Chapter 16, Problem 40P

To determine

To find:The magnitude and direction of electric field at P

Answer to Problem 40P

Net electric field at point $\mathrm{P}$ is $\frac{2\mathrm{k}\mathrm{Q}\mathrm{a}}{{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}^{3/2}}$ and direction along the negative $\mathrm{y}$ axis.

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used: $\mathrm{E}=\frac{\mathrm{K}\mathrm{q}}{{\mathrm{r}}^2}$

Calculation:

Consider the figure shown below.

  

From the figure, the horizontal components (along the x axis) cancel with each other as they opposite each other as they opposite each other and equal in magnitude, so they can’t contribute any value. Thus , net electric field is given by,

  $\mathrm{E}={\mathrm{E}}_1\sin \mathrm{\theta }+{\mathrm{E}}_2\sin \mathrm{\theta }$

By the definition of electric field,

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

Using the geometry of the above figure,

  $\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}$

From the figure,

  $\sin \mathrm{\theta }=\frac{\mathrm{a}}{\mathrm{r}}$

Substitute $\sqrt{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}$ for $\mathrm{r}$

  $\sin \mathrm{\theta }=\frac{\mathrm{a}}{\sqrt{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}}$

Hence , net electric field at point P is,

  $\mathrm{E}={\mathrm{E}}_1\sin \mathrm{\theta }+{\mathrm{E}}_2\sin \mathrm{\theta }$

Substitute $\mathrm{E}$ for ${\mathrm{E}}_1$ and ${\mathrm{E}}_2$

  $\begin{array}{cc}\mathrm{E}& =\mathrm{E}\sin \mathrm{\theta }+\mathrm{E}\sin \mathrm{\theta }\\ & =2\mathrm{E}\sin \mathrm{\theta }\end{array}$

Substitute $\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}$ for $\mathrm{r}$ and $\frac{\mathrm{a}}{\sqrt{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}}$ for $\sin \mathrm{\theta }.$

  $\begin{array}{l}\mathrm{E}=2\frac{\mathrm{k}\mathrm{Q}}{{\left(\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\right)}^2}\frac{\mathrm{a}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\\ =\frac{2\mathrm{k}\mathrm{Q}\mathrm{a}}{{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}^{3/2}}\end{array}$

Conclusion:

Therefore, net electric field at point $\mathrm{P}$ is $\frac{2\mathrm{k}\mathrm{Q}\mathrm{a}}{{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}^{3/2}}$ and direction along the negative $\mathrm{y}$ axis.