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Unformatted text preview: lim (kl9356) – 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 π ǫ a Q 2. bardbl vector E P bardbl = 2 ǫ Q A 3. bardbl vector E P bardbl = ǫ Q a 2 4. bardbl vector E P bardbl = Q 4 π ǫ a 2 5. bardbl vector E P bardbl = Q 2 ǫ A correct 6. bardbl vector E P bardbl = Q ǫ A 7. bardbl vector E P bardbl = ǫ Q A 8. bardbl vector E P bardbl = 4 π ǫ a 2 Q 9. bardbl vector E P bardbl = Q 4 π ǫ a 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 ǫ Q A S , since Φ TOTAL ≡ contintegraldisplay vector E · d vector A, so E = Q 2 ǫ 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 = 5 Q ǫ A 2. bardbl vector E P bardbl = 3 Q 2 ǫ A 3. bardbl vector E P bardbl = 0 lim (kl9356) – Homework 4 – Weathers – (22202) 2 4. bardbl vector E P bardbl = Q ǫ 5. bardbl vector E P bardbl = 2 Q ǫ A correct 6. bardbl vector E P bardbl = 4 Q ǫ A 7. bardbl vector E P bardbl = Q 3 ǫ A 8. bardbl vector E P bardbl = Q ǫ A 9. bardbl vector E P bardbl = 3 Q ǫ A 10. bardbl vector E P bardbl = Q 2 ǫ A Explanation: According to the result of part 1, the...
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Charge, Magnetism, Work

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