Phys102 Lecture 4 Phys102 Lecture 4-1 Gauss s Law Key Points Electric Flux Gauss s Law Applications of Gauss s Law References SFU Ed: 22-1,2,3. 6 th Ed: 16-10,+.
Electric Flux Electric flux: The direction of area if defined as the normal direction. Electric flux through an area is proportional to the total number of field lines crossing the area.
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.
Flux through a closed surface: (Just a new notation) out (+) in (-)
22-2 Gauss s Law The total flux through a closed surface is equal to the charge enclosed divided by e 0 : Total E A This can be used to find the electric field in situations with a high degree of symmetry. To use Gauss s law to calculate E, we need to choose a closed surface such that it is easy to calculate the summation. Specifically, E is constant on the surface. Sometimes the surface can consists of a few parts, on each part E should be a constant or 0.
For a point charge, we choose a spherical surface as our Gaussian surface Therefore, Non-calculus version: Q e 0 E A E Solving for E gives the result we expect from Coulomb s law: 4 r 2 On every point of the GS, E = constant, and E // A
Gauss s Law and the source of E-field QEncl E A Total If there is a positive charge enclosed, the total outward flux is positive. This means the positive charge is are the source of the E-field lines. If there is a negative charge enclosed, the total outward flux is negative, meaning there is a net inward flux. This means the negative charge is a sink of the E-field lines. I-clicker question 4-1: The total flux of magnetic field through a closed surface should be: A) Always positive. B) Always negative. C) Always zero. D) Can be any value. E) No idea. e 0
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?
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 ). I-clicker question 4-2: This question is the same as part C of the previous question. A) True. B) False.
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.
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.
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/ε 0 where σ is the surface charge density on the conductor s surface at that point.
Procedure for Gauss s law problems: 1. Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). 2. Draw the surface. 3. Use the symmetry to find the direction of E. E 4. Evaluate the flux in terms of E. 5. Calculate the enclosed charge. 6. Solve for the field.
A conceptual question about Gauss s law QEncl E A Total Gauss s law only needs the enclosed charge inside the Gaussian surface. Any charges outside the GS do not appear in the formula. Why? (i-clicker question 4-3) A) Gauss s law doesn t apply when there are charges outside the Gaussian surface (GS). B) Charges outside the Gaussian surface do not contribute to the electric field inside the GS. C) We choose the GS surface such that the charges outside does not appear in the formula. D) Because of the symmetry. E) Charges outside the GS do not contribute to the total flux through the GS. e 0