Chapter 16, Problem 37P

(a)

To determine

To find: The electric field at origin.

Answer to Problem 37P

  $\mathrm{E}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{{\mathrm{l}}^2}$

Explanation of Solution

Given:

The two charges +Q and +Q are placed at two points A and B, as shown in the figure.

  

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

  $|{\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}|=\sqrt{{\mathrm{E}}_1^2+{\mathrm{E}}_2^2+2{\mathrm{E}}_1{\mathrm{E}}_2\cos \mathrm{\theta }}$

Calculation:

Since the two charges placed at A and B are positive, the directions of ${\mathrm{E}}_{\mathrm{A}}$ and ${\mathrm{E}}_{\mathrm{B}}$ are shown in the figure

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\\ {\mathrm{E}}_{\mathrm{B}}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\end{array}$

Here,

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}\mathrm{x}}=0,\\ {\mathrm{E}}_{\mathrm{A}\mathrm{y}}=-\frac{1}{4\mathrm{\pi }{\in }_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\end{array}$

And

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{B}\mathrm{x}}=\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\cos {30}^{°}\\ =\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

And

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{B}\mathrm{y}}=-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\sin {30}^{°}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

Thus, ${\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{A}\mathrm{x}}+{\mathrm{E}}_{\mathrm{B}\mathrm{x}}$

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{x}}=0+\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

And

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{A}\mathrm{y}}+{\mathrm{E}}_{\mathrm{B}\mathrm{y}}{\mathrm{E}}_{\mathrm{y}}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}+\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\frac{\mathrm{Q}}{2{\mathrm{l}}^2}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{3\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

The magnitude of electric field is

  $\begin{array}{l}\mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2}\\ \begin{array}{l}=\sqrt{{\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}]}^2+\left[\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\left(\frac{\mathrm{Q}}{{\mathrm{l}}^2}+\frac{\mathrm{Q}}{2{\mathrm{l}}^2}\right)\right]}\hfill \\ \mathrm{E}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{{\mathrm{l}}^2}\hfill \end{array}\end{array}$

The angle between the electric fields is

  $\begin{array}{l}\mathrm{\theta }={\tan }^{-1}\left[\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right]\\ ={\tan }^{-1}\left[\begin{array}{c}\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{3\mathrm{Q}}{2{\mathrm{l}}^2}\\ \frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}\right]\\ ={\tan }^{-1}(\sqrt{3})\\ ={\tan }^{-1}(1.732)\\ \mathrm{\theta }={240}^{°}\end{array}$

(b)

To determine

To find: The electric field at origin if charge at B is negative

Answer to Problem 37P

  $\mathrm{E}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}$

Explanation of Solution

Given:

The two charges + Q and - Q are placed at two points A and B.

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

  $|{\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}|=\sqrt{{\mathrm{E}}_1^2+{\mathrm{E}}_2^2+2{\mathrm{E}}_1{\mathrm{E}}_2\cos \mathrm{\theta }}$

Calculation:

Here, the charge at B is reversed to − Q .

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}\mathrm{x}}=0\\ \text{and}\\ -{\mathrm{E}}_{\mathrm{A}\mathrm{y}}=-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\\ {\mathrm{E}}_{\mathrm{B}\mathrm{x}}=\left|\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{-\mathrm{Q}}{{\mathrm{l}}^2}\right|\cos {30}^{°}\\ =\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

And

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}\mathrm{x}}=0\\ \text{So,}\\ -{\mathrm{E}}_{\mathrm{A}\mathrm{y}}=-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\\ {\mathrm{E}}_{\mathrm{B}\mathrm{x}}=\left|\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{-\mathrm{Q}}{{\mathrm{l}}^2}\right|\cos {30}^{°}\\ =\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

  $\begin{array}{l}\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2}=\mathrm{E}\\ {\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{A}\mathrm{x}}+{\mathrm{E}}_{\mathrm{B}\mathrm{x}}\\ =0+\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\\ =\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\\ {\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{A}\mathrm{y}}+{\mathrm{E}}_{\mathrm{B}\mathrm{y}}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}+\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{2{\mathrm{l}}^2}\\ =-\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}$

The magnitude of electric field is $\mathrm{E}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2}$

  $\begin{array}{l}\mathrm{E}=\sqrt{{\left(\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2\mathrm{l}}\right)}^2+{\left(\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\frac{\mathrm{Q}}{2{\mathrm{l}}^2}\right)}^2}\\ \mathrm{E}=\frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{{\mathrm{l}}^2}\end{array}$

The angle between the electric field is

  $\begin{array}{l}\mathrm{\theta }={\tan }^{-1}\left[\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right]\\ ={\tan }^{-1}\left[\begin{array}{c}\frac{-1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\mathrm{Q}}{2{\mathrm{l}}^2}\\ \frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\cdot \frac{\sqrt{3}\mathrm{Q}}{2{\mathrm{l}}^2}\end{array}\right]\\ ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\\ \mathrm{\theta }=330°\end{array}$