Gauss's Law II

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Gauss’s law II If a lightning bolt strikes a car (unless it is a convertible), electrons would quickly (in the order of nano seconds) move to the surface, so it is almost an ideal conductive shell that you can hide inside during a lightning storm!

Reading: Mazur 24.7-25.2 Deriving Gauss’s law Applying Gauss’s law Electric potential energy Electrostatic work

LECTURE 7 1 24.7 Deriving Gauss’s law Gauss’s law states that the flux through a closed surface is proportional to the enclosed charge, �, such that

� Φ = �

LECTURE 7 2 Question: 24.7-1 (Flux from point charge) Consider a spherical Gaussian surface of radius � with a point charge � at the center. � ⃗ The flux through a surface is given by Φ = ∮ � ⋅ ��. Using this definition for flux and Coulomb's law, calculate the flux through the Gaussian surface in terms of �, �, � and any constants.

LECTURE 7 3 Question: 24.7-1 (Flux from point charge) answer Consider a spherical Gaussian surface of radius � with a point charge � at the center. � ⃗ The flux through a surface is given by Φ = ∮ � ⋅ ��. Using this definition for flux and Coulomb's law, calculate the flux through the Gaussian surface in terms of �, �, � and any constants. The electric field is always normal to the Gaussian surface and has equal magnitude at all points on the surface. Therefore � Φ = � ⋅ ��⃗ = �� = � �� = � 4�� = � 4�� = 4�� From Gauss’s law Φ = Therefore Φ = = 4��, and � =

LECTURE 7 4 Reading quiz (Problem 24.70) There is a Gauss's law for gravity analogous to Gauss's law for electricity. If the electric flux through a closed surface is proportional to the enclosed charge, what is the gravitational flux proportional to? A. acceleration B. energy C. force D. mass Write an equation for Gauss's law for gravity for the mass �. Express your answer in terms of the variable �, the gravitational constant �, and �.

Φ = −4�� Φ = ∮ �⃗ ⋅ ��⃗ = ∮ −�� = −� ∮ �� = −� 4�� = − � 4�� = −4��

LECTURE 7 5 Question: 24.7-2 (Flux from uniformly charged rod)

An infinitely long charged rod has uniform charge density of λ, and passes through a cylinder (gray). The cylinder in case 2 has twice the radius and half the length compared to the cylinder in case 1. What is the ratio of flux through the cylinder in case 1 over the flux through the cylinder in case 2, ,? ,

LECTURE 7 6 Question: 24.7-2 (Flux from uniformly charged rod) answer

Gauss’s law: Φ = Case 1 has twice the charge enclosed as case 2.

LECTURE 7 7 Question: 24.7-3 (Gaussian surface near 2 charges)

LECTURE 7 8 Question: 24.7-3 (Gaussian surface near 2 charges) answer

The figure shows field lines for a configuration of charges. The Gaussian surface is shown in red. If the left charge were removed, which of the following statements would be correct. A. The flux through the surface would remain the same. B. The electric field at the surface would remain the same. C. None of the above. D. Not enough information provided. The charge enclosed has not changed, so the flux through the surface has not changed. The charge distribution has changed, so the electric field has changed.

LECTURE 7 9 24.8 Applying Gauss’s law Gauss’s law allows you to calculate the electric field for charge distributions that exhibit spherical, cylindrical, or planar symmetry.

LECTURE 7 10 Question: 24.8-1 (Gaussian surface choice)

You are told to use Gauss‘s Law to calculate the electric field at a distance � away from a charged cube of dimension �. Select all of the following Gaussian surfaces suited for this purpose? A. A B. B C. C D. This field cannot be calculated using these surfaces

LECTURE 7 11 Question: 24.8-1 (Gaussian surface choice) answer

LECTURE 7 12 Question 24.8-2 A thin spherical metal shell of radius R carries a uniformly distributed charge q. It is surrounded by a concentric thin spherical metal shell of radius 2R that carries a uniformly distributed charge –q. For which of the following regions enclosed by a concentric spherical Gaussian surface of radius r is the electric field zero? Answer all that apply. A. r < R B. R < r < 2R C. r > 2R

LECTURE 7 13 Question 24.8-2 answer For which of the following regions enclosed by a concentric spherical Gaussian surface of radius r is the electric field zero? Answer all that apply. A. r < R B. R < r < 2R C. r > 2R q Since the charge distribution and the Gaussian surface have spherical symmetry, electric field is constant over and perpendicular to the Gaussian surface. For r < R, no charge is enclosed by the Gaussian surface. -q For R < r < 2R, Φ = = ∮ � ⋅ ��⃗ = � ∮ �� = �4�� For r > 2R, net charge of zero is enclosed by the Gaussian surface.

LECTURE 7 14 25.1 Electric potential energy The electric potential energy is associated with the relative configuration of charged objects. Electric potential energy decreases for ◦ decreased separation of oppositely charged objects ◦ increased separation of like charged objects.

LECTURE 7 15 Reading quiz (Problem 25.01)

Consider an electric dipole in a uniform electric field. (Let the system comprise both the electric dipole and the sources of the uniform electric field.) What orientation of the dipole has the greatest electric potential energy? when the dipole moment and the electric field are antiparallel What orientation has the least? when the dipole moment and the electric field are parallel

When the dipole is parallel to the electric field, the dipole is most stable, minimizing the potential energy energy.

LECTURE 7 16 25.2 Electrostatic work Electrostatic work is the work by the electrostatic field. The electrostatic work done on a charged particle as it moves depends only on the positions of the endpoints and is independent of the path taken. The electrostatic work done on a charged particle is proportional to the charge carried by that particle. The potential difference between point A and point B in an electrostatic field is equal to the negative of the electrostatic work done on a charged particle as it moves from A to B divided by the charge of the particle.

LECTURE 7 17 Reading quiz (problem 25.08) In an electrostatic field, path 1 between points A and B is twice as long as path 2. The electrostatic work done on a negatively charged particle that moves from A to B along path 1 is �. How much work is done on this particle if it later goes from A to B along path 2?

Express your answer in terms of �.

� = � The electrostatic work done on a charged particle as it moves depends only on the positions of the endpoints and is independent of the path taken.

LECTURE 7 18 Question: 25.2-1 (Electrostatic work)

A positively charged particle is moved in the electric field of the large, stationary, positively charged object in the figure. Rank the electrostatic work done on the particle along these three paths A→B, A→C, C→B. Rank these from lowest to highest and compare the work to 0 J.

LECTURE 7 19 Question: 25.2-1 (Electrostatic work) answer

�→ = �→ < �→ = 0 J No work is done if the displacement is perpendicular to the force. The parallel displacement along paths A→B and C→B is the same, and in both cases it is antiparallel to the constant force.

LECTURE 7 20 Question: 25.2-2 (Sign of potential difference)

Does the sign of the potential difference between points A and B depend on if we moved a positive or negative particle from points A to B? A. Yes B. No

LECTURE 7 21 Question: 25.2-2 (Sign of potential difference) answer

Does the sign of the potential difference between points A and B depend on if we moved a positive or negative particle from points A to B? A. Yes B. No

For a positive particle, �→ is negative, and for a negative particle, �→ is positive. The potential difference between A and B is the negative of the electrostatic work divided by the charge of the particle, so in both cases it is positive. The potential is a description of space, so it does not depend on the charge of the particle.

LECTURE 7 22