Two point charges are placed on the x axis. The first charge, q_1 = 8.00 nC, is

Question

Two point charges are placed on the x axis. The first charge, q_1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q_2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis. Calculate the electric field at point A, located at coordinates (0m, 12.0 m).

E_Ax, E_ Ay=0,0.300 N/C Correct

I need help in Part B

Part B) An unknown additional charge q_3 is now placed at point B, located at coordinates (0 m, 15.0 m). Find the magnitude and sign of q_3 needed to make the total electric field at point A equal to zero.

Express your answer in nanocoulombs to three significant figures

Explanation / Answer

The field due to the first charge points in the -x direction Fx1 = k*q/r^2 = 9.0x10^9*8.00x10^-9/16^2 = 0.281N/C And Fx2 is due to the second charge points in the +x direction Fx2 =9.0x10^9*6.00x10^-9/9^2 = 0.667N/C So the field is 0.667 - 0.281 = 0.386 So the ordered pair is (0.386N/C, 0) B) Field at A =0 so vertical component of force must be zero vertical component due to q1 =k*q1/r^2 cos53.13 = 0.1080 N vertical component due to q2 =k*q2/r^2 cos36.86 = 0.192 N field due to Q3 = Kq3/r^2 = must be equal to 0.1080 N+0.192 N =0.3 so q3 = 0.3 nC ..........Ans plsz rate me 1st

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