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Step 6 of 8 A Electric field due to first charge is E,卡 kq Substitute 9x10, N-m2 /C2 for k, 6.00x10 С for q and 0.5 m for 5 , (9x10, N-㎡/C)(6.00x10° c) (0.5 m) E, = 216 N/C And, Electric field due to second charge is, kq Substitute 9 x 10, N-m /C fork, 6.00x10° C for q and 04 m for不 (9x10 N·m/C*)(6.00 x10 C) 0.4 m 337.5 NC Both electric fields act away from the point and make an angle θ with them.

21.47 In a rectangular coordinate system a positive point charge q 6.00 x 10 9 C is placed at the point x +0.150 m, y = 0, and an identical point charge is placed at x =-0.150 m, y 0. Find the a- and y-components, the magnitude, and the direction of the electric field at the following points: (a) the origin: (b)x 0.300 m, y 0; (c) 0.150 m, y0.400 m; (d) x = 0, y = 0.200 m.

in Part c) I am not sure how they came up with 0.5 for r for E1. will you please give a clear details as you solve it, and please just go ahead and solve 47 so I can check my answers. Thank you.

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Answer #1

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