Chapter 22, Problem 78P

To determine

The magnitude and the direction of the electric field in the z plane at $x=1.50\text{m, y=0}\text{.50m}$ .

The electric field at point P is $E=241\text{kN/C}$ pointing at $\theta \text{=220}°$ .

Given:

  

The charges are placed as shown in the figure. The first plane at $x=2.0\text{m}$ . The line charge passing through the origin at an angle of $\text{45}°$ with the x axis in the x-y plane. The spherical shell centered at point $\text{(1}\text{.5m,0}\text{.5m)}$ in the x-y plane.

The charge densities are

  $\sigma =2\mu {\text{nC/m}}^2$

  $\lambda =4\mu {\text{nC/m}}^2$

  $\rho =-6.0\mu {\text{C/m}}^3$

Formula Used:

Electric field

  $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

E is the electric field.

  $\sigma$ is the surface charge density.

  $\widehat{r}$ is the unit vector in the direction normal to the charged plane.

  ${\epsilon }_o$ is the permittivity of free space.

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

  $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

Electric field at point P due to sphere.

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}\text{=}\frac{4\pi }{3}kr\text{'}\rho \widehat{r}\text{'}$

  $r\text{'}$ Represents the distance from the center of the sphere to P. constructing a Gaussian sphere of radius $r\text{'}$ centered at $(1,0)$

From the figure

  $\overrightarrow{r}\text{'}=-0.5\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.5<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $r\text{'}=\sqrt{{(-0.5)}^2+{(-0.5)}^2}=0.707\text{m}$

  $\widehat{r}\text{'}=-0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Substituting the values

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=\frac{4\pi }{3}(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(0.707\text{m)(-6}\text{.00}\mu {\text{C/m}}^3)\left[0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\right]$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-112.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{Q}}_{sphere}=\sigma A_{sphere}$

  $=4\pi \sigma R^2$

  ${\text{Q}}_{sphere}=4\pi (-3.0\mu {\text{C/m}}^2){(1.0\text{m)}}^2$

  ${\text{Q}}_{sphere}=-37.30\mu \text{C}$

  $\widehat{r}=0.9285\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0.3714\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=\frac{(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(-37.70\mu \text{C)}}{{(1.616\text{m)}}^2}\widehat{r}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-129.8\text{kN/C)(0}\text{.9285<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0.3714\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-120.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+\text{(-48}\text{.22kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Electric field at point P due to plane 1

Substituting values in the formula $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

  ${\overrightarrow{E}}_{plane}=\frac{\text{2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{(-<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  ${\overrightarrow{E}}_{plane}=(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point P due to line charge.

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=\frac{2k\lambda }{r}\widehat{r}$

From the figure

  $\begin{array}{l}\overrightarrow{r}\text{'}=0.5\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.5<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\\ \text{r'=}\sqrt{{(0.5)}^2+{(-0.5)}^2}=0.707\text{m}\\ \text{<mover accent="true"><mi>r</mi> <mo>^</mo></mover> '=}0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\end{array}$

Substituting the values

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=\frac{2(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(4.0\mu \text{C/)}}{0.707\text{m}}(0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})$

  $\begin{array}{l}\\ {\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=(71.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-71.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\end{array}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

  $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}=(71.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-71.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-112.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

  $\overrightarrow{E}=(-154\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(-184.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $\text{vector magnitude E=}\sqrt{(x^2+y^2)}$

  $\text{E=}\sqrt{{(-154.0\text{kN/C)}}^2+{(-184.9\text{kN/C)}}^2}$

  $\text{E=241kN/C}$

  $\text{vector direction }\theta {\text{=tan}}^{-1}\left(\frac{x}{y}\right)$

  $\theta ={\tan }^{-1}\left(\frac{\text{-154kN/C}}{\text{-184}\text{.9kN/C}}\right)=220°$

Conclusion:

The electric field $E=241\text{kN/C}$ pointing at $\theta \text{=220}°$ .

Answer to Problem 78P

The electric field at point P is $E=241\text{kN/C}$ pointing at $\theta \text{=220}°$ .

Explanation of Solution

Given:

  

The charges are placed as shown in the figure. The first plane at $x=2.0\text{m}$ . The line charge passing through the origin at an angle of $\text{45}°$ with the x axis in the x-y plane. The spherical shell centered at point $\text{(1}\text{.5m,0}\text{.5m)}$ in the x-y plane.

The charge densities are

  $\sigma =2\mu {\text{nC/m}}^2$

  $\lambda =4\mu {\text{nC/m}}^2$

  $\rho =-6.0\mu {\text{C/m}}^3$

Formula Used:

Electric field

  $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

E is the electric field.

  $\sigma$ is the surface charge density.

  $\widehat{r}$ is the unit vector in the direction normal to the charged plane.

  ${\epsilon }_o$ is the permittivity of free space.

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

  $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

Electric field at point P due to sphere.

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}\text{=}\frac{4\pi }{3}kr\text{'}\rho \widehat{r}\text{'}$

  $r\text{'}$ Represents the distance from the center of the sphere to P. constructing a Gaussian sphere of radius $r\text{'}$ centered at $(1,0)$

From the figure

  $\overrightarrow{r}\text{'}=-0.5\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.5<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $r\text{'}=\sqrt{{(-0.5)}^2+{(-0.5)}^2}=0.707\text{m}$

  $\widehat{r}\text{'}=-0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Substituting the values

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=\frac{4\pi }{3}(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(0.707\text{m)(-6}\text{.00}\mu {\text{C/m}}^3)\left[0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\right]$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-112.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{Q}}_{sphere}=\sigma A_{sphere}$

  $=4\pi \sigma R^2$

  ${\text{Q}}_{sphere}=4\pi (-3.0\mu {\text{C/m}}^2){(1.0\text{m)}}^2$

  ${\text{Q}}_{sphere}=-37.30\mu \text{C}$

  $\widehat{r}=0.9285\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0.3714\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=\frac{(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(-37.70\mu \text{C)}}{{(1.616\text{m)}}^2}\widehat{r}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-129.8\text{kN/C)(0}\text{.9285<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0.3714\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{sphere}=(-120.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+\text{(-48}\text{.22kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Electric field at point P due to plane 1

Substituting values in the formula $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_o}\widehat{r}$

  ${\overrightarrow{E}}_{plane}=\frac{\text{2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{(-<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  ${\overrightarrow{E}}_{plane}=(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point P due to line charge.

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=\frac{2k\lambda }{r}\widehat{r}$

From the figure

  $\begin{array}{l}\overrightarrow{r}\text{'}=0.5\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.5<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\\ \text{r'=}\sqrt{{(0.5)}^2+{(-0.5)}^2}=0.707\text{m}\\ \text{<mover accent="true"><mi>r</mi> <mo>^</mo></mover> '=}0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\end{array}$

Substituting the values

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=\frac{2(8.988\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2)(4.0\mu \text{C/)}}{0.707\text{m}}(0.707\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> -0}\text{.707<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})$

  $\begin{array}{l}\\ {\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_{line}=(71.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-71.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}\end{array}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_{line}+{\overrightarrow{E}}_{shpere}$

  $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}=(71.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-71.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+(-112.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

  $\overrightarrow{E}=(-154\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(-184.9\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $\text{vector magnitude E=}\sqrt{(x^2+y^2)}$

  $\text{E=}\sqrt{{(-154.0\text{kN/C)}}^2+{(-184.9\text{kN/C)}}^2}$

  $\text{E=241kN/C}$

  $\text{vector direction }\theta {\text{=tan}}^{-1}\left(\frac{x}{y}\right)$

  $\theta ={\tan }^{-1}\left(\frac{\text{-154kN/C}}{\text{-184}\text{.9kN/C}}\right)=220°$

Conclusion:

The electric field $E=241\text{kN/C}$ pointing at $\theta \text{=220}°$ .