Question

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?

Short answer

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

Step-by-step solution

The given data

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,$\mathrm{R}=10.0\times {10}^{-2}\text{m}$

Understanding the concept of potential energy of the system           

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.

Calculation of the charge of the third particle 

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 $5.68\times {10}^{-6}\text{C}$.