HW 4 solution

# HW 4 solution - song(shs546 Homework 4 weathers(22202 This...

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song (shs546) – Homework 4 – weathers – (22202) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points A uniFormly charged conducting plate with area A has a total charge Q which is positive. The fgure below shows a cross-sectional view oF the plane and the electric feld lines due to the charge on the plane. The fgure is not drawn to scale. E E + Q P ±ind the magnitude oF the feld 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 e²ects can be ignored. 1. b v E P b = 4 π ǫ 0 a 2 Q 2. b v E P b = Q 4 π ǫ 0 a 3. b v E P b = ǫ 0 Q A 4. b v E P b = 4 π ǫ 0 a Q 5. b v E P b = ǫ 0 Q a 2 6. b v E P b = 2 ǫ 0 Q A 7. b v E P b = Q 4 π ǫ 0 a 2 8. b v E P b = Q ǫ 0 A 9. b v E P b = Q 2 ǫ 0 A correct Explanation: Basic Concepts Gauss’ Law, electrostatic properties oF conductors. Solution: Consider the Gaussian surFace shown in the fgure. E + Q E S Due to the symmetry oF the problem, there is an electric ³ux 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 ǫ 0 Q A S , since Φ TOTAL c v E · d v A, so E = Q 2 ǫ 0 A . 002 (part 2 of 2) 10.0 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 fnd the magnitude oF the electric feld at a point P in the gap. 1. b v E P b = Q ǫ 0 2. b v E P b = 2 Q ǫ 0 A correct 3. b v E P b = Q 3 ǫ 0 A

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song (shs546) – Homework 4 – weathers – (22202) 2 4. b v E P b = 5 Q ǫ 0 A 5. b v E P b = 3 Q ǫ 0 A 6. b v E P b = 0 7. b v E P b = 3 Q 2 ǫ 0 A 8. b v E P b = Q 2 ǫ 0 A 9. b v E P b = 4 Q ǫ 0 A 10. b v E P b = Q ǫ 0 A Explanation: According to the result of part 1, the electric Feld generated by plate #1 at P is E 1 = Q 2 ǫ 0 A directed along the positive x -axis.
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HW 4 solution - song(shs546 Homework 4 weathers(22202 This...

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