10c-<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C-<0.09>=f(1.33)
fnet=-2.5x1013N
A 10C charge is positioned at x=0m, a -5C charge is positioned at x=0.06m and a 2C charge is at 0.09m. What is the net force on the -5C charge? answer is (Fnet = -2.5x1013 N)
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