Two point charges (q_1 = -3.8 mu C and q_2 = 8.2 mu C) are fixed along the x-axi

Question

Two point charges (q_1 = -3.8 mu C and q_2 = 8.2 mu C) are fixed along the x-axis, separated by a distance d = 6.4 cm. Point P is located at (x, y) = (d, d). What is E_x(P), the value of the x-component of the electric field produced by q_1 and q_2 at point P? What is E_y(P), the value of the y-component of the electric field produced by q_1 and q_2 at point P? A third point charge q_3 = 2.9 mu C is now positioned along the y-axis at a distance d = 6.4 cm from q_1 as shown. What is E_x(P), the x-component of the field produced by all 3 charges at point P?

Explanation / Answer

Well, you know that E=F/q which eventually will get to kQ(source)/r^2

so by using this equation you simply plug in what is given. if i am not mistaken about the picture, P is right above q2 so that won't affect the x-component.

So, you will have Ex(P)=(9E9)*(-3.8E-6)/(.064^2+.064^2).
this will give you the magnitude of the electric field produced by q1. all thats left is to multiply your answer by a sin(45) and voila, there's your answer!
This should be your final answer (-2952032.70 N/C)

for the second part you do the same, use the formula but this time q2 will affect it because you are looking for the y-component.

Ey(P)=(9E9)*(8.2E-6)/(.064^2) which equals out to 18017578.13 N/C.
to get the final answer just use the answer from part A and subtract it from the above answer and you are all set to go!
This should be the final answer for the second part. (15065545.43 N/C)

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