P23_064 - = 3 q 4 d 2 + 5 q 4 d 2 cos 45 = 6 . 536 q 4 d 2...

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64. From symmetry, the only two pairs of charges which produce a non-vanishing ±eld ~ E net are: pair 1, which is in the middle of the two vertical sides of the square (the + q , 2 q pair); and pair 2, the +5 q , 5 q pair. We denote the electric ±elds produced by each pair as ~ E 1 and ~ E 2 , respectively. We set up a coordinate system as shown to the right, with the origin at the center of the square. Now, x y q + q 5 q 5 q ~ E 1 ~ E 2 ~ E net . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....................................................... . . . . . . . . . . . . . E 1 = 1 4 πε 0 µ q d 2 + 2 q d 2 = 3 q 4 πε 0 d 2 and E 2 = k · 5 q ( 2 d ) 2 + 5 q ( 2 d ) 2 ¸ = 5 q 4 πε 0 d 2 . Therefore, the components of ~ E net are given by E x = E 1 x + E 2 x = E 1 + E 2 cos 45
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Unformatted text preview: = 3 q 4 d 2 + 5 q 4 d 2 cos 45 = 6 . 536 q 4 d 2 , and E y = E 1 y + E 2 y = E 2 sin 45 = 5 q 4 d 2 sin 45 = 3 . 536 q 4 d 2 . Thus, the magnitude of ~ E net is E = q E 2 x + E 2 y = p (6 . 536) 2 + (3 . 536) 2 q 4 d 2 = 7 . 43 q 4 d 2 , and ~ E net makes an angle with the positive x axis, where = tan 1 E y E x = tan 1 3 . 536 6 . 536 = 28 . 4 ....
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