Chapter 16, Problem 18P

To determine

To find: the charge on two points

Answer to Problem 18P

The charge $\mathrm{q}$ is $5.54\times {10}^{-4}\text{C}$ and $5.54\times {10}^{-6}\text{C}$ another charges $5.54\times {10}^{-6}\text{C}$ and $5.54\times {10}^{-4}\text{C}$

Explanation of Solution

Given:

Total charge $=560\mathrm{\mu }\mathrm{C}\text{placed}\text{at }1.10\mathrm{m},\text{ Force}=22.8\mathrm{N}$

Formula used:

It is given that total charge of two point charges is positive, so both charges must be positive. As electrostatic force between them is repulsive, so both charges must be of same sign. Let $\mathrm{Q}$ be the total charge of two point charges. Then if one charge is of magnitude $\mathrm{q}$ then other charge must be of magnitude $(\mathrm{Q}-\mathrm{q})$

The coulomb's expression for electrostatic force between the charges is given by,

  $\mathrm{F}=\frac{\mathrm{k}\mathrm{q}(\mathrm{Q}-\mathrm{q})}{{\mathrm{r}}^2}$

Here, $\mathrm{k}$ is electrostatic constant $\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right),\mathrm{r}$ is distance between the point charges.

Calculation:

On re-arrange this equation,

  ${\mathrm{q}}^2-\mathrm{Q}\mathrm{q}+\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}=0$

Compare this equation with general quadratic equation,

  $\mathrm{a}{\mathrm{x}}^2+\mathrm{b}\mathrm{x}+\mathrm{c}=0$

The roots of this equation,

  $\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}$

The values resembles to, $\mathrm{x}=\mathrm{q};\mathrm{a}=1;\mathrm{b}=-\mathrm{Q};\mathrm{c}=\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}$

Thus, roots of the quadratic equation: ${\mathrm{q}}^2-\mathrm{Q}\mathrm{q}+\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}=0$

  $\mathrm{q}=\frac{-\mathrm{Q}\pm \sqrt{{\mathrm{Q}}^2-4\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}}}{2}$

Conversion of units of charge from micro coulomb's to coulomb's as,

  $\begin{array}{cc}\mathrm{Q}& =560\mathrm{\mu }\text{C}\\ & =(560\mathrm{\mu }\text{C})\left(\frac{{10}^{-6}\text{C}}{\mathrm{\mu }\text{C}}\right)\\ & =560\times {10}^{-6}\text{C}\end{array}$

To find the magnitude of the charge, substitute $560\times {10}^{-6}\text{C},22.8\text{N}$ 22.8 $\text{F},1.10\text{m}$ 1.10 for $\mathrm{r},$

  $\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)$ for $\mathrm{k}$ in the equation $\mathrm{q}=\frac{-\mathrm{Q}\pm \sqrt{{\mathrm{Q}}^2-4\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}}}{2}$

  $\begin{array}{l}\mathrm{q}=\frac{-\mathrm{Q}\pm \sqrt{{\mathrm{Q}}^2-4\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}}}{2}\\ =\frac{\left(560\times {10}^{-6}\text{C}\right)\pm \sqrt{{\left(560\times {10}^{-6}\text{C}\right)}^2-4\frac{(22.8\text{N}){(1.10\text{m})}^2}{\left(8.988\times {10}^{\circ }\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)}}2}{2}\\ =5.54\times {10}^{-4}\text{Cand}5.54\times {10}^{-6}\text{C}\end{array}$

Corresponding values of another charge is,

  $\mathrm{Q}-\mathrm{q}=5.54\times {10}^{-6}\text{C}$ and $5.54\times {10}^{-4}\text{C}$

Conclusion:

Therefore, charge $\mathrm{q}$ is $5.54\times {10}^{-4}\text{C}$ and $5.54\times {10}^{-6}\text{C}$ another charges $5.54\times {10}^{-6}\text{C}$ and $5.54\times {10}^{-4}\text{C}$