Question

Please answer the following question in detail step.

A point charge of +6 micro-coulombs is placed at x- +0.27 m, y- 0 m and a point charge of -3 micro-coulombs is placed at x 0 m, y +0.48 m. What is the angle of the total electric field produced by both charges at the origin? In degrees measured counter-clockwise from the +x axis?

Answer 1

E_1^vector = k a_1/x^2 (-x^^) = 9 times 10^9 times 6 times 10^-6/(0.27)^2 (-x^^) = -740.74 times 10^-3 x^^ E_1^vector = -0.740 N/C x^^ E_x^vector = k a_2/y^2 (-y^^) = -9 times 10^9 times 3 times 10^-6/(0.48)^2 (-y^^) = 0.117 N/C y^^ Net Field (E^vector) = E_1^vector + E_2^vector = -0.740 x^^+ 0.117 y^^ Magnitude = squareroot (0.740)^2 + (0.117)^2 N/C = 0.749 N/C Angle (theta) = pi + tan^-1(0.117/- 0.740) = 180 degrees + (- 8.984) degrees = 171.016 degrees counter clockwise from + X