1. Chapter 22 Gauss’s Law HW 3: Chapter 22: Pb.1, Pb.6, Pb.24,
Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces.
HW 3: Chapter 22: Pb.1, Pb.6, Pb.24,
Pb.27, Pb.35, Pb.46 Due Monday, Feb. 13

2. 22-3 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 r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).
Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).
b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).

3. 22-3 Applications of Gauss’s Law
Example 22-5: Nonuniformly charged solid sphere.
Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.
Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05.
b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05.

4. 22-3 Applications of Gauss’s Law
Example 22-6: Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends.
Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R.

5. 22-3 Applications of Gauss’s Law
Example 22-7: Infinite plane of charge.
Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane.
Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0.

6. 22-3 Applications of Gauss’s Law
Example 22-8: Electric field near any conducting surface.
Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by
E = σ/ε0
where σ is the surface charge density on the conductor’s surface at that point.
Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0.

7. 22-3 Applications of Gauss’s Law
The difference between the electric field outside a conducting plane of charge and outside a non-conducting plane of charge can be thought of in two ways:
The field inside the conductor is zero, so the flux is all through one end of the cylinder.
The non-conducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.

8. 22-3 Applications of Gauss’s Law
Procedure for Gauss’s law problems:
Identify the symmetry, and choose a Gaussian surface that takes advantage of it (with surfaces along surfaces of constant field).
Draw the surface.
Use the symmetry to find the direction of E.
Evaluate the flux by integrating.
Calculate the enclosed charge.
Solve for the field.

9. Chapter 23 Electric Potential
Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

10. 23.1 Electrostatic Potential Energy
•Gravitational Potential Energy
•Electric Potential Energy
•UE can result from either an attractive force or a repulsive force
•Ug is due to an attractive force + - r r
•Ug 0 when r ∞
•UE 0 when r ∞

11. 23.1 Electrostatic Potential Energy
•Gravitational Ug
Attractive force
•Electric UE
Attractive force
•Electric UE
Repulsive force - + + + or - -
> 0
< 0
Bigger when farther
Bigger when farther
Bigger when closer U r U r U r

12. Work and Electrostatic Potential energy- 1 FE 1 FE + 2 2 + +
∆U > 0
Work done BY conservative force
∆U < 0 U r U r 2
Move in the direction of force -->positive work 1 1
Move opposite the direction of force --> negative work 2