Electric field due to Two point charges, homework help

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Jan 18th, 2016

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Part A:

The electric field from q1 is given by:

F1 = kq1 / r12
F1 = 8.9875 * 109 * (8.00 * 10-9) / 202
F1 = 0.18 N/C

Likewise, for q2:

F2 = kq2 / r22
F2 = 8.9875 * 109 * (6.00 * 10-9) / 152
F2 = 0.24 N/C

To find the x and y components, just use trigonometry (note that the direction of the field is away from the charges, since the charges are positive, so q1‘s x component will be negative, q2‘s x component will be positive, and both y components will be positive:

F1, x = F1 * cos(θ1)
F1, x = 0.18 * -cos(36.87)
F1, x = -0.144 N/C

F1, y = F1 * sin(θ1)
F1, y = 0.18 * sin(36.87)
F1, y = 0.108 N/C

F2, x = F2 * cos(θ2)
F2, x = 0.24 * cos(53.13)
F2, x = 0.144 N/C

F2, y = F2 * sin(θ2)
F2, y = 0.24 * sin(53.13)
F2, y = 0.192 N/C

So the sum of the x components is 0 (+0.144 from q2, -0.144 from q1. The sum of the y components is 0.300 (0.144 + 0.192):

0,0.300 N/C



Part B:

The sum of the x components is -0.288(-0.144 from q2, -0.144 from q1. The sum of the y components is 0.300 (0.144 - 0.192):

-0.288, -0.048

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jan 18th, 2016

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