# ch23-p035 - 35. (a) To calculate the electric field at a...

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distributed uniformly over both sides of the original plate, with half being on the side near the field point. Thus, 6 42 2 6.0 10 C 4.69 10 C/m . 2 2(0.080 m) q A σ × == = × The magnitude of the field is 42 7 12 2 2 0 4.69 10 C/m 5.3 10 N/C. 8.85 10 C / N m E σ ε × == ×⋅ The field is normal to the plate and since the charge on the plate is positive, it points away from the plate. (b) At a point far away from the plate, the electric field is nearly that of a point particle with charge equal to the total charge on the plate. The magnitude of the field is 22 0 /4 / Eq r k qr πε == , where r is the distance from the plate. Thus,
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## This note was uploaded on 09/07/2010 for the course PHYS 247 taught by Professor Zv during the Fall '10 term at The Petroleum Institute.

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