P23_067 - absolute values): ~ E net x = 4 k Q 125 a 2 ~ E...

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67. The distance from Q to P is 5 a , and the distance from q to P is 3 a . Therefore, the magnitudes of the individual electric Felds are, using Eq. 23-3 (writing 1 / 4 πε 0 = k ), | ~ E Q | = k | Q | 25 a 2 , | ~ E q | = k | q | 9 a 2 . We note that ~ E q is along the y axis (directed towards ± y in accordance with the sign of q ), and ~ E Q has x and y components, with ~ E Qx = ± 4 5 | ~ E Q | and ~ E Qy = ± 3 5 | ~ E Q | (signs corresponding to the sign of Q ). Consequently, we can write the addition of components in a simple way (basically, by dropping the
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Unformatted text preview: absolute values): ~ E net x = 4 k Q 125 a 2 ~ E net y = 3 k Q 125 a 2 + k q 9 a 2 (a) Equating ~ E net x and ~ E net y , it is straightforward to solve for the relation between Q and q . We obtain Q = 125 9 q 14 q . (b) We set ~ E net y = 0 and Fnd the necessary relation between Q and q . We obtain Q = 125 27 q 4 . 6 q ....
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