Chapter 24 Coulomb’s Law and Gauss’ Law

Coulomb’s Law

F = (k q₁ q₂ )/ r² k = 9 x 10⁹ for standard metric units (MKS system)

This calculation gives the magnitude of the electrical force.

Coulomb determined the mathematical form of this law in 1785. The value of the constant depends on the system of units being used. The constant was determined in the early 1800’s with the use of electrical circuits and magnetic force.

page 2 We will use MKS units

We will discuss amperes and watts in the chapter that discusses electrical circuits.

The more formal representation of k is k = 1 / (4 πε_o ) ε_o = 8.85 x 10^-12

We will work some examples with Coulomb’s Law.

You should repeat the work done in the text in example 2. What questions do you have on example 2?

Work Exercise 1

Pages 24-6 to 24-9 will be covered by the 162 class only.

page 10 The Electric Field

CASE A Suppose that a 2 m C charge is fixed at the origin and a -3 m C charge is fixed at the 5 cm position on the X axis. Calculate the force on a 4 m C charge located at the 9 cm position on the X axis.

As the number of charges increases and/or the charges are not along one coordinate axis, the difficulty of electrostatic problems increases.

CASE B Suppose that a 2 m C charge is fixed at the origin and a -3 m C charge is fixed at the 5 cm position on the Y axis. Calculate the force on a 4 m C charge located at the 9 cm position on the X axis.

A similar difficulty arises with gravitational problems. The use of the "field" concept can simplify the calculations of the electric force. In gravitational problems, weight = mass * gravitational field, W = mg. In electrical problems, Force = charge * electric field F = q E .

Experimental determination of E.

Suppose that a small "test" charge is placed at a point in space. The test charge does not significantly disturb the positions of any other charges in the problem. The force on the test charge is measured. The amount of charge on the test charge is known. Is it possible to calculate the value for E at this point in space? The length of the arrow is proportional the magnitude of F and E.

Calculation of E

Compare F = q_test E and F = (k q_test q₂ )/ r² If these are combined can you solve for E in symbols?

The net electric Field E is the vector sum of the individual E’s. E = E₁ + E₂

Calculate the value of E at a point 9 cm on the X axis for each of the following:

Suppose that a 2 m C charge is fixed at the origin and a -3 m C charge is fixed at the 5 cm position on the X axis.

Suppose that a 2 m C charge is fixed at the origin and a -3 m C charge is fixed at the 5 cm position on the Y axis.

For both cases above, calculate the force on a 4 m C charge located at the 9 cm position on the X axis.

page 12 Electric Field Lines

Electric Field lines do not exist. In this chapter, they are a convenience used to help us imagine the path of a moving charged particle. Electric field lines always begin on a positive charge and end on a negative charge.

You should visit the web site (http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html ) and view electric field lines for some charge distributions. Read the instructions before you click any buttons. Remove all check marks except the one for Electric Field Lines.

Suppose a positive charge is fixed at the origin. Describe the path of motion for a positive charge released from X = 1 cm, Y = 0 cm; X = 0 cm, Y = 1 cm; X = -1 cm, Y = 0 cm; X = 0 cm, Y = -1 cm.

Suppose a positive charge is fixed at the origin and a negative charge is fixed at 8cm on the X axis. A positive charge is released from rest at the position X = 1 cm, Y = 1 cm. Describe the path of its motion.

14 Continuity Equation for Electric Fields

You can skip this page. We will use flux and Gauss’s Law to find the electric field.

15 Flux

The concept of fluid flow was a help to scientists in the early 1800’s as the theories of electricity and magnetism were developed. The electric flux, Φ, is calculated by multiplying the value of the electric field by the area the field is passing through. Φ = E A E and A must be perpendicular. If they are not perpendicular then Φ = E A cos θ, where θ is the angle between the electric vector and the area vector. The area vector is a vector that is perpendicular to the area. The concept of electric flux is shown in Figure 17.

The electric flux due to a certain charge is constant. As the book shows, if the area is larger then the value of the electric field must be smaller in order to maintain a constant flux value. Since the area gets larger by a factor of r² , the electric field must decrease with distance. E is proportional to 1/ r² , and E = Φ/(4 π r²) . This is in agreement with the way we calculated E using results from Coulomb’s Law.

In order that this flux method of finding E gives the same result as the Coulomb’s Law method the size of the flux must be equal to the charge inside the area divided by ε_o Φ = q_inside/ ε_o .

17 Negative Charge

A positive charge creates a positive flux value. The E field lines are pointed away from the charge and the angle between E and the area vector is 0 degrees.

A negative charge creates a negative flux value. The E field lines are pointed toward the charge and the angle between E and area vector is 180 degrees. Recall: cos(180 degrees) = -1 .

Conserved Field Lines

Electric field lines start at positive charges and terminate at negative charges. There is no change in the number of E field lines in the space between the charges.

A Mapping Convention

The number of field lines is proportional to the charge. We will not take the time to count the number of field lines. I will try to use more field lines when the charge value is larger.

Speculate as to where the two field lines that exit Figure 21 to the right finally terminate.

18 Summary

Field line rules: E field lines start and point away from positive charges.

E field lines end and point toward negative charges.

The field lines show the path a positive test charge would travel if released from rest.

20 Gauss’ Law

Gauss’ Law is used to derive an equation that allows the magnitude of the electric field to be calculated as a function of position in space. We will use Gauss’ Law to easily and quickly derive the equation. The equation will then allow us to easily calculate the value of the electric force.

Gauss’ Law Φ = q_inside/ ε_o

Recall that Φ = E A cos θ The area of interest is an area on a three dimensional surface. This surface is called a "Gaussian Surface." We will select the shape of the three dimensional surface to ensure that θ (the angle between E and the area vector) is 0 or 90 degrees. We will also set up the geometry of the surface such that E has a constant value on each part of the surface where the angle is 0 degrees. The only shapes we will use are the sphere, cylinder and cube. Problems that do not meet these conditions are beyond the scope of this course.

Example: Find the formula for the electric field for the case of a single positive charge, q.

Solution: We know that the electric field lines leave the charge in all directions. What shape can we put around the charge such that the angle of the electric field lines to the area vector is 0 or 90 degrees?

The flux is equal to E (4 π r² ) cos (0 degrees) . Gauss’ Law then tells us that E (4 π r² ) = q/ ε_o

Solving for E produces: E = q/ (ε_o 4 π r² ) or E = kq/ r²

page 21 Electric Field for a Line Charge Let the line of charge be infinitely long with λ Coulombs/meter. * What Gaussian shape should be used? sphere or cylinder or cube?

Are there any parts of the shape on which the angle between E and the area vector is 90 degrees?

If so, ignore those parts in writing the expression for the flux.

Is E a constant on the other portions of the area? If so write the expression for the flux.

Write the expression for the amount of charge that is inside the Gaussian surface.

If the cylinder has a length L the amount of charge is λ L .

Make the substitutions into Gauss’ Law and solve for E.

page 22 The topics on this page were discussed above.

page 23 - 25 Gauss’ Law for the Gravitational Field

We will skip this topic. The PHY162 class might discuss this topic.

page 26 Summary of the method for Solving Gauss’ Law Problems

What questions do you have on the method? We will work examples to find E using Gauss’ Law.

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