Q7P - Fundamentals Of Physics

Question

In Fig. 22-35, the four particles form a square of edge length $\mathbf{a}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{cm}$ and have charges,${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{nC}$,${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{nC}\mathbf{,}{\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{nC}$, and ${\mathbf{q}}_{\mathbf{4}}\mathbf{=}\mathbf{+}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{nC}$. In unit-vector notation, what net electric field do the particles produce at the square’s center?

Step-by-Step Solution

Answer

The net electric field that the particles produced at the center of the square is$(1.02\times {10}^5\hspace{0.33em}\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$

Step 1: The given data

1.  The charges of the four-particle at the corners of the square,${\mathrm{q}}_1=+10\hspace{0.33em}\mathrm{nC}, {\mathrm{q}}_2=-20\hspace{0.33em}\mathrm{nC}, {\mathrm{q}}_3=+20\hspace{0.33em}\mathrm{nC} \mathrm{and} {\mathrm{q}}_4=-10\hspace{0.33em}\mathrm{nC}.$
2.  The edge length of the square, $\mathrm{a}=5\hspace{0.33em}\mathrm{cm}\frac{ 1\hspace{0.33em}\mathrm{m}}{100 \mathrm{cm}}=0.05\hspace{0.33em}\mathrm{m}$

Step 2: Understanding the concept of electric field

Using the concept of the electric field at a given point, we can get its charge value from the given formula.

 

Formulae:

The magnitude of the electric field,  $\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$                                                  (1)

where R = The distance of field point from the charge q  = charge of the particle

According to the superposition principle, the electric field at a point due to more than one charge, 

 $$\begin{gathered} \overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{{\mathrm{E}}_{\mathrm{i}}} \\ =\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{{\mathrm{r}}^2}_{\mathrm{i}}}\widehat{{\mathrm{r}}_{\mathrm{i}}} \end{gathered}$$                                                                        (2)

Step 3: Calculation of the net charge at the center of the square

Using the equation (i), the value of the x-component of the net electric field at the center of the square due to all the charges are given as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[-\frac{\left|{\mathrm{q}}_1\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}+\frac{\left|{\mathrm{q}}_2\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}+\frac{\left|{\mathrm{q}}_3\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}-\frac{\left|{\mathrm{q}}_4\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}\right]\cos {45}^{°} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{2}{{\mathrm{a}}^2}\right)\left[-\left|{\mathrm{q}}_1\right|+\left|{\mathrm{q}}_2\right|+\left|{\mathrm{q}}_3\right|-\left|{\mathrm{q}}_4\right|\right]\frac{1}{\sqrt{2}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\sqrt{2}}{{\mathrm{a}}^2}\left[-10\mathrm{nC}+20\mathrm{nC}-20\mathrm{nC}-10\mathrm{nC}\right] \\ =0 \end{gathered}$$

Similarly, the y-component of the net electric field using equation (1) is given as:

 $$\begin{gathered} {\mathrm{E}}_{\mathrm{y}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[-\frac{\left|{\mathrm{q}}_1\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}+\frac{\left|{\mathrm{q}}_2\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}+\frac{\left|{\mathrm{q}}_3\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}-\frac{\left|{\mathrm{q}}_4\right|}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}\right]\sin {45}^{°} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{2}{{\mathrm{a}}^2}\right)\left[-\left|{\mathrm{q}}_1\right|+\left|{\mathrm{q}}_2\right|+\left|{\mathrm{q}}_3\right|-\left|{\mathrm{q}}_4\right|\right]\frac{1}{\sqrt{2}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\sqrt{2}}{{\mathrm{a}}^2}\left[-10\mathrm{nC}+20\mathrm{nC}+20\mathrm{nC}-10\mathrm{nC}\right] \\ =\frac{\left(8.99\times {10}^9 \mathrm{N}.\frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}\right)\sqrt{2}}{{\left(0.050 \mathrm{m}\right)}^2}\left[2.0\times {10}^{-8} \mathrm{C}\right] \\ =1.02\times {10}^5 \mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the net electric field at the center of the square is given using equation (ii) as:

$$\begin{gathered} \overrightarrow{\mathrm{E}}={\mathrm{E}}_{\mathrm{x}}\hat{\mathrm{j}}+{\mathrm{E}}_{\mathrm{y}}\hat{\mathrm{j}} \\ =0+(1.02\times {10}^5\hspace{0.33em}\mathrm{N}/\mathrm{C})\hat{\mathrm{j}} \\ =(1.02\times {10}^5\hspace{0.33em}\mathrm{N}/\mathrm{C})\hat{\mathrm{j}} \end{gathered}$$

Hence, the value of the net electric field is $(1.02\times {10}^5\hspace{0.33em}\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$