HW6Solutions

# HW6Solutions - HW6 Solutions Notice numbers may change...

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HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely long, concentric cable in cross section. The inner conductor has a linear charge density of λ = 6.00 nC/m and the outer conductor has no net charge. (a) Find the electric field for all values of R , where R is the perpendicular distance from the common axis of the cylindrical system. (Use R as necessary.) (b) What are the surface charge densities on the inside and the outside surfaces of the outer conductor? Solution:

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Tipler 22.P.071 Two identical square parallel metal plates each have an area of 500 cm 2 . They are separated by 1.50 cm. They are both initially uncharged. Now a charge of +1.50 nC is transferred from the plate on the left to the plate on the right and the charges then establish electrostatic equilibrium. (Neglect edge effects.) (a) What is the electric field between the plates at a distance of 0.25 cm from the plate on the right? (b) What is the electric field between the plates a distance of 1.00 cm from the plate on the left? (c) What is the electric field just to the left of the plate on the left? (d) What is the electric field just to the right of the plate on the right? Solution: The left plate has negative charge -1.50 nC and the right one has positive charge + 1.50 nC.

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The field due to each surface charge distribution can be calculated by the Gauss’ law applied to a cylinder that crosses each plane. For each plane: ES+ES = σ S/ ε 0 E = σ /2 ε 0 The field for the 2 plates sums inside the 2 plates and cancels outside. (a) and (b) the field is uniform inside the plates so in
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## This note was uploaded on 02/26/2011 for the course PHYS 205 taught by Professor Marko during the Spring '11 term at York University.

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HW6Solutions - HW6 Solutions Notice numbers may change...

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