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Homework 02-problems - le(jml2943 Homework 02 de...

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le (jml2943) – Homework 02 – de lozanne – (56785) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points 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. bardbl vector E P bardbl = ǫ 0 Q a 2 2. bardbl vector E P bardbl = Q 4 π ǫ 0 a 3. bardbl vector E P bardbl = 4 π ǫ 0 a Q 4. bardbl vector E P bardbl = Q 2 ǫ 0 A 5. bardbl vector E P bardbl = ǫ 0 Q A 6. bardbl vector E P bardbl = Q 4 π ǫ 0 a 2 7. bardbl vector E P bardbl = 2 ǫ 0 Q A 8. bardbl vector E P bardbl = Q ǫ 0 A 9. bardbl vector E P bardbl = 4 π ǫ 0 a 2 Q 002(part2of2)10.0points 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 = 3 Q 2 ǫ 0 A 2. bardbl vector E P bardbl = Q 3 ǫ 0 A 3. bardbl vector E P bardbl = 2 Q ǫ 0 A 4. bardbl vector E P bardbl = 4 Q ǫ 0 A 5. bardbl vector E P bardbl = Q ǫ 0 A 6. bardbl vector E P bardbl = Q ǫ 0 7. bardbl vector E P bardbl = 0 8. bardbl vector E P bardbl = 3 Q ǫ 0 A 9. bardbl vector E P bardbl = Q 2 ǫ 0 A 10. bardbl vector E P bardbl = 5 Q ǫ 0 A 003 10.0points A large flat sheet of charge has a charge per unit area of 7 . 7 μ C / m 2 . Find the electric field intensity just above the surface of the sheet, measured from its midpoint.
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