Chapter 16, Problem 39P

To determine

To find: the ratio of $\raisebox{1ex}{${\mathrm{Q}}_1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Q}}_2$}\right.$

Answer to Problem 39P

  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

The expression for the electric field is,

  $\mathrm{E}=\frac{\mathrm{k}\mathrm{Q}}{{\mathrm{r}}^2}$

Here, $\mathrm{k}$ is the coulomb's constant, $\mathrm{Q}$ is the charge, and $\mathrm{r}$ is the distance.

Calculation:

The point where the electric field is $0,$ is at a distance ${\mathrm{r}}_1=\frac{\mathrm{l}}{3}$ from ${\mathrm{Q}}_1$

And at a distance ${\mathrm{r}}_2=\mathrm{l}-\frac{\mathrm{l}}{3}$ or, ${\mathrm{r}}_2=\frac{2\mathrm{l}}{3}$ from ${\mathrm{Q}}_2$ .

If the electric field is zero, then

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

Substitute $\frac{\mathrm{l}}{3}$ for ${\mathrm{r}}_1$ and $\frac{2\mathrm{l}}{3}$ for ${\mathrm{r}}_2$ in the above equation.

  $\frac{{\mathrm{Q}}_1}{{(\mathrm{l}/3)}^2}=\frac{{\mathrm{Q}}_2}{{(2\mathrm{l}/3)}^2}$

Rewrite the above expression for $\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}$ as follows:

  $\begin{array}{l}\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}=\frac{{(\mathrm{l}/3)}^2}{{(2\mathrm{l}/3)}^2}\\ =\frac{1}{4}\end{array}$

Conclusion:

Therefore, the ratio of the two charges is $\frac{1}{4}$ .