Three electrical charges +8m(8m-3n)Q, -3n(8m-3n)Q and +13mnQ are respectively situated at the vertices A, B and C of a triangle ABC with AB=15L; AC=13L and BC < 14L. m and n are positive real numbers such that m > (3*n)/8.

Determine the precise length of the side BC such that when the charge +13mnQ (located at C) is shifted to the circum-centre of the triangle ABC; the Net Electric Potential Energy of the new arrangement is equal to zero.

NOTE:
Definition of Net Electric Potential Energy is given here.

In conformity with the above definition, if Qa,Qb and Qc respectively denote the charges of particles a, b and c and the respective separation between the particles a & b, b & c and a & c are R1, R2 and R3, then the Net Electric Potential Energy (U) of the arrangement is given by:

U = k*(Qa*Qb/R1) + k*(Qb*Qc/R2) + k*(Qa*Qc/R3), where k is Coulomb's constant.

(In reply to Solution by Bractals)

I wish to acknowledge the superb methodology together with   nice formatting employed by Bractals which culminated in a comprehensive analytical solution to the problem under reference.

Posted by K Sengupta on 2006-06-12 23:11:43