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

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.

Ithink it is c ,i hope i you