What are the strength and direction of the electric field at the position indica

Question

What are the strength and direction of the electric field at the position indicated by the dot?

Specify the direction as an angel above or below horizontal.

I know the answer is E = 3975 N/C @ 9.32º above horizontal. But I would like SPECIFIC, step by step instuction on how to find it including all formula. Most importantly, I want to know WHY, when find the angle of the lower 1.0nC charge, you use 90-tan-1 instead of just tan-1 and HOW to find the final angle of 9.32. Thanks.

Explanation / Answer

KQ = for 1nc charge = 9 *10^9 x 1*10^-9 = 9

Electric field strength E = kQ / r^2

Electric field direction is radially away from + charges

At the dot:

r1 = 5cm = 0.05m

E1 = 9 / (0.05)^2 N/C = 3600 N/C direction towards x axis

E1 = 3600i ( here is no y component )

r2 = sqrt(5^2 + 10^2) cm = 11.18 cm = 0.112 m)

E2 = 9 / (0.112)^2 = 717.47 N/C

theta = 90 - tan^-1(5/10) = 63.43 deg

E2 = 320.92i + 641.7j

E = Ex + Ey

E = 3920.92 i + 641.7j

|E| = sqrt(Ex^2 + Ey^2)

|E| = 3973.08 N/C

theta = tan^-1(Ey/Ex) = 9.3 degree

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