Physics Lesson 15.1.4 - Appendix - Charge Density, Electric Field and Gauss Law

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Welcome to our Physics lesson on Appendix - Charge Density, Electric Field and Gauss Law, this is the fourth lesson of our suite of physics lessons covering the topic of Electric Current. Current Density, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Appendix - Charge Density, Electric Field and Gauss Law

Now that we explained the meaning of charge density in all possible forms (linear, surface and volume charge density), let's recall Gauss Law and extend its explanation further by taking into consideration the new concepts. Remember that Gauss Law involves electric flux Φ and its general mathematical form is

while the formula derived directly from definition of electric flux is

where E is the electric field, A is the area vector and θ is the angle between the area vector and electric field vector.

a. Transforming Gauss Law using the surface charge density concept

Let's consider a charged conductor of surface area A. Its surface charge density obviously is σ = Q/A where Q is the charge. From Gauss Law, we know that

For simplicity, let's consider the field lines as parallel to the area vector, i.e. cos θ = 1 because θ = 0⁰.

Given that Q = σ ∙ A, we can write

Simplifying the area A from both sides, we obtain for the electric field in terms of surface charge density

The above formula means that the magnitude of electric field outside a conductor is proportional to the surface charge density on the conductor. This is one of the most important assertions in Electromagnetism.

b. Transforming Gauss Law using the linear charge density concept. Cylindrical symmetry

Let's consider a long and very thin bar carrying a uniform linear charge density, λ. We want to find an expression for the electric field E in terms of linear charge density and distance r from the bar. For this, we must consider a cylinder of radius r and height h, and then, find the electric field produced on the lateral surface of it.

Again, the area vector A is parallel to the direction of electric field E. This means cos θ = 1.

From the definition of electric flux, we have

From Gauss Law, we have

Therefore, combining the two above equations, we obtain

Simplifying h from both sides and rearranging in order to isolate E, we obtain

Example 3

What is the electric field produced by a long bar of linear charge density λ = 600 μC/cm at a distance of 4 m from the bar?

Solution 3

Clues:

λ = 600 μC/cm = 60000 μC/m = 0.06 C/m = 6 × 10-2 C/m
r = 4 m
ϵ0 = 8.85 × 10-12 F/m
E = ?

From the equation of electric field in terms of linear charge density, we have

You have reached the end of Physics lesson 15.1.4 Appendix - Charge Density, Electric Field and Gauss Law. There are 4 lessons in this physics tutorial covering Electric Current. Current Density, you can access all the lessons from this tutorial below.

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