Question

Figure 21-34ashows charged particles 1 and 2 that are fixed in place on an x-axis. Particle 1 has a charge with a magnitude of$\mathbf{\left|}{\mathbf{q}}_{\mathbf{1}}\mathbf{\right|}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8.00}\mathbf{e}$. Particle 3 of charge${\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.00}\mathbf{e}$is initially on the x-axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x-axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3

changes. Figure 21-34bgives the xcomponent of that net force as a function of the position xof particle 3.The scale of the x-axis is setby ${\mathbf{x}}_{\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.80}\mathbf{\text{m}}$. Theplot has an asymptote of${\mathbf{F}}_{\mathbf{2}\mathbf{net}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.5}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathbf{\text{\hspace{0.17em}N}}$as$\mathit{x}\mathbf{}\mathbf{\to }\mathit{\infty }$. As a multiple of eand including the sign, what is the charge${\mathbf{q}}_{\mathbf{2}}$of particle 2?

Short answer

The value of the charge ${\mathrm{q}}_2$of particle 2 as a multiple of e, the electronic charge, is $13\mathrm{e}$.

Step-by-step solution

Stating the given data

1. Charges of particles 1 and 3:$\left|{\mathrm{q}}_1\right|=8.00\mathrm{e}$and ${\mathrm{q}}_3=+8.00\mathrm{e}$.
2. The scale of the x axis is set by${\mathrm{x}}_{\mathrm{s}}=0.80\text{\hspace{0.17em}m}$ and ${\mathrm{F}}_{2\left(\mathrm{net}\right)}=1.5\times {10}^{-25}\text{N}$.

Understanding the concept of Coulomb’s law

Using the concept of Coulomb's law, the value of the charge of the second particle is found using the net value of the force acting on it.

Formula:

The magnitude of the electrostatic force between any two particles is $\mathrm{F}=\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$… (i)

Calculation of the charge of the second particle

Since the graph crosses zero, q₁ must be positive-valued:${\mathrm{q}}_1=+8.00\mathrm{e}$. We can note that it crosses zero at $\mathrm{r}=0.40\mathrm{m}$. Now the asymptotic value of the force yields the magnitude and sign of${\mathrm{q}}_2$as follows, using equation (i):

$\begin{array}{c}{\mathrm{q}}_2=\left(\frac{{\mathrm{F}}_{2\mathrm{net}}}{{\mathrm{kq}}_1}\right){\mathrm{r}}^2\\ =\left(\frac{1.5\times {10}^{-25}}{9\times {10}^9\times 8\mathrm{e}}\right)\\ =2.086\times {10}^{-18}\text{C}\\ =13\mathrm{e}\end{array}$

Hence, the value of the charge is $13e$.