Question
A 60uC charge is placed at the origin. A second -20uC charge is placed 1m to the right of the first charge. What is the electric field at point (0.5,0.5)?
Answers
E = k Q/ d^2
E1 = [10^-5 k (6) / 1^2] i + 0 j
E2 = +10^-5 k(2) / (.5 sqrt 2)^2] (i - j) = +4*10^-5 k i - 4*10^-5 k j
add the i components and add the j components
E1 = [10^-5 k (6) / 1^2] i + 0 j
E2 = +10^-5 k(2) / (.5 sqrt 2)^2] (i - j) = +4*10^-5 k i - 4*10^-5 k j
add the i components and add the j components
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