Chapter 5, Problem 42P

(a)

To determine

Find the magnitude and direction of the electric field in the glass and in the enclosed air gap when the field is normal to the glass surfaces.

Answer to Problem 42P

The electric field in the glass is $E_{\text{glass}}=705.9\text{V}/\text{m}$ and the angle is ${\theta }_2=0$. And the electric field in the air is $E_{\text{air}}=6000\text{V}/\text{m}$ and the angle is ${\theta }_3=0$.

Explanation of Solution

Calculation:

Refer to Figure given in the textbook.

Given, the relative permittivity of glass is ${\epsilon }_r=8.5$ and the relative permittivity of oil is ${\epsilon }_r=3$, and the uniform electric field strength in oil is $2\text{kV}/\text{m}$.

The given Figure is modified as shown in Figure 1.

Refer to Figure 1. The normal component of electric field $E_{1n}$ is the same as the electric field strength in the oil. That is,

$E_{1n}=E_{\text{oil}}=2\text{kV}/\text{m}$

A uniform electric field of strength $2\text{kV}/\text{m}$ in the horizontal direction exists in the oil. Therefore,

$E_{1t}=0$

Consider the general expression for dielectric-dielectric interface.

$\begin{array}{c}{E}_{2t}=E_{1t}\\ =0\left\{\because E_{1t}=0\right\}\end{array}$

Consider the general expression for dielectric-dielectric interface.

$D_{1n}=D_{2n}$

The above equation becomes,

$\begin{array}{l}{\epsilon }_{r\left(\text{oil}\right)}{E}_{1n}={\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}\\ E_{2n}=\frac{{\epsilon }_{r\left(\text{oil}\right)}{E}_{1n}}{{\epsilon }_{r\left(\text{glass}\right)}}\end{array}$

Substitute 3 for ${\epsilon }_{r\left(\text{oil}\right)}$, $2\times {10}^3$ for $E_{1n}$, and 8.5 for ${\epsilon }_{r\left(\text{glass}\right)}$ in the above equation.

$\begin{array}{c}{E}_{2n}=\frac{\left(3\right)\left(2\times {10}^3\right)}{\left(8.5\right)}\\ =705.9\text{V}/\text{m}\end{array}$

Find the electric field strength in the glass.

$E_{\text{glass}}=\sqrt{{\left(E_{2t}\right)}^2+{\left(E_{2n}\right)}^2}$

Substitute 0 for $E_{2t}$ and 705.9 for $E_{2n}$ in the above equation.

$\begin{array}{c}{E}_{\text{glass}}=\sqrt{{\left(0\right)}^2+{\left(705.9\right)}^2}\\ =705.9\text{V}/\text{m}\end{array}$

As the tangential component is zero, the angle is zero. That is, ${\theta }_2=0$.

Consider the general expression for dielectric-dielectric interface.

$\begin{array}{c}{E}_{3t}=E_{2t}\\ =0\left\{\because E_{2t}=0\right\}\end{array}$

Consider the general expression for dielectric-dielectric interface.

$D_{2n}=D_{3n}$

The above equation becomes,

$\begin{array}{l}{\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}={\epsilon }_{r\left(\text{air}\right)}{E}_{3n}\\ E_{3n}=\frac{{\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}}{{\epsilon }_{r\left(\text{air}\right)}}\end{array}$

Substitute 1 for ${\epsilon }_{r\left(\text{air}\right)}$, $705.9$ for $E_{2n}$, and 8.5 for ${\epsilon }_{r\left(\text{glass}\right)}$ in the above equation.

$\begin{array}{c}{E}_{3n}=\frac{\left(8.5\right)\left(705.9\right)}{\left(1\right)}\\ =6000\text{V}/\text{m}\end{array}$

Find the electric field strength in the air.

$E_{\text{air}}=\sqrt{{\left(E_{3t}\right)}^2+{\left(E_{3n}\right)}^2}$

Substitute 0 for $E_{3t}$ and 6000 for $E_{3n}$ in the above equation.

$\begin{array}{c}{E}_{\text{air}}=\sqrt{{\left(0\right)}^2+{\left(6000\right)}^2}\\ =6000\text{V}/\text{m}\end{array}$

As the tangential component is zero, the angle is zero. That is, ${\theta }_3=0$.

Conclusion:

Thus, the electric field in the glass is $E_{\text{glass}}=705.9\text{V}/\text{m}$ and the angle is ${\theta }_2=0$. And the electric field in the air is $E_{\text{air}}=6000\text{V}/\text{m}$ and the angle is ${\theta }_3=0$.

(b)

To determine

Find the magnitude and direction of the electric field in the glass and in the enclosed air gap when the field in the oil makes an angle of $75°$ with a normal to the glass surfaces.

Answer to Problem 42P

The electric field in the glass is $E_{\text{glass}}=1940.5\text{V}/\text{m}$ and the angle is ${\theta }_2=84.6°$. And the electric field in the air is $E_{\text{air}}=2478.6\text{V}/\text{m}$ and the angle is ${\theta }_3=51.2°$.

Explanation of Solution

Calculation:

Refer to Figure given in the textbook.

The given Figure is modified as shown in Figure 2.

Given, the field in the oil makes an angle of $75°$ with a normal to the glass surfaces.

Consider the general expressio for normal component of $E_{1n}$.

$\begin{array}{c}{E}_{1n}=E_{\text{oil}}\cos \left(75°\right)\left\{\because \theta =75°\right\}\\ =E_{\text{oil}}\left(0.2588\right)\end{array}$

Substitute $2\times {10}^3$ for $E_{\text{oil}}$ in above equation.

$\begin{array}{c}{E}_{1n}=\left(2\times {10}^3\right)\left(0.25882\right)\\ =517.63\text{V}/\text{m}\end{array}$

Similarly,

$\begin{array}{c}{E}_{1t}=E_{\text{oil}}\sin \left(75°\right)\\ =E_{\text{oil}}\left(0.965925\right)\end{array}$

Substitute $2\times {10}^3$ for $E_{\text{oil}}$ in above equation.

$\begin{array}{c}{E}_{1t}=\left(2\times {10}^3\right)\left(0.965925\right)\\ =1931.85\text{V}/\text{m}\end{array}$

For dielectric-dielectric interface,

$E_{2t}=E_{3t}=1931.85\text{V}/\text{m}$

For dielectric-dielectric interface,

$\begin{array}{l}{\epsilon }_{r\left(\text{oil}\right)}{E}_{1n}={\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}\\ E_{2n}=\frac{{\epsilon }_{r\left(\text{oil}\right)}{E}_{1n}}{{\epsilon }_{r\left(\text{glass}\right)}}\end{array}$

Substitute 3 for ${\epsilon }_{r\left(\text{oil}\right)}$, 517.63 for $E_{1n}$, and 8.5 for ${\epsilon }_{r\left(\text{glass}\right)}$ in the above equation.

$\begin{array}{c}{E}_{2n}=\frac{\left(3\right)\left(517.63\right)}{\left(8.5\right)}\\ =182.7\text{V}/\text{m}\end{array}$

Find the electric field strength in the glass.

$E_{\text{glass}}=\sqrt{{\left(E_{2t}\right)}^2+{\left(E_{2n}\right)}^2}$

Substitute 1931.85 for $E_{2t}$ and 182.7 for $E_{2n}$ in the above equation.

$\begin{array}{c}{E}_{\text{glass}}=\sqrt{{\left(1931.85\right)}^2+{\left(182.7\right)}^2}\\ =1940.5\text{V}/\text{m}\end{array}$

Find the value of angle ${\theta }_2$.

${\theta }_2={\tan }^{-1}\left(\frac{E_{2t}}{E_{2n}}\right)$

Substitute 1931.85 for $E_{2t}$ and 182.7 for $E_{2n}$ in the above equation.

$\begin{array}{c}{\theta }_2={\tan }^{-1}\left(\frac{1931.85}{182.7}\right)\\ =84.6°\end{array}$

For dielectric-dielectric interface,

$E_{3t}=E_{2t}=1931.85\text{V}/\text{m}$

For dielectric- dielectric interface,

$\begin{array}{l}{\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}={\epsilon }_{r\left(\text{air}\right)}{E}_{3n}\\ E_{3n}=\frac{{\epsilon }_{r\left(\text{glass}\right)}{E}_{2n}}{{\epsilon }_{r\left(\text{air}\right)}}\end{array}$

Substitute 1 for ${\epsilon }_{r\left(\text{air}\right)}$, 182.7 for $E_{2n}$, and 8.5 for ${\epsilon }_{r\left(\text{glass}\right)}$ in the above equation.

$\begin{array}{c}{E}_{3n}=\frac{\left(8.5\right)\left(182.7\right)}{\left(1\right)}\\ =1552.9\text{V}/\text{m}\end{array}$

Find the electric field strength in the air.

$E_{\text{air}}=\sqrt{{\left(E_{3t}\right)}^2+{\left(E_{3n}\right)}^2}$

Substitute 1931.85 for $E_{3t}$ and 1552.9 for $E_{2n}$ in the above equation.

$\begin{array}{c}{E}_{\text{air}}=\sqrt{{\left(1931.85\right)}^2+{\left(1552.9\right)}^2}\\ =2478.6\text{V}/\text{m}\end{array}$

Find the value of angle ${\theta }_3$.

${\theta }_3={\tan }^{-1}\left(\frac{E_{3t}}{E_{3n}}\right)$

Substitute 1931.85 for $E_{3t}$ and 1552.9 for $E_{2n}$ in the above equation.

$\begin{array}{c}{\theta }_3={\tan }^{-1}\left(\frac{1931.85}{1552.9}\right)\\ =51.2°\end{array}$

Conclusion:

Thus, the electric field in the glass is $E_{\text{glass}}=1940.5\text{V}/\text{m}$ and the angle is ${\theta }_2=84.6°$. And the electric field in the air is $E_{\text{air}}=2478.6\text{V}/\text{m}$ and the angle is ${\theta }_3=51.2°$.