91Ó°ÊÓ

Draw the given figure as below.

Here, q₁ and q₂ are fixed charges.

The charge, ${\mathrm{q}}_3=+6.0\mathrm{μ°ä}=+6.0×{10}^{-6}\mathrm{C}$

Use the electric potential energy for the final three particle system,

$\begin{array}{rcl}{\mathbf{U}}_{\mathbf{net}}& \mathbf{=}& {\mathbf{U}}_{\mathbf{1}}\mathbf{+}{\mathbf{U}}_{\mathbf{2}}\mathbf{+}{\mathbf{U}}_{\mathbf{3}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\left[\frac{q_1{q}_2}{d}+\frac{q_2{q}_3}{\left(1.5d\right)}+\frac{q_1{q}_3}{\left(2.5d\right)}\right]\end{array}$

If two like charges (two protons or two electrons) are brought towards each other, the potential energy of the system increases. If two unlike charges i.e. a proton and an electron are brought towards each other, the electric potential energy of the system decreases.

Formulae:

The electric potential is define by,

${\mathrm{U}}_{12}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}$

The bet potential is define by,

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{net}}& =& {\mathrm{U}}_1+{\mathrm{U}}_2+{\mathrm{U}}_3\end{array}$

Where, U is electric potential energy, R is distance between the point charges, q is charge, ${\mathrm{Î\mu }}_0$is the permittivity of free space, and k is the Coulomb’s constant having a value

The net potential energy is,

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{net}}& =& {\mathrm{U}}_1+{\mathrm{U}}_2+{\mathrm{U}}_3\\ & =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}[\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}+\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{\left(1.5\mathrm{d}\right)}+\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{\left(2.5\mathrm{d}\right)}]\end{array}$

According to the given condition,

$\begin{array}{rcl}\frac{1}{4{\mathrm{Ï€Î\mu }}_0}[\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}+\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{\left(1.5\mathrm{d}\right)}+\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{\left(2.5\mathrm{d}\right)}]& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}.\frac{1}{\mathrm{d}}\left[{\mathrm{q}}_1{\mathrm{q}}_2\right]\\ {\mathrm{q}}_1{\mathrm{q}}_2+\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{1.5\mathrm{d}}+\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{2.5\mathrm{d}}& =& {\mathrm{q}}_1{\mathrm{q}}_2\\ \frac{{\mathrm{q}}_2{\mathrm{q}}_3}{1.5\mathrm{d}}& =& -\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{2.5\mathrm{d}}\\ \frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}& =& -1.66\end{array}$

Hence, the charge ratio ${\mathrm{q}}_1/{\mathrm{q}}_2$is -1.66.