# Electric Flux. Gauss Law | iCalculator™

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14.6Electric Flux. Gauss Law

In this Physics tutorial, you will learn:

• What is area vector?
• What is flux of a quantity?
• What is electric flux and how can we calculate it?
• What is the electric flux of a point charge placed at centre of a sphere?
• What does the Gauss Law say?
• How to find the electric flux in some regular surfaces?
• How does electric flux depend on the incidence angle of the field lines?

## Introduction

Do you think the electric field depends on the number of lines crossing the surface of a charged plate?

What do you understand with the term "flux"? For example, what conclusion do you draw if somebody says, "the flux of people in the airport is large"?

The concept of "electric flux" is very important in understanding the behaviour of electric current, for which we will discuss in the next chapter. But first, we have to explain what is electric flux and how can we find it.

## The Meaning of Vector Area

It is known from geometry that area is a scalar quantity, which represents the amount of surface enclosed within the boundary lines of a figure. Area is measured in square metres (m2). For example, the area of a square of side length L is L × L = L2, the area of rectangle of sides a and b is a × b, the area of a circle of radius R is π × R2 and so on. In some cases, it is convenient to appoint a vector to a given area, in order to make it operable with other vector quantities. This trick is especially common in electromagnetism. Thus, instead of writing the surface area as A square units, we consider the corresponding area vector of A units which is perpendicular to the given surface. For example, the vector area of a rectangle of side lengths a = 5 cm and b = 4 cm has a magnitude of A = 5 cm × 4 cm = 20 cm2 (20 units) and it is perpendicular to the plane of rectangle, as shown in the figure. When area represents a positive quantity, the direction of its corresponding vector is in the outward direction, while when it represents a negative quantity, the corresponding vector is in the inward direction.

If another vector quantity is multiplied by a given area vector, we consider the angle between them as a part of multiplication. The introduction of area vector creates the possibility to apply both the dot and cross product of the two vectors, as needed. For example, if we have a vector quantity X which forms and angle θ to the area vector, as shown in the figure, From the definition of the dot (scalar) product of two vectors, we have

A × X = |A| × |X| × cos θ

and the cross (vector) product of the two given vectors (based on the definition of cross product of vectors) is

A × X = |A| × |X| × sin θ

When the angle between the two vectors is not given but is given the angle α between the vector X and the surface, we consider the complementary angle θ = 90° - α during the operations. ### Example 1

A vector V = 8 units forms an angle of 30° with the plane of a rectangle of dimensions 3 units and 5 units. The vector V extends in the outward direction. What is the magnitude of the cross product of the vector V and the area vector? ### Solution 1

The area vector A is

A = 3 × 5 = 15 square units

The angle between the two vectors is

θ = 90° - 30°
= 60°

The magnitude of cross product of A and V therefore is

A × V = |A | × |V | × sin θ
= 8 × 15 × sin 600
= 8 × 15 × √3/2
= 60√3

## What is Electric Flux?

It is convenient to explain the meaning of electric flux using the analogy with the flux of water flowing through a pipe.

Suppose a square-shaped loop placed at right angle to the flowing direction of water stream. Let A be the area of loop and v the flowing velocity of water. The rate of water flow through the loop (otherwise known as the water flux) is denoted by Φ and is calculated by

Φ = A × v

Since area is measured in [m2] and velocity in [m/s], the unit of water flux is [m3/s]. From the last unit, we can conclude that water flux represents the volume of water flowing through a loop in unit time (in every second). In simpler words, we can consider the water flux as the amount of water flowing through the loop in unit time.

For simplicity (and for a better understanding), the water stream in the above figure is represented through horizontal lines. Obviously, if water flow would be more intense, we had to write more lines in the figure.

This pattern is similar to when we represent the electric field using the field lines. Therefore, we can extend the concept of flux to represent the number of electric field flowing through a similar loop. In this way, we obtain the electric flux, which shows the amount (number) of electric field lines (i.e. the electric field E) flowing through a closed loop of area A. Mathematically, we have

ΦE = E × A Electric flux Φ is a scalar quantity because it is obtained through the dot product of two vectors, E and A. As stated erlier, we must consider the area vector in the calculations instead of the geometric area. In addition, we must consider during the operations the angle formed by the two vectors as well. Hence, the formula of electric flux becomes

Φ = E × A

Or

Φ = |E| × |A| × cos θ

When electric field lines are perpendicular to the loop plane, the two vectors E and A are parallel, i.e. they form an angle of 0° or 180° to each other. Since cos 0° = 1 and cos 180° = -1, the electric flux takes its maximum or minimum values for these angles.

The unit of electric flux is

[V/m] × [m2 ] = [V × m]

### Example 2

Two parallel plates are connected to a 3V battery as shown in the figure. A 5 cm × 8 cm metal loop is inserted between the plates at an angle of 30° to the vertical direction. What is the electric flux passing through the loop?

### Solution 2

We must consider the vector area instead of the surface area of the loop. It forms a 30° angle to the direction of electric field lines as shown in the figure. We calculate the electric field E using the formula

E = ΔV/d
= 3V/0.1m
= 30 V/m

The area A of the loop is

A = 8 cm × 5 cm
= 40 cm2
= 0.004 m2

The angle between the electric field lines and area vector is wide (obtuse). However, for simplicity, we can assume its corresponding acute angle but with negative sign. Thus, giving that cos 30° = √3/2 = 0.86, we obtain for the electric flux Φ:

Φ = -|E | × |A | × cos θ
= -30 V/m × 0.004 m2 × 0.86
≈ - 0.1 V × m

## Electric Flux of a Charged Sphere. Gauss Law

Let's compute the electric flux across a spherical surface of radius R that contains a nearly point charge Q at its centre. As we have explained earlier, electric field is the same in all points on the surface of the sphere. It is produced by the point charge Q and the field lines spread outwards (when Q is positive) from the charge to the surface of the sphere, in the perpendicular direction to the surface at any point. Therefore, the perpendicular component of the electric field summed across the entire surface is simply the electric field at a distance R from the charge multiplied by the area of the surface. In this case, the field vector is in the direction of area vector at every point. This means the angle between them is zero and cos 0° = 1. Thus,

Φ = |E | × |A | × cos θ
= E × A × cos 00
= E × A
= 1/4πϵ0 × Q/R2 × 4πR2

Thus, we obtain for the electric flux Φ of a charged sphere:

Φ = Q/ϵ0

Here we have made use of the equation of the area of a sphere A = 4πR2 and that of electric field of a point charge E = Q/4πϵ0R2.

The last formula is one of the many representations of the Gauss Law, which in simple word says:

The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface.

The sphere we considered above is called Gaussian Sphere. As you see, the electric flux does not depend on the radius of sphere but only on the amount of charge it carries at its centre.

### Example 3

A -7μC point charge is placed at centre of a sphere whose radius is equal to 50 cm. Calculate the electric flux on the surface of the sphere. ### Solution 3

The flux will be negative as the charge is negative. This means the electric field lines will be in the inward direction. From the Gauss Law, we see that electric flux of a sphere is independent from the radius R. Thus, we have

Φ = Q/ϵ0
= -7 × 10-6 C/8.85 × 10-12 C2/N × m2
= -7.9 × 105 V × m

## Electric Flux of a Charged Cube

Let's consider the electric flux across a cube of side length a when it is inserted inside a uniform electric field E. Assume the electric field lines are perpendicular to one of the cube faces. The four faces parallel to the field have zero flux as no field line crosses them. This is also clear when considering the electric flux formula, in which the cosine between the field lines and the area vectors is zero as they are perpendicular (cos 90° = 0).

From the two remaining faces, it is easy to see that in the left face there is an inwards flux while in the right face the flux is in the outward direction. Therefore, the total flux is zero as the area vectors in these two faces are opposite. This means for one of them the angle must be taken as zero (outward) while for the other face (inward) we must take the angle as 180°. Therefore, we obtain for the total electric flux flowing through the cube

Φtot = Φ1 + Φ2
= |E | × |A | × cos 1800 +|E | × |A | × cos 00
= |E | × |A | × (-1) + |E | × |A | × 1
= -|E | × |A | + |E | × |A |
= 0

In general, when electric field is at angle θ to the area vector, we obtain for the total electric flux flowing through a cube

Φtot = Φ1 + Φ2
= |E | × |A | × cos θ + |E | × |A | × cos (180 - θ)
= 0

### Example 4

What is the net electric flux for the triangular prism shown in the figure if the prism is placed inside a 20 V/m uniform electric field whose lines are horizontal? ### Solution 4

Electric field lines cross only two of the three rectangular faces of prism, because the bottom face and the two bases are parallel to the electric field lines.

First, we find the missing side of the prism using the Pythagorean Theorem. We have

c2 = 32 + 42 = 25

Thus, c = √25 = 5 cm.

The electric flux crossing the 4 cm × 10 cm face is in the inward direction, so we take it as negative, while that on the 5 cm × 10 cm face is in the outward direction, so we take it as positive.

The area vector of the left face is in the opposite direction of the electric field vector, so the angle between them is 180° (cos 180° = - 1). In addition, the magnitude of area vector on the left side (at the input) is

Ai = 4 cm × 10 cm
= 40 cm2
= 4 × 10-3 m2

Therefore, the inward flux is

Φi = |E | × |Ai | × cos 1800
= 20 V/m × 4 × 10-3 m2 × (-1)
= -8 × 10-2 V × m
= -0.08 V × m

The magnitude of area vector for the right face of prism (output) is

Ao = 5 cm × 10 cm
= 50 cm2
= 5 × 10-3 m2

Furthermore, the electric field lines are not in the direction of the area vector but they form an angle θ equal to that of the upper vertex (the sides are respectively perpendicular) as shown in the figure. We need the cosine of the angle θ to work out the flux, i.e.

= 3 cm/5 cm
= 0.6

Therefore, the outward flux is

Φo = |E | × |Ao | × cos θ
= 20 V/m × 5 × 10-3 m2 × 0.6
= 6 × 10-2 V × m
= 0.06 V × m

Hence, the net electric flux flowing through the prism is

Φnet = Φi + Φo
= -0.08 V × m + 0.06 V × m
= -0.02 V/m

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