Chapter 16, Problem 33P

To determine

To find:The electric field at center

Answer to Problem 33P

  $2.35\times {10}^6\text{N}/\text{C}$

Explanation of Solution

Given:

  $\begin{array}{l}\mathrm{a}=52.5\text{cm}\\ {\mathrm{q}}_1=45\text{μC}\\ {\mathrm{q}}_2=-27\text{μC}\\ {\mathrm{q}}_3=-27\text{μC}\end{array}$

Formula used:

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

Calculation:

The directions of electric fields of four charges which lie at four corners of a square are shown below figure:

  

In the above figure, a is the length of the each side of the square.

From the above figure, the directions of electric fields due to the two charges $-27.0\mathrm{\mu }\text{C}$ atcorners B and D are opposite. The magnitude of charges and distances from the point P aresame. So, the electric fields due to these two charges at corners B and D will be canceled out.

The distance of the center of the square (point P ) has equal distance from each corner. Using

the Pythagoras theorem, we can calculate the distance of point Pfrom each corner as follows.

  $\begin{array}{cc}{(\mathrm{A}\mathrm{P})}^2& =\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{a}}{2}\right)}^2}\\ & =\frac{\mathrm{a}}{\sqrt{2}}\end{array}$

Convert the units for charge from $+45.0\mathrm{\mu }\text{CtoC}$ :

  $\begin{array}{cc}+45.0\mathrm{\mu }\text{C}& =(+45.0\mathrm{\mu }\text{C})\left(\frac{1.0\times {10}^{-}\text{C}}{1.0\mathrm{\mu }\text{C}}\right)\\ & =+45.0\times {10}^{-6}\text{C}\end{array}$

Convert the units for charge $-27.0\mathrm{\mu }\text{CtoC}$ :

  $\begin{array}{cc}-27.0\mathrm{\mu }\text{C}& =(-27.0\mathrm{\mu }\text{C})\left(\frac{1.0\times {10}^{-6}\text{C}}{1.0\mathrm{\mu }\text{C}}\right)\\ & =-27.0\times {10}^{-6}\text{C}\end{array}$

Convert the units for length of the each side from 52.5 cm to m:

  $\begin{array}{cc}\mathrm{a}& =52.5\text{cm}\\ & =(52.5\text{cm})\left(\frac{1.0\text{m}}{100\text{cm}}\right)\\ & =0.525\text{m}\end{array}$

Let ${\mathrm{E}}_1$ be the magnitude of electric field due the charge at corner $\mathrm{A}$ and ${\mathrm{E}}_2$ be the magnitudeof electric field due the charge at corner $\mathrm{C}.{\mathrm{E}}_1$ and ${\mathrm{E}}_2$ Can be written as

  ${\mathrm{E}}_1=\mathrm{k}\frac{45.0\times {10}^{-6}\text{C}}{{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^2}$

  $\text{Substitute}9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\text{for}\mathrm{k}\text{and}0.525\text{mfor}\mathrm{a}\text{in above equation, solve for}{\mathrm{E}}_1$

  $\begin{array}{cc}{\mathrm{E}}_1& =\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(45.0\times {10}^{-6}\text{C}\right)}{{\left(\frac{0.525\text{m}}{\sqrt{2}}\right)}^2}\\ & =2.94\times {10}^6\text{N}/\text{C}\end{array}$

The magnitude of electric field due the charge at corner C :

  ${\mathrm{E}}_2=\mathrm{k}\frac{27.0\times {10}^{-6}\text{C}}{{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^2}$

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for k and 0.525 m for a in above equation, solve for ${\mathrm{E}}_2$

  $\begin{array}{cc}{\mathrm{E}}_2& =\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(27.0\times {10}^{-6}\text{C}\right)}{{\left(\frac{0.525\text{m}}{\sqrt{2}}\right)}^2}\\ & =1.76\times {10}^6\text{N}/\text{C}\end{array}$

The electric field due to the change at corner A:

  ${\overrightarrow{\mathrm{E}}}_1={\mathrm{E}}_1\cos {45}^{\circ }\widehat{\mathrm{i}}+{\mathrm{E}}_1\sin {45}^{\circ }\widehat{\mathrm{j}}$

Substitute $2.94\times {10}^6\text{N}/\text{C}$ for E in above equation, solve for ${\overrightarrow{\mathrm{E}}}_1$ .

  $\begin{array}{cc}{\overrightarrow{\mathrm{E}}}_1& =\left(2.94\times {10}^6\text{N}/\text{C}\right)\cos {45}^{\circ }\widehat{\mathrm{i}}+\left(2.94\times {10}^6\text{N}/\text{C}\right)\sin {45}^{\circ }\widehat{\mathrm{j}}\\ & =\left(2.078\times {10}^6\text{N}/\text{C}\right)(\widehat{\mathrm{i}}+\widehat{\mathrm{j}})\end{array}$

The electric field due to the charge at corner B :

  ${\overrightarrow{\mathrm{E}}}_2={\mathrm{E}}_2\cos {45}^{\circ }\widehat{\mathrm{i}}+{\mathrm{E}}_2\sin {45}^{\circ }\widehat{\mathrm{j}}$

Substitute $1.76\times {10}^6\text{N}/\text{C}$ for ${\mathrm{E}}_2$ in above equation ${\overrightarrow{\mathrm{E}}}_1$

  $\begin{array}{cc}{\overrightarrow{\mathrm{E}}}_2& =\left(1.76\times {10}^6\text{N}/\text{C}\right)\cos {45}^{\circ }\widehat{\mathrm{i}}+\left(1.76\times {10}^6\text{N}/\text{C}\right)\sin {45}^{\circ }\widehat{\mathrm{j}}\\ & =\left(1.24\times {10}^6\text{N}/\text{C}\right)(\widehat{\mathrm{i}}+\widehat{\mathrm{j}})\end{array}$

The net electric field due to all charges at center of square is only the net electric field due to the

charges at corner A and C . The net electric field due the charges at corners B and D will be

cancel out.

Therefore, the net electric field at center of the square:

  ${\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}={\overrightarrow{\mathrm{E}}}_1+{\overrightarrow{\mathrm{E}}}_2$

Substitute $\left(2.078\times {10}^6\text{N}/\text{C}\right)(\mathrm{i}+\mathrm{j})\text{for}\overrightarrow{{\mathrm{E}}_1}\text{ and}\left(1.24\times {10}^6\text{N}/\text{C}\right)(\mathrm{i}+\mathrm{j})\text{for }{\overrightarrow{\mathrm{E}}}_2$ in above equation,

  $\begin{array}{cc}{\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}& =\left(2.078\times {10}^6\text{N}/\text{C}\right)(\widehat{\mathrm{i}}+\widehat{\mathrm{j}})+\left(1.24\times {10}^6\text{N}/\text{C}\right)(\widehat{\mathrm{i}}+\widehat{\mathrm{j}})\\ & =\left(3.318\times {10}^6\text{N}/\text{C}\right)(\widehat{\mathrm{i}}+\widehat{\mathrm{j}})\end{array}$

Find the magnitude of ${\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}$

  $\begin{array}{cc}\left|{\overrightarrow{\mathrm{E}}}_{\mathrm{n}\mathrm{e}\mathrm{t}}\right|& =\sqrt{{\left(3.318\times {10}^6\text{N}/\text{C}\right)}^2+{\left(3.318\times {10}^6\text{N}/\text{C}\right)}^2}\\ & =4.69\times {10}^6\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the magnitude of net electric field at center of the square is $4.70\times {10}^6\text{N}/\text{C}$ . It has

direction ${45}^{\circ }$ towards the corner .