Chapter 24, Problem 20PQ

Figure P24.20 shows three charged spheres arranged along the y axis.

a. What is the electric field at x = 0, y = 3.00 m?

b. What is the electric field at x = 3.00 m, y = 0?

FIGURE P24.20

To determine

The electric field at $x=0,y=3.00\text{m}$.

Answer to Problem 20PQ

The electric field at $x=0,y=3.00\text{m}$ is $+3.33\times {10}^3\widehat{j}\text{N/C}$.

Explanation of Solution

The electric field at $x=0,y=3.00\text{m}$ will be the sum of the electric field due to all three charged spheres.

Write the formula for the electric field of a spherical source.

    $\overrightarrow{E}\left(r\right)=\frac{kq}{r^2}\widehat{r}$

Here, $\overrightarrow{E}\left(r\right)$ is the electric field, $k$ is the coulomb’s constant, $q$ is the charge of the source, and $r$ is the distance of the point from the center.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $3.00\text{m}-\text{0}\text{.750}\text{m}$ for $r$, $2.50\mu \text{C}$ for $q$ to find the electric field due to $2.50\mu \text{C}$ charge in the $+y$ direction.

  $\begin{array}{c}{\overrightarrow{E}}_1=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.50\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.00\text{m}-\text{0}\text{.750}\text{m}\right)}^2}\widehat{j}\\ =\left(4.44\times {10}^3\text{N/C}\right)\widehat{j}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $3.00\text{m}$ for $r$, $-4.80\mu \text{C}$ for $q$ to find the electric field due to $-4.80\mu \text{C}$ charge in the $+y$ direction.

  $\begin{array}{c}{\overrightarrow{E}}_2=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-4.80\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.00\text{m}\right)}^2}\widehat{j}\\ =-\left(4.79\times {10}^3\text{N/C}\right)\widehat{j}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $3.22\text{m}$ for $r$, $4.25\mu \text{C}$ for $q$ to find the electric field due to $4.25\mu \text{C}$ charge in the $+y$ direction.

  $\begin{array}{c}{\overrightarrow{E}}_3=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(4.25\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.22\text{m}\right)}^2}\widehat{j}\\ =\left(3.33\times {10}^3\text{N/C}\right)\widehat{j}\end{array}$

Write the expression to find the electric field at $x=0,y=3.00\text{m}$.

  ${\overrightarrow{E}}_R={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3$

Here, ${\overrightarrow{E}}_R$ is the electric field at $x=0,y=3.00\text{m}$, ${\overrightarrow{E}}_1$ is the electric field due to $2.50\mu \text{C}$ charge, ${\overrightarrow{E}}_2$ is the electric field due to $-4.80\mu \text{C}$ charge, ${\overrightarrow{E}}_3$ is the electric field due to $4.25\mu \text{C}$ charge.

Conclusion:

Substitute $\left(4.44\times {10}^3\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_1$, $-\left(4.79\times {10}^3\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_2$ and $\left(3.33\times {10}^3\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_3$ to find the electric field at $x=0,y=3.00\text{m}$.

  $\begin{array}{c}{\overrightarrow{E}}_R=\left(4.44\times {10}^3\text{N/C}\right)\widehat{j}-\left(4.79\times {10}^3\text{N/C}\right)\widehat{j}+\left(3.33\times {10}^3\text{N/C}\right)\widehat{j}\\ =+3.33\times {10}^3\widehat{j}\text{N/C}\end{array}$

Therefore, the electric field at $x=0,y=3.00\text{m}$ is $+3.33\times {10}^3\widehat{j}\text{N/C}$.

To determine

The electric field at $x=3.00\text{m},y=0$.

Answer to Problem 20PQ

The electric field at $x=3.00\text{m},y=0$ is $\left(1.70\times {10}^3\text{N/C}\right)\widehat{i}-\left(2.61\times {10}^2\text{N/C}\right)\widehat{j}$.

Explanation of Solution

The figure below show the system,

The electric field at $x=3.00\text{m},y=0$ will be the sum of the electric field due to all three charged spheres.

Write the formula for the electric field of a spherical source.

    $\overrightarrow{E}\left(r\right)=\frac{kq}{r^2}\widehat{r}$

Here, $\overrightarrow{E}\left(r\right)$ is the electric field, $k$ is the coulomb’s constant, $q$ is the charge of the source, and $r$ is the distance of the point from the center.

Write value of ${\theta }_1$ form the diagram.

  $\begin{array}{c}{\theta }_1={\tan }^{-1}\left(\frac{75}{300}\right)\\ =14.0°\end{array}$

Write value of ${\theta }_3$ form the diagram.

  $\begin{array}{c}{\theta }_3={\tan }^{-1}\left(\frac{22}{300}\right)\\ =4.19°\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, ${\left(3.00\text{m}\right)}^2\text{+}{\left(\text{0}\text{.750}\text{m}\right)}^2$ for $r$, $2.50\mu \text{C}$ for $q$, $\cos 14.0°\widehat{i}-\sin 14.0°\widehat{j}$ for $\widehat{r}$ to find the electric field due to $2.50\mu \text{C}$ charge.

  $\begin{array}{c}{\overrightarrow{E}}_1=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.50\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.00\text{m}\right)}^2\text{+}{\left(\text{0}\text{.750}\text{m}\right)}^2}\left(\cos 14.0°\widehat{i}-\sin 14.0°\widehat{j}\right)\\ =\left(2.28\times {10}^3\text{N/C}\right)\widehat{i}-\left(5.69\times {10}^2\text{N/C}\right)\widehat{j}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $3.00\text{m}$ for $r$, $-4.80\mu \text{C}$ for $q$ to find the electric field due to $-4.80\mu \text{C}$ charge.

  $\begin{array}{c}{\overrightarrow{E}}_2=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-4.80\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.00\text{m}\right)}^2}\widehat{i}\\ =-\left(4.79\times {10}^3\text{N/C}\right)\widehat{i}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, ${\left(3.00\text{m}\right)}^2\text{+}{\left(\text{0}\text{.220}\text{m}\right)}^2$ for $r$, $4.25\mu \text{C}$ for $q$, $\cos 4.19°\widehat{i}+\sin 4.19°\widehat{j}$ for $\widehat{r}$ to find the electric field due to $4.25\mu \text{C}$ charge.

  $\begin{array}{c}{\overrightarrow{E}}_3=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(4.25\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(3.00\text{m}\right)}^2\text{+}{\left(\text{0}\text{.220}\text{m}\right)}^2}\left(\cos 4.19°\widehat{i}+\sin 4.19°\widehat{j}\right)\\ =\left(4.21\times {10}^3\text{N/C}\right)\widehat{i}+\left(3.09\times {10}^2\text{N/C}\right)\widehat{j}\end{array}$

Write the expression to find the electric field at $x=3.00\text{m},y=0$.

  ${\overrightarrow{E}}_R={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3$

Here, ${\overrightarrow{E}}_R$ is the electric field at $x=3.00\text{m},y=0$, ${\overrightarrow{E}}_1$ is the electric field due to $2.50\mu \text{C}$ charge, ${\overrightarrow{E}}_2$ is the electric field due to $-4.80\mu \text{C}$ charge, ${\overrightarrow{E}}_3$ is the electric field due to $4.25\mu \text{C}$ charge.

Conclusion:

Substitute $\left(2.28\times {10}^3\text{N/C}\right)\widehat{i}-\left(5.69\times {10}^2\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_1$, $-\left(4.79\times {10}^3\text{N/C}\right)\widehat{i}$ for ${\overrightarrow{E}}_2$ and $\left(4.21\times {10}^3\text{N/C}\right)\widehat{i}+\left(3.09\times {10}^2\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_3$ to find the electric field at $x=3.00\text{m},y=0$.

  $\begin{array}{c}{\overrightarrow{E}}_R=\left[\left(2.28\times {10}^3\text{N/C}\right)\widehat{i}-\left(5.69\times {10}^2\text{N/C}\right)\widehat{j}\right]+\\ \left[-\left(4.79\times {10}^3\text{N/C}\right)\widehat{i}\right]+\left[\left(4.21\times {10}^3\text{N/C}\right)\widehat{i}+\left(3.09\times {10}^2\text{N/C}\right)\widehat{j}\right]\\ =\left(1.70\times {10}^3\text{N/C}\right)\widehat{i}-\left(2.61\times {10}^2\text{N/C}\right)\widehat{j}\end{array}$

Therefore, the electric field at $x=3.00\text{m},y=0$ is $\left(1.70\times {10}^3\text{N/C}\right)\widehat{i}-\left(2.61\times {10}^2\text{N/C}\right)\widehat{j}$.