Particle 1 (with a charge of $\mathbf{+}\mathbf{5}\mathbf{.0}\mathbf{}\mathbf{\mu }\mathbf{\text{C}}$) and particle 2 (with a charge of$\mathbf{+}\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{\mu }\mathbf{\text{C}}$) are fixed in place with separation$\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}\mathbf{}\mathbf{\text{cm}}$on the xaxis shown in Fig. 24-58a. Particle 3 can be moved along the xaxis to the right of particle 2. Figure 24-58bgives the electric potential energy Uof the three-particle system as a function of the xcoordinate of particle 3. The scale of the vertical axis is set by${\mathbf{U}}_{\mathbf{S}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.0}\mathbf{}\mathbf{\text{J}}$.

What is the charge of particle 3?

1. Charge of particle 1,${\mathrm{q}}_1=5.0\times {10}^{-6}\text{C}$
2. Charge of particle 2,${\mathrm{q}}_2=3.0\times {10}^{-6}\text{C}$
3. Separation of the charges 1 and 2,$\mathrm{R}=4.0\times {10}^{-2}\text{m}$
4. The scale of the vertical axis,${\mathrm{U}}_{\mathrm{s}}=5\text{J}$
5. Separation of the charges 2 and 3,$\mathrm{R}=10.0\times {10}^{-2}\text{m}$

Using the concept of the potential energy of the system, we can get the distance between the charged particles by substituting the given values.

Formula:

The potential energy of two charged system is given by, $\mathrm{U}=\frac{{\mathrm{KQ}}_1{\mathrm{Q}}_2}{\mathrm{R}}$ (i)

Here, K is the coulomb constant. ${\mathrm{Q}}_1,{\mathrm{Q}}_2$are charges in the system and R is the distance between the charges.

The potential energy of the system is given as:

$\mathrm{U}={\mathrm{U}}_{12}+{\mathrm{U}}_{23}+{\mathrm{U}}_{13}\mathrm{......}\dots \dots \left(\mathrm{a}\right)$

Here ${\mathrm{U}}_{12}$is the potential energy due to interaction of charge ${\mathrm{q}}_1$and role="math" localid="1661923037181" ${\mathrm{q}}_2,{\mathrm{U}}_{23}$is the potential energy due to interaction of charge${\mathrm{q}}_2$and $q_3$and ${\mathrm{U}}_{13}$is the potential energy due to interaction of charge $q_1$and$q_2$

Substituting the values $5.0\times {10}^{-6}\mathrm{C},3.0\times {10}^{-6}\mathrm{C}$for ${\mathrm{Q}}_1,{\mathrm{Q}}_2$respectively and $4.0\times {10}^{-2}\mathrm{m}$for $\mathrm{R}$in equation (i), we get the potential energy of the system for particles 1 and 2 as:

${\mathrm{U}}_{12}=\frac{\mathrm{K}(5.0\times {10}^{-6}\text{C})(3.0\times {10}^{-6}\text{C})}{4\times {10}^{-2}\text{m}}$

Substituting the values $3.0\times {10}^{-6}\mathrm{C}$,${\mathrm{q}}_3$for ${\mathrm{Q}}_2,{\mathrm{Q}}_3$respectively in equation (i) and using the graph to substitute role="math" localid="1661923903653" $10\times {10}^{-2}\mathrm{m}$for $\mathrm{R}$, we get the potential energy as:

${\mathrm{U}}_{23}=\frac{\mathrm{K}(3.0\times {10}^{-6}\text{C}){\mathrm{q}}_3}{10\times {10}^{-2}\text{m}}$

Now, substituting the values$5.0\times {10}^{-6}\mathrm{C},{\mathrm{q}}_3$for${\mathrm{Q}}_1,{\mathrm{Q}}_3$respectively and use graph to substitute $(10+4)\times {10}^{-2}$m for R in equation (i), we get the potential energy as:

${\mathrm{U}}_{13}=\frac{\mathrm{K}(5.0\times {10}^{-6}\text{C}){\mathrm{q}}_3}{(10+4)\times {10}^{-2}\text{m}}$

Now, substituting the above values in equation (ii), we get the total potential energy as follows:

$\mathrm{U}=\frac{\mathrm{K}(5.0\times {10}^{-6}\text{C})(3.0\times {10}^{-6}\text{C})}{4\times {10}^{-2}\text{m}}+\frac{\mathrm{K}(3.0\times {10}^{-6}\text{C}){\mathrm{q}}_3}{10\times {10}^{-2}\text{m}}+\frac{\mathrm{K}(5.0\times {10}^{-6}\text{C}){\mathrm{q}}_3}{(10+4)\times {10}^{-2}\text{m}}$

Now from graph the net potential energy is 0 when x is 10 m, hence

$\begin{array}{c}\frac{\mathrm{K}(5.0\times {10}^{-6})(3.0\times {10}^{-6})}{4\times {10}^{-2}}+\frac{\mathrm{K}(3.0\times {10}^{-6}){\mathrm{q}}_3}{10\times {10}^{-2}}+\frac{\mathrm{K}(5.0\times {10}^{-6}){\mathrm{q}}_3}{(10+4)\times {10}^{-2}}=0\\ {\mathrm{q}}_3\left(\frac{3.0\times {10}^{-6}}{10.0\times {10}^{-2}}+\frac{5.0\times {10}^{-6}}{10.0\times {10}^{-2}+4.0\times {10}^{-2}}\right)=-\frac{(5.0\times {10}^{-6})(3.0\times {10}^{-6})}{4.0\times {10}^{-2}}\\ {\mathrm{q}}_3\left(\frac{3.0}{10.0}+\frac{5.0}{14.0}\right)=-\frac{(5.0)(3.0\times {10}^{-6})}{4.0}\\ {\mathrm{q}}_3=\frac{-3.75\times {10}^{-6}}{0.66}\text{C}\\ =5.68\times {10}^{-6}\text{C}\end{array}$

Hence, the charge of particle 3 is $5.68\times {10}^{-6}\text{C}$.