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An Astute Evaluation (Posted on 2006-05-17) Difficulty: 5 of 5
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.

  Submitted by K Sengupta    
Rating: 5.0000 (1 votes)
Solution: (Hide)
The required length of the side BC = 4L.

EXPLANATION:

Let BC= xL (say) where x < 14.
Let rL be the length of the circum-radius of the triangle ABC.
Now, s = (15+13+x)/2 = 14 + x/2; so that:
Area of triangle ABC (A) = sqrt ((196 –(x^2)/4) ((x^2)/4 - 1))

Hence, r =(15*13*x)/(4*A)
= 195*x/(4*sqrt(196 – (x^2)/4) * sqrt ((x^2)/4 - 1))…………..(#)

Let the circum-centre of the triangle ABC be located at O.
Then, we observe that the Electrical Potential Energy of each of the combinations
A-B,A-O and B-O are as follows :

P(A-O) = 104* (m^2)* n *(8m-3n)*(Q^2)/ (k*rL)

P(A-B) = -24mn((8m-3n)^2)*(Q^2)/ (k*15L)

P(B-O) = -39* m*(n^2)(8m-3n)*(Q^2)/ (k*rL)

where k=4*Pi*e0

Accordingly,
P(A-O)+ P(A-B)+ P(B-O) =0 gives,
13mn*((8m-3n)^2) /r – 24mn*((8m-3n)^2 /15 = 0
or, 13mn /r = 24mn /15 giving r = 65/8 ( since m and n are
both real numbers)

Consequently, from (#), we obtain:

195*x/(4*sqrt(196 – (x^2)/4) * sqrt ((x^2)/4 - 1) = 65/8
Simplifying, we obtain:

x^4 – 212*(x^2) +3136 = 0
or, x^2 = 196,16
or, x = 14,4 (ignoring the negative roots which are inadmissible)

However, we know that x < 14 and accordingly, x must be equal to 4.

Consequently, the required length of the side BC = 4L.

------------------------------------------------------------------

NOTE: Also refer to Bractals' method in solving the problem which is provided in this location .

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AcknowledgementK Sengupta2006-06-12 23:11:43
SolutionSolutionBractals2006-05-17 11:40:50
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