Concept explainers
(a)
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.
(a)
Answer to Problem 46P
The electric field in the glass is
Explanation of Solution
Calculation:
Refer to Figure given in the textbook.
Given, the relative permittivity of glass is
The given Figure is modified as shown in Figure 1.
Refer to Figure 1. The normal component of electric field
A uniform electric field of strength
Consider the general expression for dielectric-dielectric interface.
Consider the general expression for dielectric-dielectric interface.
The above equation becomes,
Substitute 3 for
Find the electric field strength in the glass.
Substitute 0 for
As the tangential component is zero, the angle is zero. That is,
Consider the general expression for dielectric-dielectric interface.
Consider the general expression for dielectric-dielectric interface.
The above equation becomes,
Substitute 1 for
Find the electric field strength in the air.
Substitute 0 for
As the tangential component is zero, the angle is zero. That is,
Conclusion:
Thus, the electric field in the glass is
(b)
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
(b)
Answer to Problem 46P
The electric field in the glass is
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
Consider the general expressio for normal component of
Substitute
Similarly,
Substitute
For dielectric-dielectric interface,
For dielectric-dielectric interface,
Substitute 3 for
Find the electric field strength in the glass.
Substitute 1931.85 for
Find the value of angle
Substitute 1931.85 for
For dielectric-dielectric interface,
For dielectric- dielectric interface,
Substitute 1 for
Find the electric field strength in the air.
Substitute 1931.85 for
Find the value of angle
Substitute 1931.85 for
Conclusion:
Thus, the electric field in the glass is
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Chapter 5 Solutions
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