Question

In Fig. 21-29a, three positively charged particles are fixed on an x-axis. Particles Band Care so close to each other that they can be considered to be at the same distance from particle A. The net force on particle Adue to particles Band Cis $\mathbf{2}\mathbf{.014}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathbf{\text{N}}$in the negative direction of the x-axis. In Fig. 21- 29b, particle Bhas been moved to the opposite side of Abut is still at the same distance from it. The net force on Ais now$\mathbf{2.877}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{24}}\mathbf{\text{N}}$ in the negative direction of the x-axis. What is the ratio ${\mathit{q}}_{\mathbf{C}}\mathbf{/}{\mathit{q}}_{\mathbf{B}}$?

Short answer

The value of the ratio of the charges${\mathrm{q}}_{\mathrm{C}}/{\mathrm{q}}_{\mathrm{B}}$is$1.33$

Step-by-step solution

The given data

The net force on particle A is$2.014\times {10}^{-23}\mathrm{N}$in the negative direction of the x-axis.

The net force on A is$2.877\times {10}^{-23}N$in the negative direction of the x-axis.

Understanding the concept of Coulomb’s law

Using the same concept of Coulomb's law, we can get the proportionality relation of charge and force. Hence, this is used to find the required ratio of the charges.

Formula:

The magnitude of the electrostatic force between any two particles,

$\mathrm{F}=\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (1)

Calculation of the ratio  qC/qB

Since the forces involved are proportional to q, we see that the essential difference between the two situations is ${\mathrm{F}}_{\mathrm{a}}\propto {\mathrm{q}}_{\mathrm{B}}+{\mathrm{q}}_{\mathrm{C}}$ (when those two charges are on the same side) versus ${\mathrm{F}}_{\mathrm{b}}\propto -{\mathrm{q}}_{\mathrm{B}}+{\mathrm{q}}_{\mathrm{C}}$ (when they are on opposite sides). Setting up ratios considering the equation (1), we have the given relation:

$\begin{array}{c}\frac{{\mathrm{F}}_{\mathrm{a}}}{{\mathrm{F}}_{\mathrm{b}}}=\frac{{\mathrm{q}}_{\mathrm{B}}+{\mathrm{q}}_{\mathrm{C}}}{-{\mathrm{q}}_{\mathrm{B}}+{\mathrm{q}}_{\mathrm{C}}}\\ \frac{2.014\times {10}^{-23}\mathrm{N}}{-2.877\times {10}^{-23}\mathrm{N}}=\frac{1+{\mathrm{q}}_{\mathrm{C}}/{\mathrm{q}}_{\mathrm{B}}}{-1+{\mathrm{q}}_{\mathrm{C}}/{\mathrm{q}}_{\mathrm{B}}}\\ -7=\frac{1+{\mathrm{q}}_{\mathrm{C}}/{\mathrm{q}}_{\mathrm{B}}}{-1+{\mathrm{q}}_{\mathrm{C}}/{\mathrm{q}}_{\mathrm{B}}}\end{array}$

Therefore,

$\begin{array}{c}\frac{{\mathrm{q}}_{\mathrm{C}}}{{\mathrm{q}}_{\mathrm{B}}}=\frac{7+1}{7-1}\\ =1.33\end{array}$

Hence, the value of the required ratio is $1.33$