Question

Three charges 2q, 4q and 2q are to be placed on a 1.0 m long straight wire. 

Determine the positions where the charges should to be placed so that the potential 

energy of the system is a minimum.

Answer 1

Consider the sketch below

$U= k\frac{q_1q_2}{r_{12}}+k\frac{q_2q_3}{r_{23}}+k\frac{q_3q_1}{r_{31}}$

$U= k[\frac{2q \times 2q}{x}+\frac{2q \times4q}{1-x}+\frac{4q \times2q}{1}]$

$U= k[\frac{4q^2}{x}+\frac{8q^2}{1-x}+\frac{8q^2}{1}]$

$U= kq^2[\frac{4}{x}+\frac{8}{1-x}+\frac{8}{1}]$

Now potential energy of the system depends on the value of x.

$U= [\frac{4}{x}+\frac{8}{1-x}+\frac{8}{1}]$

Minimizing differentiating w.r.t x

$U'= [-\frac{4}{x^2}+\frac{8}{(1-x)^2}]$

for U' to be minimum, it should be equal to zero.

$U'= [-\frac{4}{x^2}+\frac{8}{(1-x)^2}]=0$

$\frac{8}{(1-x)^2}=\frac{4}{x^2}$

$x=-1-\sqrt{2},\:x=\sqrt{2}-1$

Now, $x=-1-\sqrt{2}$  is not applicable hence we will take $\:x=\sqrt{2}-1$

Therefore x = 0.4142 m

Hence between 2q and 2q, the distance is 0.4142 m and the distance between second 2q and 4q is 0.5858 m