Chapter 16, Problem 19P

To determine

To find: the charge and placement of the third charge

Answer to Problem 19P

The charge should be of magnitude $0.40{\mathrm{Q}}_0$ and a distance 0.371 from $-{\mathrm{Q}}_0$ towards.

Explanation of Solution

Given:

Two charges $-3{\mathrm{Q}}_{\mathrm{o}}$ and $-{\mathrm{Q}}_0$

Formula used:

According to the Coulomb's law, the force between two charges is directly proportional to the product of two charges and inversely proportional to the distance between them, and is given by the following relation:

  $\mathrm{F}=\frac{{\mathrm{k}}_{\mathrm{e}}{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Here, $\mathrm{F}$ is the force, ${\mathrm{k}}_{\mathrm{e}}$ is the Coulomb's constant, ${\mathrm{q}}_1$ is the first charge, ${\mathrm{q}}_2$ is the second charge, and $\mathrm{r}$ is the distance between them.

Calculation:

The following figure shows the third charge is placed in between two charges to become equilibrium as shown below:

  

From the above figure:

Here, $-3{\mathrm{Q}}_{\mathrm{o}}$ and $-{\mathrm{Q}}_0$ are two charges and the third charge $\mathrm{q}$ is placed between the two charges to become equilibrium. The force ${\mathrm{F}}_{-\mathrm{Q}.}$ is the force on charge $\mathrm{q}$ due to charge $-{\mathrm{Q}}_{\mathrm{o}}$ and the force ${\mathrm{F}}_{-30.}$ is on charge $\mathrm{q}$ due to charge ${\mathrm{F}}_{-30.}$

The magnitude of force ${\mathrm{F}}_{-0.}$ on charge $\mathrm{q}$ due to charge $-{\mathrm{Q}}_0$ as follows:

  $\left|{\mathrm{F}}_{-0.}\right|=\frac{{\mathrm{k}}_{\mathrm{e}}\mathrm{q}{\mathrm{Q}}_{\mathrm{e}}}{{\mathrm{x}}^2}$

The magnitude of force ${\mathrm{F}}_{-3\mathrm{Q}}$ is on charge $\mathrm{q}$ due to charge $-3{\mathrm{Q}}_{\mathrm{e}}$ as follows:

  $\left|{\mathrm{F}}_{-3{\mathrm{Q}}_0}\right|=\frac{3{\mathrm{k}}_{\mathrm{c}}\mathrm{q}{\mathrm{Q}}_0}{{(\mathrm{l}-\mathrm{x})}^2}$

From the given problem:

Due to placing the third charge in-between two charges to become equilibrium and is given by the following relation:

  $\left|{\mathrm{F}}_{-0.}\right|=\left|{\mathrm{F}}_{-3{\mathrm{Q}}_0}\right|$

  $\frac{{\mathrm{k}}_{\mathrm{c}}\mathrm{q}{\mathrm{Q}}_{\mathrm{e}}}{{\mathrm{x}}^2}=\frac{3{\mathrm{k}}_{\mathrm{e}}\mathrm{q}{\mathrm{Q}}_{\mathrm{e}}}{{(\mathrm{l}-\mathrm{x})}^2}$

  ${(\mathrm{l}-\mathrm{x})}^2=3{\mathrm{x}}^2$

Solve the equation ${(\mathrm{l}-\mathrm{x})}^2=3{\mathrm{x}}^2$ and solve for $\mathrm{x}$ in terms of $/$ as follows:

  ${(\mathrm{l}-\mathrm{x})}^2=3{\mathrm{x}}^2$

  $\begin{array}{cc}(\mathrm{l}-\mathrm{x})& =\sqrt{3}\mathrm{x}\\ \mathrm{l}-\mathrm{x}& =\sqrt{3}\mathrm{x}\\ \mathrm{x}& =0.366\mathrm{l}\\ & =0.37\mathrm{l}\end{array}$

  $\text{The figure shows force on charge}-{\mathrm{Q}}_0\text{ due to charges }\mathrm{q}\text{ and}-3{\mathrm{Q}}_0\text{ as shown below:}$

  

Calculate the third charge in terms of ${\mathrm{Q}}_{\mathrm{e}}$ as follows:

The magnitude of the force on charge $-{\mathrm{Q}}_{\mathrm{e}}$ due to charge $\mathrm{q}$ as follows:

  $\left|{\mathrm{F}}_{\mathrm{q}}\right|=\frac{{\mathrm{k}}_{\mathrm{e}}\mathrm{q}{\mathrm{Q}}_{\mathrm{e}}}{{\mathrm{x}}^2}$

The magnitude of the force on charge $-{\mathrm{Q}}_0$ due to charge $-3{\mathrm{Q}}_{\mathrm{e}}$ as follows:

  $\left|{\mathrm{F}}_{-30.}\right|=\frac{3{\mathrm{k}}_{\mathrm{c}}{\mathrm{Q}}_0{\mathrm{Q}}_0}{{\mathrm{l}}^2}$

The relation of above forces is,

  $\left|{\mathrm{F}}_{\mathrm{q}}\right|=\left|{\mathrm{F}}_{-3\text{Q}}\right|$

  $\frac{{\mathrm{k}}_{\mathrm{c}}\mathrm{q}{\mathrm{Q}}_{\mathrm{e}}}{{\mathrm{x}}^2}=\frac{3{\mathrm{k}}_{\mathrm{c}}{\mathrm{Q}}_{\mathrm{e}}{\mathrm{Q}}_{\mathrm{e}}}{{\mathrm{l}}^2}$

  $\mathrm{q}=\frac{3{\mathrm{Q}}_{\mathrm{e}}{\mathrm{x}}^2}{{\mathrm{l}}^2}$

Substitute 0.371 for $\mathrm{x}$ in the equation $\mathrm{q}=\frac{3{\mathrm{Q}}_0{\mathrm{x}}^2}{{\mathrm{l}}^2}$

  $\mathrm{q}=\frac{3{\mathrm{Q}}_0{(0.366\mathrm{l})}^2}{{\mathrm{l}}^2}$

  $\mathrm{q}=0.40{\mathrm{Q}}_0$

Conclusion:

Therefore, the charge should be of magnitude $0.40{\mathrm{Q}}_0$ and a distance 0.371 from $-{\mathrm{Q}}_0$ towards.