# ch22_p63 - y axis is zero; the net field component along...

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symmetry, we see the net field component along the x axis is zero; the net field component along the y axis points upward. With θ = 60 ° , E net y = 2 Q sin θ 4 πε o a 2 . Since sin(60 ° ) = 3 /2 , we can write this as E net = kQ 3 /a 2 (using the notation of the constant k defined in Eq. 21-5). Numerically, this gives roughly 47 N/C. (b) From symmetry, we see in this case that the net field component along the
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Unformatted text preview: y axis is zero; the net field component along the x axis points rightward. With θ = 60 ° , E net x = 2 Q cos θ 4 πε o a 2 . Since cos(60 ° ) = 1/2, we can write this as E net = kQ/a 2 (using the notation of Eq. 21-5). Thus, E net ≈ 27 N/C. 6 3. (a) We refer to the same figure to which problem 63 refers (but without “ q ”). From...
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## This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.

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