This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Patel, Kinal – Homework 4 – Due: Feb 1 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A uniformly charged conducting plate with area A has a total charge Q which is positive. The figure below shows a cross-sectional view of the plane and the electric field lines due to the charge on the plane. The figure is not drawn to scale. E E + Q P Find the magnitude of the field at point P, which is a distance a from the plate. Assume that a is very small when compared to the dimensions of the plate, such that edge effects can be ignored. 1. k ~ E P k = Q ² A 2. k ~ E P k = Q 2 ² A correct 3. k ~ E P k = 4 π ² aQ 4. k ~ E P k = 2 ² QA 5. k ~ E P k = ² QA 6. k ~ E P k = ² Qa 2 7. k ~ E P k = Q 4 π ² a 2 8. k ~ E P k = Q 4 π ² a 9. k ~ E P k = 4 π ² a 2 Q Explanation: Basic Concepts Gauss’ Law, electrostatic properties of conductors. Solution: Consider the Gaussian surface shown in the figure. E + Q E S Due to the symmetry of the problem, there is an electric flux only through the right and left surfaces and these two are equal and σ = Q A . If the cross section of the surface is S , then Gauss’ Law states that Φ TOTAL = 2 E S = 1 ² Q A S , since Φ TOTAL ≡ I ~ E · d ~ A, so E = Q 2 ² A . 002 (part 2 of 2) 10 points Two uniformly charged conducting plates are parallel to each other. They each have area A . Plate #1 has a positive charge Q while plate #2 has a charge- 3 Q . + Q #1- 3 Q #2 P x y Using the superposition principle find the magnitude of the electric field at a point P in the gap. 1. k ~ E P k = Q ² A 2. k ~ E P k = 3 Q ² A Patel, Kinal – Homework 4 – Due: Feb 1 2008, 7:00 pm – Inst: Weathers 2 3. k ~ E P k = 4 Q ² A 4. k ~ E P k = 3 Q 2 ² A 5. k ~ E P k = 5 Q ² A 6. k ~ E P k = Q ² 7. k ~ E P k = 2 Q ² A correct 8. k ~ E P k = Q 3 ² A 9. k ~ E P k = Q 2 ² A 10. k ~ E P k = 0 Explanation: According to the result of part 1, the electric field generated by plate #1 at P is E 1 = Q 2 ² A directed along the positive x-axis....
View Full Document
This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.
- Spring '08