Chapter 22, Problem 77P

a.

To determine

The magnitude and the direction of the electric field at $x=0.40\text{m}$ .

The electric field $E=203.6\text{kN/C}$ pointing at $\theta \text{=56}\text{.3}°$ from the x axis.

Given:

  

The charges are placed as shown in the figure. The first plane at $y=-0.6\text{m}$ . The second plane at $x=\text{1}\text{.0}\text{m}$ . The spherical shell centered at the intersection of the planes at point $\text{(1}\text{.0,-0}\text{.6)}$ in the x-y plane.

The surface charge densities are

  ${\sigma }_1=3.0\mu {\text{nC/m}}^2$

  ${\sigma }_2=-2.0\mu {\text{nC/m}}^2$

  ${\sigma }_3=-3.0\mu {\text{C/m}}^3$

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 $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

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

Electric field at point 1 due to sphere.

As the point is inside the sphere the electric field is zero.

  ${\overrightarrow{E}}_{shpere}=0$

Electric field at point 1 due to plane 1

Substituting values

  ${\overrightarrow{E}}_1=\frac{\text{3}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\overrightarrow{E}}_1=(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The electric field at point 1 due to plane 2.

  ${\overrightarrow{E}}_2=\frac{\text{-2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}(-\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  $E_2=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Substituting

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +0}$

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $E=\sqrt{(x^2+y^2)}$

  $E=\sqrt{{(112.9\text{kN/C)}}^2+{(169.4\text{kN/C)}}^2}$

  $\text{E=203}{\text{.6kN/C}}^2$

Direction:

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

  $\theta ={\tan }^{-1}\left(\frac{169.4\text{kN/C}}{112.9\text{kN/C}}\right)=56.31°$

Conclusion:

The electric field $E=203.6\text{kN/C}$ pointing at $\theta \text{=56}\text{.3}°$ from the x axis.

b.

The magnitude and the direction of the electric field at $x=2.50\text{m}$ .

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.

Given:

  

The charges are placed as shown in the figure. The first plane at $y=-0.6\text{m}$ . The second plane at $x=\text{1}\text{.0}\text{m}$ . The spherical shell centered at the intersection of the planes at point $\text{(1}\text{.0,-0}\text{.6)}$ in the x-y plane.

The surface charge densities are

  ${\sigma }_1=3.0\mu {\text{nC/m}}^2$

  ${\sigma }_2=-2.0\mu {\text{nC/m}}^2$

  ${\sigma }_3=-3.0\mu {\text{C/m}}^3$

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 $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

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

Electric field at point 1 due to sphere.

  $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover> =}\left(\frac{k{Q}_{sphere}}{r^2}\right)\widehat{r}$ where Q is the charge in the sphere.

Where $\widehat{r}$ is a unit vector pointing from $(1.0\text{m,-0}\text{.6m) to (2}\text{.50m,0)}$

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

  $=4\pi \sigma R^2$

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

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

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

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

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

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

Electric field at point 1 due to plane 1

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

  ${\overrightarrow{E}}_1=\frac{\text{3}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\overrightarrow{E}}_1=(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The electric field at point 1 due to plane 2.

  ${\overrightarrow{E}}_2=\frac{\text{-2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}(\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  $E_2=-(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Substituting

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-120.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+\text{(-48}\text{.22kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $\overrightarrow{E}=(-233.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(121.2\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $E=\sqrt{(x^2+y^2)}$

  $E=\sqrt{{(-233.5\text{kN/C)}}^2+{(121.2\text{kN/C)}}^2}$

  $E=\text{263kN/C}$

Direction:

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

  $\theta ={\tan }^{-1}\left(\frac{\text{121}\text{.2kN/C}}{\text{-233}\text{.5kN/C}}\right)=153°$

Conclusion:

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.

Answer to Problem 77P

The electric field $E=203.6\text{kN/C}$ pointing at $\theta \text{=56}\text{.3}°$ from the x axis.

Explanation of Solution

Given:

  

The charges are placed as shown in the figure. The first plane at $y=-0.6\text{m}$ . The second plane at $x=\text{1}\text{.0}\text{m}$ . The spherical shell centered at the intersection of the planes at point $\text{(1}\text{.0,-0}\text{.6)}$ in the x-y plane.

The surface charge densities are

  ${\sigma }_1=3.0\mu {\text{nC/m}}^2$

  ${\sigma }_2=-2.0\mu {\text{nC/m}}^2$

  ${\sigma }_3=-3.0\mu {\text{C/m}}^3$

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 $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

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

Electric field at point 1 due to sphere.

As the point is inside the sphere the electric field is zero.

  ${\overrightarrow{E}}_{shpere}=0$

Electric field at point 1 due to plane 1

Substituting values

  ${\overrightarrow{E}}_1=\frac{\text{3}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\overrightarrow{E}}_1=(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The electric field at point 1 due to plane 2.

  ${\overrightarrow{E}}_2=\frac{\text{-2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}(-\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  $E_2=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Substituting

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +0}$

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $E=\sqrt{(x^2+y^2)}$

  $E=\sqrt{{(112.9\text{kN/C)}}^2+{(169.4\text{kN/C)}}^2}$

  $\text{E=203}{\text{.6kN/C}}^2$

Direction:

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

  $\theta ={\tan }^{-1}\left(\frac{169.4\text{kN/C}}{112.9\text{kN/C}}\right)=56.31°$

Conclusion:

The electric field $E=203.6\text{kN/C}$ pointing at $\theta \text{=56}\text{.3}°$ from the x axis.

b.

To determine

The magnitude and the direction of the electric field at $x=2.50\text{m}$ .

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.

Given:

  

The charges are placed as shown in the figure. The first plane at $y=-0.6\text{m}$ . The second plane at $x=\text{1}\text{.0}\text{m}$ . The spherical shell centered at the intersection of the planes at point $\text{(1}\text{.0,-0}\text{.6)}$ in the x-y plane.

The surface charge densities are

  ${\sigma }_1=3.0\mu {\text{nC/m}}^2$

  ${\sigma }_2=-2.0\mu {\text{nC/m}}^2$

  ${\sigma }_3=-3.0\mu {\text{C/m}}^3$

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 $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

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

Electric field at point 1 due to sphere.

  $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover> =}\left(\frac{k{Q}_{sphere}}{r^2}\right)\widehat{r}$ where Q is the charge in the sphere.

Where $\widehat{r}$ is a unit vector pointing from $(1.0\text{m,-0}\text{.6m) to (2}\text{.50m,0)}$

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

  $=4\pi \sigma R^2$

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

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

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

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

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

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

Electric field at point 1 due to plane 1

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

  ${\overrightarrow{E}}_1=\frac{\text{3}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\overrightarrow{E}}_1=(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The electric field at point 1 due to plane 2.

  ${\overrightarrow{E}}_2=\frac{\text{-2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}(\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  $E_2=-(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Substituting

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-120.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+\text{(-48}\text{.22kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $\overrightarrow{E}=(-233.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(121.2\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $E=\sqrt{(x^2+y^2)}$

  $E=\sqrt{{(-233.5\text{kN/C)}}^2+{(121.2\text{kN/C)}}^2}$

  $E=\text{263kN/C}$

Direction:

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

  $\theta ={\tan }^{-1}\left(\frac{\text{121}\text{.2kN/C}}{\text{-233}\text{.5kN/C}}\right)=153°$

Conclusion:

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.

Answer to Problem 77P

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.

Explanation of Solution

Given:

  

The charges are placed as shown in the figure. The first plane at $y=-0.6\text{m}$ . The second plane at $x=\text{1}\text{.0}\text{m}$ . The spherical shell centered at the intersection of the planes at point $\text{(1}\text{.0,-0}\text{.6)}$ in the x-y plane.

The surface charge densities are

  ${\sigma }_1=3.0\mu {\text{nC/m}}^2$

  ${\sigma }_2=-2.0\mu {\text{nC/m}}^2$

  ${\sigma }_3=-3.0\mu {\text{C/m}}^3$

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 $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Calculations:

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

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

Electric field at point 1 due to sphere.

  $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover> =}\left(\frac{k{Q}_{sphere}}{r^2}\right)\widehat{r}$ where Q is the charge in the sphere.

Where $\widehat{r}$ is a unit vector pointing from $(1.0\text{m,-0}\text{.6m) to (2}\text{.50m,0)}$

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

  $=4\pi \sigma R^2$

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

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

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

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

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

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

Electric field at point 1 due to plane 1

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

  ${\overrightarrow{E}}_1=\frac{\text{3}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\overrightarrow{E}}_1=(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The electric field at point 1 due to plane 2.

  ${\overrightarrow{E}}_2=\frac{\text{-2}\text{.0}\mu {\text{nC/m}}^2}{2(8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2)}(\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>})$

  $E_2=-(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Substituting in the equation

The resultant electric field at point is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_{shpere}$

Substituting

  $\overrightarrow{E}=(112.9\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(169.4\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover> +}(-120.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+\text{(-48}\text{.22kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $\overrightarrow{E}=(-233.5\text{kN/C)<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}(121.2\text{kN/C)<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

The magnitude of the electric field is

  $E=\sqrt{(x^2+y^2)}$

  $E=\sqrt{{(-233.5\text{kN/C)}}^2+{(121.2\text{kN/C)}}^2}$

  $E=\text{263kN/C}$

Direction:

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

  $\theta ={\tan }^{-1}\left(\frac{\text{121}\text{.2kN/C}}{\text{-233}\text{.5kN/C}}\right)=153°$

Conclusion:

The electric field $E=263\text{kN/C}$ pointing at $\theta \text{=153}°$ from the x axis.