quiz1 - k e 2q/a 2 + k e 3q/( √ 2a) 2 (1/ √ 2) where...

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Name: ________________________________ ID#: __________________________________ Physics 7D Disc. TA: ____ __________ ________________ Quiz Ch.23 Version A Date: _______________ Hour:_____________ Four point charges are at the corners of a square of side a (a=10 cm) as shown in the figure. (a) Determine the magnitude and direction of the electric field at the location of charge q (q=1.69*10 -19 C) . (b) What is the resultant force on q ? All versions are similar, with the same symmetry, but different magnitudes of the charges at the upper left, lower left, and lower right. We can use symmetry to notice that the electric field at the point charge q has equal x and y components. Hence, we only need to calculate one of them, for example the x component. The field is due to three charges, only two of which give an x component. The x component is given by
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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.

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