Question

A +2.2 x 10-9 C charge is on the axis ar x= 1.5 m, a +5.4 x 10-9 C charge is on the x-axis at x= 2.0 m, and a+3.5 x 10-9 C charge is at the origin. find the net force on the charge at the origin

Answer 1

The net force on the charge at the origin is -1.2×10-8

Explanation:

Solving the problem,

• Draw the x-axis and the locations of the given three charges.

• The forces applied on the charge at the origin and there are two of them, and since all the changes are positive, all the forces are repulsive.
• we have the formula, F = kq1Q/r².
• F1 = kq1Q/r²1 = (9.0*109Nm²/C²)(2.2*10^-9C)(3.5*10^-9C)/(1.5m)² = 31*10-9N = 3.1*10-8N.  F1 points to the right (+x direction).

• F2 = kq2Q/r²2 = (9.0*109Nm²/C²)(5.4*10^-9C)(3.5*10^-9C)/(2.0m)² = 43*10^-9N = 4.3*10^-8N.
• F2 points to the left (-x direction).
• To find the net force we have to subtract the force F1 and force F2 .
• The net force is F(origin) = F1 - F2 = -1.2×10-8N.