Chapter 16, Problem 21P

(a)

To determine

To find: the charge

Answer to Problem 21P

The two charges are $67.95\text{μC}$ and $22.05\text{μC}$

Explanation of Solution

Given:

Total charge $\mathrm{Q}=90.0\mathrm{\mu }\mathrm{C}$

Repulsive force $\text{F}=12.0\text{N}$

Distance $\mathrm{r}=1.06\text{m}$

Formula used:

  $\mathrm{F}=\frac{\mathrm{k}{\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)}{{\mathrm{r}}^2}$

Calculation:

Total charge of two conducting spheres $\mathrm{Q}=90.0\mathrm{\mu }\mathrm{C}$

Distance between two charges $\mathrm{r}=1.06\text{m}$

Repulsive force between them $\text{F}=12.0\text{N}$

Let two charges be ${\text{q}}_1=\text{q}\mathrm{\mu }\text{C}$ and ${\text{q}}_2=(90-\mathrm{q})\mathrm{\mu }\mathrm{C}$

Apply Colombian’s force for two charges

  $\mathrm{F}=\frac{\mathrm{k}{\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)}{{\mathrm{r}}^2}$

Or, $\frac{\mathrm{F}{\mathrm{r}}^2}{\mathrm{k}}={\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)$

Or, $\frac{12\times {1.06}^2}{9.0\times {10}^9}={\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)$

Or, $1.498\times {10}^{-9}={\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)$

  $\text{Or,}{\mathrm{q}}_1^2-90\mathrm{\mu }\mathrm{C}{\mathrm{q}}_1+1.498\times {10}^{-9}{\text{C}}^2=0$

Or, ${\mathrm{q}}_1=\frac{90\mathrm{\mu }\mathrm{C}\pm \sqrt{{(90\mathrm{\mu }\mathrm{C})}^2-4\times 1.498\times {10}^{-9}{\mathrm{C}}^2}}{2}$

Or, ${\mathrm{q}}_1=\frac{90\times {10}^{-6}\mathrm{C}\pm \sqrt{{\left(90\times {10}^{-6}\mathrm{C}\right)}^2-4\times 1.498\times {10}^{-9}{\mathrm{C}}^2}}{2}$

Or, ${\mathrm{q}}_1=\frac{90\times {10}^{-6}\mathrm{C}\pm 45.91\times {10}^{-6}\mathrm{C}}{2}$

  ${\mathrm{q}}_1=67.95\times {10}^{-6}\text{C}$ or $22.05\times {10}^{-6}\text{C}$

So, ${\mathrm{q}}_2=22.05\times {10}^{-6}\text{C}$ or $67.95\times {10}^{-6}\text{C}$

Conclusion:

Hence two charges are $67.95\text{μC}$ and $22.05\text{μC}$

(b)

To determine

To find: the two charges

Answer to Problem 21P

The two charges are $+104.0\text{μC}$ and $-14.0\text{μC}$

Explanation of Solution

Given:

Total charge $\mathrm{Q}=90.0\mathrm{\mu }\mathrm{C}$

Repulsive force $\text{F}=12.0\text{N}$

Distance $\mathrm{r}=1.06\text{m}$

Formula used:

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

Calculation:

If the force between the charges is attractive, the charges must be of opposite sign

Let two charges be ${\text{q}}_1=\text{q}\mathrm{\mu }\text{C}$ and ${\text{q}}_2=-(90-\mathrm{q})\mathrm{\mu }\mathrm{C}$

Apply Colombian’s force for two charges

  $\mathrm{F}=-\frac{\mathrm{k}{\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)}{{\mathrm{r}}^2}$

  ${\text{Or}}_1-\frac{{\text{Fr}}^2}{\mathrm{k}}={\mathrm{q}}_1\left(90\mathrm{\mu }\text{C}-{\mathrm{q}}_1\right)$

  ${\text{Or}}_{,}-\frac{12\times {1.06}^2}{9.0\times {10}^9}={\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)$

  ${\text{Or}}_1-1.498\times {10}^{-9}={\mathrm{q}}_1\left(90\mathrm{\mu }\mathrm{C}-{\mathrm{q}}_1\right)$

  $\text{Or},{\mathrm{q}}_1^2-90\mathrm{\mu }\mathrm{C}{\mathrm{q}}_1-1.498\times {10}^{-9}{\text{C}}^2=0$

Or, ${\mathrm{q}}_1=\frac{90\mathrm{\mu }\mathrm{C}\pm \sqrt{{(90\mathrm{\mu }\mathrm{C})}^2+4\times 1.498\times {10}^{-9}{\mathrm{C}}^2}}{2}$

Or, ${\mathrm{q}}_1=\frac{90\times {10}^{-6}\mathrm{C}\pm \sqrt{{\left(90\times {10}^{-6}\mathrm{C}\right)}^2+4\times 1.498\times {10}^{-9}{\mathrm{C}}^2}}{2}$

Or, ${\mathrm{q}}_1=\frac{90\times {10}^{-6}\mathrm{C}\pm 118.71\times {10}^{-6}\mathrm{C}}{2}$

  ${\mathrm{q}}_1=+104.0\times {10}^{-6}\text{C}$ or $-14.0\times {10}^{-6}\text{C}$

So, ${\mathrm{q}}_2=-14.0\times {10}^{-6}\text{C}$ or $+104.0\times {10}^{-6}\text{C}$

Conclusion:

Hence two charges are $+104.0\text{μC}$ and $-14.0\text{μC}$