A charge of 5.13C is at the origin and a second chargeof - 2.90C is on the oppos

Question

A charge of 5.13C is at the origin and a second chargeof - 2.90C is on the opposite x-axis 0.920 m from the origin.The coulomb constant is 8.99 N*m2 /C2.
1) Find the magnitude of the field at a point P on the y-axis0.680m from the origin. Answer in units of N/C. For this question ialready have an answer. I just thought it might be needed forreference. 2) determine the direction of this electric field (as an anglebetween -180 degrees and 180 degrees measured from the positivex-axis, with counterclockwise positive). Answer in units ofdegree. Can please someone help i can't figure out this answer. Thankyou!!!! points!!!! A charge of 5.13C is at the origin and a second chargeof - 2.90C is on the opposite x-axis 0.920 m from the origin.The coulomb constant is 8.99 N*m2 /C2.
1) Find the magnitude of the field at a point P on the y-axis0.680m from the origin. Answer in units of N/C. For this question ialready have an answer. I just thought it might be needed forreference. 2) determine the direction of this electric field (as an anglebetween -180 degrees and 180 degrees measured from the positivex-axis, with counterclockwise positive). Answer in units ofdegree. Can please someone help i can't figure out this answer. Thankyou!!!! points!!!!

Explanation / Answer

1)the electric field due to charge at origin E1 = k * (q1/r1^2) k = (1/4o) = 9 * 10^9 Nm^2/C^2,q1 =5.13 C = 5.13 * 10^-6 C and r1 = 0.680 m the electric field due to charge at x = 0.920 m E2 = k * (q2/r2^2) q2 = -2.90 C = -2.90 * 10^-6 C and r2^2 =[(0.920 - 0)^2 + (0 - 0.680)^2] the magnitude of the field at a point P is E = (Ex^2 + Ey^2)^(1/2) Ex = E1 * cos1 + E2 * cos2 Ey = E1 * sin1 + E2 * sin2 1 = 90o and tan2 = [(0 - 0.680)/(0.920- 0)] = -(0.680/0.920) or 2 = -36.4o 2)the direction of this electric field tan = (Ey/Ex) or = tan-1(Ey/Ex)

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