Two point charges are placed on the x-axis as follows: charge q_1 = 3.96 nC is l

Question

Two point charges are placed on the x-axis as follows: charge q_1 = 3.96 nC is located at x = 0.200 m. and charge q_2 = 5.02 nC is at x = -0.304 m. What is the magnitude of the total force exerted by these two charges on a negative point charge = -6.03 nC that is placed at the origin? What is the direction of the total force exerted by these two charges on a negative point charge q_3 = -6.03 nC that is placed at the origin? to the +x direction to the -x direction perpendicular to the x-axis the force is zero

Explanation / Answer

One charge q1= 3.96 nC located at y1= 0.200m ,

Second charge q2= 5.02nC at the origin (y=0) .

Third charge q3 = 6.03 nC placed between q1 and q2 at y3 = -0.304m

The force exerted by q1 will be attractive and force exerted by q2 will be repulsive .Thus both the forces will be in

same direction ,both towards q1(both forces in negative y axis direction)

Distance between q3 and q1 = r1 = 0.304 - 0.200 = 0.104

Distance between q3 and q2 = r2 = 0.304 - 0 = 0.304

Total force = F = f1 + f2

Total force = F = (1/4pieo)q3 [ (q1/ (r1)^2) + (q2/ (r2)^2) ]

F =(9)*6.03[((3.96 x 10^-9) / 0.104^2) + (5.02 x 10^-9) / 0.304^2) ]

F = (9 x 6.03 x 10^-9)*(366 + 54.32)

F = 2.28 x 10^-5 N

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