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Unformatted text preview: k e 2q/a 2 + k e 3q/( √ 2a) 2 (1/ √ 2) where the first term is from the charge 2q at the upper left and the second term is from the charge 3q, with the factor of 1/ √ 2 required to get the x component of the total field. The charge 2q at the lower right does not produce an x component of the field at the position of the charge q. Hence, the total field is given by E = k e q/a 2 [ ( 2 + 3/(2 √ 2) ) i + ( 2 + 3/(2 √ 2) ) j ] where I use i and j for unit vectors in the x and y directions, respectively. To get the force on the charge, you simply multiply the field by the charge q, giving F = k e q 2 /a 2 [ ( 2 + 3/(2 √ 2) ) i + ( 2 + 3/(2 √ 2) ) j ] These solutions in terms of a, q, k e are sufficient, and numerical solutions can be obtained by substituting the given values of a and q and the known value of k e ....
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This note was uploaded on 08/22/2009 for the course PHYS 7A taught by Professor Guerra during the Spring '08 term at UC Irvine.
 Spring '08
 GUERRA
 Physics, Charge

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