Chapter 16, Problem 34P

To determine

To find: The electric field at corner of square

Answer to Problem 34P

  $3.87\times {10}^4\text{N}/\text{C}$

Explanation of Solution

Given:

  ${\mathrm{q}}_1={\mathrm{q}}_2={\mathrm{q}}_3=2.25\times {10}^{-6}\mathrm{C},\mathrm{a}=1\mathrm{m}$

Formula used:

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

Calculation:

  

The electric field at the upper right corner of the square is the vector sum of the fields due to

other three charges.

  ${\overrightarrow{\mathrm{E}}}_{\mathrm{A}}={\overrightarrow{\mathrm{E}}}_{\mathrm{B}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{C}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{D}}$

Here, ${\overrightarrow{\mathrm{E}}}_{\mathrm{A}},{\overrightarrow{\mathrm{E}}}_{\mathrm{B}},{\overrightarrow{\mathrm{E}}}_{\mathrm{c}}$ and ${\overrightarrow{\mathrm{E}}}_{\mathrm{D}}$ are the electric fields at point $\text{A},\text{B},\text{C}$ and $\text{D}$ respectively.

Let $\mathrm{r}$ represent the length of a side of square, and let $\mathrm{q}$ be the charge present at each of three occupied corners.

Electric field due to charge q presenting at a D is,

  ${\mathrm{E}}_{\mathrm{D}}=\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$

x -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{D}\mathrm{x}}=\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$

y -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{D}\mathrm{y}}=0$

Electric field due to charge q presenting at a C is,

  ${\mathrm{E}}_{\mathrm{c}}=\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}$

x -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{C}\mathrm{x}}=\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\cos {45}^{\circ }$

y -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{C}\mathrm{y}}=\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\sin {45}^{\circ }$

Electric field due to charge q presenting at a B is,

  ${\mathrm{E}}_{\mathrm{B}}=\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$

x -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{B}\mathrm{x}}=0$

y -component of this electric field is,

  ${\mathrm{E}}_{\mathrm{B}\mathrm{y}}=\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$

The x, y components together,

  ${\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{B}\mathrm{x}}+{\mathrm{E}}_{\mathrm{\alpha }}+{\mathrm{E}}_{\mathrm{D}\mathrm{x}}$

Substitute 0 for ${\mathrm{E}}_{\mathrm{B}{\mathrm{x}}^{\prime }}\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\cos {45}^{\circ }$ for ${\mathrm{E}}_{\mathrm{C}\mathrm{x}}$ , and $\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$ for ${\mathrm{E}}_{\mathrm{D}\mathrm{x}}$

  $\begin{array}{cc}{\mathrm{E}}_{\mathrm{x}}& =0+\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\cos {45}^0+\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\\ & =\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)\end{array}$

And.

  ${\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{B}\mathrm{y}}+{\mathrm{E}}_{\mathrm{C}\mathrm{y}}+{\mathrm{E}}_{\mathrm{D}\mathrm{y}}$

Substitute $\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}$ for ${\mathrm{E}}_{\mathrm{B}{\mathrm{y}}^{\prime }},\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\sin {45}^{°}$ for ${\mathrm{E}}_{\mathrm{G}}$ and 0 for ${\mathrm{E}}_{\mathrm{D}\mathrm{y}}$

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{y}}=\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}+\mathrm{k}\frac{\mathrm{q}}{{(\sqrt{2}\mathrm{r})}^2}\sin {45}^{°}+0\\ =\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)\end{array}$

Net electric field acting at a point A due to remaining charges,

  ${\mathrm{E}}_{\mathrm{A}}=\sqrt{{\mathrm{E}}_{\mathrm{x}}^2+{\mathrm{E}}_{\mathrm{y}}^2}$

Substitute $\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)$ for ${\mathrm{E}}_{\mathrm{x}}$ and $\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)$ for ${\mathrm{E}}_{\mathrm{y}}$

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}}=\sqrt{{\left(\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)\right)}^2+{\left(\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)\right)}^2}\\ =\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\left(\sqrt{2}+\frac{1}{2}\right)\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\mathrm{k},2.25\times {10}^{-6}\text{C}$ for $\mathrm{q},$ and $1.0\text{m}$ for r in the equation $\begin{array}{l}{\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\left(\sqrt{2}+\frac{1}{2}\right)\\ {\mathrm{E}}_{\mathrm{A}}=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(2.25\times {10}^{-6}\text{C}\right)}{{(1.0\text{m})}^2}\left(\sqrt{2}+\frac{1}{2}\right)\\ =3.87\times {10}^4\text{N}/\text{C}\end{array}$

Direction of this field is given by,

  $\begin{array}{l}\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{E}}_{\mathrm{y}}}{{\mathrm{E}}_{\mathrm{x}}}\right)\\ ={\tan }^{-1}\left(\frac{\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)}{\mathrm{k}\frac{\mathrm{q}}{{\mathrm{r}}^2}\left(1+\frac{1}{2\sqrt{2}}\right)}\right)\\ ={45}^{°}\end{array}$

Conclusion:

Therefore, direction of this with horizontal direction is ${45}^{°}$ .