Two point charges are located on the y axis as follows: charge q 1 = -1.40 nC at

Question

Two point charges are located on the y axis as follows: charge q1 = -1.40 nC at y1 = -0.650 m , and charge q2 = 2.10 nC at the origin (y = 0). What is the magnitude of the net force exerted by these two charges on a third charge q3 = 8.00 nC located at y3 = 0.300 m ?

Part A: Find the magnitude

Part B: What is the direction of the net force? A) The net force is in + y direction. B) The net force is in - y direction. C)The net force is in + x direction. D) he net force is in - x direction.

Please help!!

Explanation / Answer

q1 = -1.40 nC at y1 = -0.650 m

q2 = 2.10 nC at y2 = 0

q3 = 8.00 nC at y3 = 0.300 m

The force exerted by q1 will be attractive and force exerted by q2 will be repulsive. Thus both the forces will be in opposite direction. Force by q1 will be in -ve y direction and Force by q2 will be in +ve y direction.

Distance between q3 and q1 = r1 = -0.650- 0.300=0.950m

Distance between q3 and q2 = r2 = 0.300 - 0 = 0.300m

Total force = F = F1 + F2

F = K*q3/((q1/r1^2) + (q2/r2^2))

F = 9*10^9*8*10^-9((-1.40*10^-9/0.95^2) + (2.10*10^-9/0.3^2))

F = 1.56*10^-6 N

F is positive it means repulsive force was higher so force direction will be positive y direction.

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