# ass8 - SOLUTIONS ASSIGNMENT 8 22.51 IDENTIFY:The net...

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SOLUTIONS, ASSIGNMENT 8 22.51. I DENTIFY : The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the plate. S ET U P : The electric field due to the sheet of charge on each surface is 0 /2 E σ = P and is directed away from the surface. E XECUTE : (a) For the conductor the charge sheet on each surface produces fields of magnitude 0 /2 P and in the same direction, so the total field is twice this, or 0 / P . (b) At points inside the plate the fields of the sheets of charge on each surface are equal in magnitude and opposite in direction, so their vector sum is zero. At points outside the plate, on either side, the fields of the two sheets of charge are in the same direction so their magnitudes add, giving 0 / E = P . E VALUATE : Gauss’s law can also be used directly to determine the fields in these regions. 23.1.I DENTIFY : Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9).

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ass8 - SOLUTIONS ASSIGNMENT 8 22.51 IDENTIFY:The net...

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