Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q'. separated by a distance d is \F] = k12. where K = data, and 60 = 8.854 x 10-12 C2/(N·m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, qi = -12.5 nC, is located at X1 = -1.670 m ; the second charge, q2 = 40.0 nC, is at the origin (x = 0).

What is (F_net3)_x, the x-component of the net force exerted by these two charges on a third charge q3 = 50.0 nC placed between q1 and q2 at x3 = -1.185 mm ?

Your answer may be positive or negative, depending on the direction of the force.

Express your answer numerically in newtons to three significant figures.

Answer 1

the force is 3.67* 10^-5 N directed along negative x-direction.

Now F - Now I -12.Snc 5000 the force on aine qz by a = K (12.5 X 50 X10-18) KX 625 X10 48 -1.670 - |-}85 - (0.485) 2 0.2352 (-we-xside) = k(265 7:03) X 10-18 N is directed away 23 force on 9 by q = f32= K (40 x 50 x 10-18) = x/1424.27)X 10% (1185)²x10-6 is directed away qz since both the force torex side) direction are in same dirchen - we will take the fire position laddition here) - so Fret = kx10-18 [2651.0341424.27] - 40 81-39 x 9 x109 x 1018 Fret 5 10=9x36731-7 N 3.67x104X10-9 N os 367x10-5 Newton -