Charles Allison © 2000 Chapter 22 Gauss’s Law.

Charles Allison © 2000 Problem 57

Charles Allison © Motion of a Charged Particle in an Electric Field The force on an object of charge q in an electric field is given by: = q Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.

Charles Allison © Electric Dipoles An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance.The dipole moment, p= Q, points from the negative to the positive charge.

Charles Allison © Electric Dipoles An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:

Charles Allison © Electric Dipoles The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r 3 dependence :

Charles Allison © 2000 Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area Electric Flux

Charles Allison © Electric Flux Example 22-1: Electric flux. Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.

Charles Allison © Electric Flux-Problem 4

Charles Allison © 2000 Flux through a closed surface : 22-1 Electric Flux E leaving the surface θ< π/2 E entering the surface θ> π/2 So the net flux out of the volume is 0 If the flux is negative there is no net flux into the volume

Charles Allison © Gauss’s Law-Problem 7 Problem 7.(II): In Fig. 22–27, two objects O 1 and O 2, have charges 1.0 μC and -2.0 μC respectively, and a third object, O3 is electrically neutral. (a) What is the electric flux through the surface A 1 that encloses all the three objects? (b) What is the electric flux through the surface A 2 that encloses the third object only?

Charles Allison © 2000 The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry Gauss’s Law

Charles Allison © Gauss’s Law For a point charge, Therefore, Solving for E gives the result we expect from Coulomb’s law:

Charles Allison © Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Looking at the arbitrarily shaped surface A 2, we see that the same flux passes through it as passes through A 1. Therefore, this result should be valid for any closed surface.

Charles Allison © Gauss’s Law Conceptual Example 22-2: Flux from Gauss’s law. Consider the two Gaussian surfaces, A 1 and A 2, as shown. The only charge present is the charge Q at the center of surface A 1. What is the net flux through each surface, A 1 and A 2 ?

Charles Allison © Applications of Gauss’s Law Example 22-3: Spherical conductor. A thin spherical shell of radius r 0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere?

Charles Allison © Applications of Gauss’s Law Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r 0. Determine the electric field (a) outside the sphere (r > r 0 ) and (b) inside the sphere (r < r 0 ).

Charles Allison © Applications of Gauss’s Law Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρ E = αr 2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r 0. (b) Find the electric field as a function of r inside the sphere.