A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters,

Question

A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00 nC is at the point x = 0.600 meters, y = 0. Part A Calculate the magnitude E of the net electric field at the origin due to these two point charges. Express your answer in newtons per coulomb to three significant figures. Part B What is the direction, relative to the negative x axis, of the net electric field at the origin due to these two point charges. Express your answer in degrees to three significant figures.

Explanation / Answer

q1 = - 4nC, r1[0.600, 0.800]
q2 = +6nC, r2[0.600, 0]

x = 0.600 m, y = 0.800 m

|E1| = k q1 / (x²+y²) = 36 N/C
|E1| = k q2 / x² = 150 N/C

E1 direction relative to the positive x axis:
= Atan(y / x) = 53°

Eo(x) = E1(x) - E2(x) = |E1| cos - |E2| = - 128 N/C
Eo(y) = E1(y) + E2(y)= |E1| sen + 0 = |E1| sen = 29 N/C

magnitude Eo:

|Eo| = Eo(x)² + Eo(y)² = 131 N/C

direction relative to the negative x axis:
' = Atan(Eo(y) / Eo(x)) = 12°

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