Two point charges are placed on the x-axis as follows: one positive charge, q1 =

Question

Two point charges are placed on the x-axis as follows: one positive charge, q1 = 4.01 nC , is located to the right of the origin at x= 0.205 m , and a second positive charge, q2 = 5.01 nC , is located to the left of the origin at x= -0.304 m

What is the total force (magnitude and direction) exerted by these two charges on a negative point charge, q3 = -5.99 nC , that is placed at the origin?

Use 8.85×1012 C2/(Nm2) for the permittivity of free space. Take positive forces to be along the positive x-axis. Do not use unit vectors.

Explanation / Answer

the electrostatic force is given by:

F = kQ1Q2/R^2

total force on charge placed at origin will be

Fnet = F13 - F23

Fnet = kq1q3/r13^2 - kq2q3/r23^2

Fnet = 9*10^9*4.01*5.99*10^-18/0.205^2 - 9*10^9*5.01*5.99*10^-18/0.304^2

Fnet = 2.22*10^-6 N

Since Fnet is positive so direction of F will be along the positive x-axis

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