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Unformatted text preview: song (shs546) Homework 4 weathers (22202) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find 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 figure below shows a crosssectional 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. bardbl vector E P bardbl = 4 a 2 Q 2. bardbl vector E P bardbl = Q 4 a 3. bardbl vector E P bardbl = Q A 4. bardbl vector E P bardbl = 4 a Q 5. bardbl vector E P bardbl = Q a 2 6. bardbl vector E P bardbl = 2 Q A 7. bardbl vector E P bardbl = Q 4 a 2 8. bardbl vector E P bardbl = Q A 9. bardbl vector E P bardbl = Q 2 A correct 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 = Q A . If the cross section of the surface is S , then Gauss Law states that TOTAL = 2 E S = 1 Q A S , since TOTAL contintegraldisplay vector E d vector A, so E = Q 2 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 find the magnitude of the electric field at a point P in the gap. 1. bardbl vector E P bardbl = Q 2. bardbl vector E P bardbl = 2 Q A correct 3. bardbl vector E P bardbl = Q 3 A song (shs546) Homework 4 weathers (22202) 2 4. bardbl vector E P bardbl = 5 Q A 5. bardbl vector E P bardbl = 3 Q A 6. bardbl vector E P bardbl = 0 7. bardbl vector E P bardbl = 3 Q 2 A 8. bardbl vector E P bardbl = Q 2 A 9. bardbl vector E P bardbl = 4 Q A 10. bardbl vector E P bardbl = Q A Explanation: According to the result of part 1, the...
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 Spring '00
 Littler
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