Density

An infinitely long rod of negligible radius has a uniform charge density λ. Calculate the
electric field at a distance r from wire.
Solution:
We shall solve the problem by following the steps outlined above.
(1) An infinitely long rod possesses cylindrical symmetry.
(2) The charge density is uniformly distributed throughout the length, and the electric field
must be point radially away from the symmetry axis of the rod (Figure 4.2.6). The magnitude
of the electric field is constant on cylindrical surfaces of radius. Therefore, we choose a
coaxial cylinder as our Gaussian surface.

Figure 6 Field lines for an infinite uniformly charged rod (the symmetry axis of the rod and
the Gaussian cylinder are perpendicular to plane of the page.)
(3) The amount of charge enclosed by the Gaussian surface, a cylinder of radius and length
(Figure 7), is

Figure 7 Gaussian surface for a uniformly charged rod.

(4) As indicated in Figure 7, the Gaussian surface consists of three parts: a two ends S1 and
S2 plus the curved side wall S3. The flux through the Gaussian surface is

(15)
where we have set E3=E. As can be seen from the figure, no flux passes through the ends

since the area vectors and are perpendicular to the electric field which points in the
radial direction.

(5) Applying Gauss’s law gives E(2πrl)=λl/Є0

(16)
The result is in complete agreement with that obtained in Eq.11, using Coulomb’s law. notice
that the result is independent of the length of the cylinder, and only depends on the
inverse of the distance r from the symmetry axis. The qualitative behavior of E as a function
of r is plotted in figure 8

Figure 8 Electric field due to a uniformly charged rod as a function of r.

The expression for the magnitude of the electric field ⃗E at a distance r from the axis
λ
E=
of the rod is 2 πε 0 r . The direction of ⃗E is radially outward if the charge is
positive; and radially inward if it is negative

Example : Infinite Plane of Charge

Consider an infinitely large non-conducting plane in the xy-plane with uniform surface charge
density σ. Determine the electric field everywhere in space.
Solution:
(1) An infinitely large plane possesses a planar symmetry.
(2) Since the charge is uniformly distributed on the surface, the electric field must point
perpendicularly away from the plane,
. The magnitude of the electric field is constant on planes parallel to the non-
conducting plane.
Figure 9 Electric field for uniform plane of charge

We choose our Gaussian surface to be a cylinder, which is often referred to as a “pillbox”

(Figure 10). The pillbox also consists of three parts: two end-caps S1 and S2, and a curved
side S 3.

Figure 10 A Gaussian “ Pillbox” for calculating the electricfield due to a large plane.

(3) Since the surface charge distribution on is uniform, the charge enclosed by the Gaussian
“pillbox” is , where is the area of the end-caps.

(4) The total flux through the Gaussian pillbox flux is

(17)
Since the two ends are at the same distance om the plane, by symmetry, the magnitude f the
electric field must be the same : E1=E2=E. Hence, the total flux can be rewritten as

(18)

(5) By applying Gauss’s law, we obtain

(19)

(20)

In unit-vector notation, we have

(20)
σ
E=
The magnitude of the electric field is 2 ε 0 , and its direction is perpendicular to
and away from the sheet. Since we are considering an infinite sheet with uniform charge
density, this result holds for any point at a finite distance from the sheet.

Thus, we see that the electric field due to an infinite large non-conducting plane is uniform in
space. The result, plotted in Figure 11, is the same as that obtained in Eq. (21) using
Coulomb’s law.

Figure 11 Electric field of an infinitely large non-conducting plane.

Note again the discontinuity in electric field as we cross the plane:

(21)

Example: Spherical Shell

A thin spherical shell of radius has a charge +Q evenly distributed over its surface. And the
electric field both inside and outside the shell.

Solutions: The charge distribution is spherically symmetric, with a surface charge density

σ = Q/ A s = Q/4πa 2 is the surface area of the sphere. The electric field ⃗

E must be radially
symmetric and directed outward (Figure 12). We treat the regions r≤a and r≥a separately

Figure 12 Electric field for uniform spherical shell of charge

Case 1: r≤a

We choose our Gaussian surface to be a sphere of radius r≤a as shown in Figure 13(a).

Figure 13 Gaussian surface for uniformally charged spherical shell for (a) r <a (b) r≥a
The charge enclosed by the Gaussian surface is q enc=0 since all the charge is located on the
surface of the shell. Thus, from Gauss’s law, , we conclude

E=0

Case 2: r≥a

In this case, the Gaussian surface is a sphere of radius r≥a as shown in Figure 13(b). Since the
radius of the “Gaussian sphere” is greater than the radius of the spherical shell, all the charge
is enclosed:

q enc=Q

since the flux through the Gaussian surface is

by applying Gauss’s law, we obtain

Figure 14 Electric field as a function of r due to a uniformly charged spherical shell.

As in the case of a non-conducting charged plane, we again see a discontinuity in E as we

cross the boundry at r=a. The change, from outer to the inner surface is given by,
Example: Non-Conducting Solid Sphere

An electric charge +Q is uniformly distributed throughout a non-conducting solid sphere of

radius a. Determine the electric field everywhere inside and outside the sphere.

Solution:
The charge distribution is spherically symmetric with the charge density given by

E is radially symmetric and

where V is the volume of the sphere. In this case, the electric field ⃗
directed outward. The magnitude of the electric field is constant on spherical surfaces of
radius . The regions r≤a and r≥a shall be studied separately.

Case 1: r≤.a

We choose our Gaussian surface to be a sphere of radius r≤a as shown in Figure 15(a).

Figure 15 Gaussian surface for uniformly charged solid sphere, for (a)r≤.a , and (b)r>a

The flux through the Gaussian surface is

With charge distribution, the charge enclosed is

(25)
which is proportional to the volume enclosed by the Gaussian surface. Applying Gauss’s law
Or

(26)

Case 2: r≥a
In this case, our Gaussian surface is a sphere of radius r≥a. as shown in Figure 15(b). Since the
radius of the Gaussian surface is greater than the radius of the sphere all the charge is enclosed
in our Gaussian surface: q enc=Q . With the electric flux through the Gaussian surface given by

upon applying Gauss’s law, we obtain

E(4πr 2)=Q/Є0 (or)

The field outside the sphere is the same as if all the charges were concentrated at the center of the
sphere. The qualitative behavior of E as a function of r is plotted in Figure 16.

Figure 4.2.16 Electric field due to a uniformly charged sphere as a function of r.