Gauss’ Law: Electric Flux, Formula, Explanation, and Solved Problems

17.3 GAUSS' LAW

17.3.1 Electric Flux

1. Concept of flux

Suppose we place a flat surface of area 1 cm² in an electric field produced by an isolated point charge Q, as shown in Figure 17.9. The surface is ‘immersed’ inside a region full of electric field lines. We orientate the surface until it is perpendicular to the electric field lines. Now we have a certain number of electric field lines ‘flowing’ perpendicularly through the surface. If the surface is closed to the point charge Q, there will be many lines flowing through the surface. Conversely, if the surface is far away from Q, then there will be very few lines flowing through the surface. This concept of some ‘number of lines’ flowing through the surface is known as flux. 

Figure 17.22

2. Electric flux

(a) Surface area in electric field

Consider a small flat surface KLMNK of area Aplaced in an electric field. The surface makes an angle θwith the line JK, which is perpendicular to the electric field lines, as shown in Figure 17.22(a). JKLHJ forms the flat surface whose area $A_n$ is perpendicular to the electric field. Actually $A_n$ is the component of area A, given by

$A_n=A \ cos \ \theta$

(b) Number of field lines flowing through $A_n$

The number of field lines flowing through the normal area $A_n$ depends on the following quantities:

(i) The electric field strength E.

More field lines flow through $A_n$ in a strong electric field strength than in a weak field.

(ii) The size of the surface area $A_n$.

More field lines flow through a large area than through a small area. 

(c) The electric flux Φ

The electric flux Φ is a measure of the number of field lines that flow through the normal area $A_n$. Its value has to be determined by the values of Eand $A_n$. It is given by

$\Phi= EA_n$

or,

$\Phi= EA \ cos \ \theta$

3. Electric flux in terms of vectors

Suppose the field lines flow through a small elemental surface area at angle $\theta$ to the normal, as shown in Figure 17.22(b). We have the following vectors at the surface:

(a) The electric field strength $E$ is a vector whose direction points at an angle $\theta$ to the normal of the small surface area.

(b) We can attach a unit vector $n$ to the elemental surface area that points in a direction that is parallel to the normal. Then the small area of magnitude $\mathrm{\Delta }\mathrm{A }$may be considered to be a vector $\Delta A$ whose direction is parallel to the direction of unit vector $n$. We can express this elemental area vector as

$\Delta A=\Delta A_n$

Now the electric flux $\Phi$ can be written as a dot product of the $E$ vector and the $\Delta A$ vector, as follows:

$$\Phi = E \cdot \Delta A$$

$$$$

4. Quantity

Electric flux is a scalar quantity.

5. Unit of Φ

Electric flux has a unit of

$N.m^2.C^{-1}$

17.3.2 Electric Flux Through a Large Surface

1. Consider a large surface that is found in a non-uniform electric field. We wish to determine the electric flux that flows through this large surface.

2. Since the electric field is not uniform, the electric field strength Eon one part of the surface may be larger or smaller than the strength at another part of the surface. In other words, the electric field strength varies from point to point on the surface. As a result, the electric flux flowing through one small surface area may be larger or smaller than that flowing through another small surface of equal area. 

Figure 17.23

3. To determine the electric flux flowing through the entire surface, we divide the large surface into a very large number of tiny surfaces as shown in Figure 17.23. Each surface is so small that:

(a) it may be assumed to be flat
(b) the electric field strength at a point on it is assumed to be constant 

4. Suppose we divide the large surface into n tiny surfaces. Consider one tiny surface of area ΔA₁ where the electric field strength is E₁. Let the angle between E₁ and the normal to this surface be θ₁. The electric flux through this tiny surface is 

$\Delta \Phi_1=E_1 \Delta A_1 \ cos \ \theta_1$

The electric field strength at a point on another tiny surface of area $\Delta A_2$ has magnitude $E_2$, pointing at an angle $\theta_2$ to the normal of the second tiny surface. The electric flux $\Delta \Phi_2$ flowing through the second tiny surface is given by

$\Delta \Phi_2=E_2 \Delta A_2 \ cos \ \theta_2$

Using the same prosedure, we determine the electric flux flowing through the rest of the tiny surfaces and obtain electric flux $\Delta \Phi_3$, $\Delta \Phi_4$, $\Delta \Phi_5$,..., $\Delta \Phi_n$. The electric flux $\Phi$ flowing throught the entire large surface is given by

$\Phi=\Delta \Phi_1+\Delta \Phi_2+\Delta \Phi_3+...+\Delta \Phi_n$

$=\sum_{i=1}^{n}\Delta \Phi_i$

$=\sum_{i=1}^{n}(E_i\Delta A_i \ cos \ \theta_i)$

5. We may express the electric flux in terms of vectors, as follows:

$\Phi=\sum_{i=1}^{n}(\textbf{E}_i\Delta \textbf{A}_i)$

if we let $\Delta A\to 0$, we will obtain

$\Phi=\int_{}^{}\textbf{E}.d\textbf{A}$

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