Potential Energy of a system of particles

closer

A system consists of three particles, each of mass 4.00 g, located at the corners of an equilateral triangle with sides of 36.0 cm.

(a) Calculate the potential energy of the system.

m = .004, d = .2078

UG = Gm1m2/R
U_total = sum of U_i
U_total = Gm1m2/R + Gm1m3/R + Gm2m3/R
U_total = [((6.67300E-11)(.004)(.004))/(.2078)] * 3
U_total = 1.5414E-14

Final answer is incorrect. Any ideas?

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Redbelly98

How did you get d = .2078 ? That looks wrong to me, though I haven't seen your attachment yet.