high voltage techniques refraction of electric fields
TRANSCRIPT
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HIGH VOLTAGE TECHNIQUES Refraction of Electric Fields Assistant Professor Suna BOLAT
Eastern Mediterranean University
Department of Electric & Electronic Engineering
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Non-parallel configuration
What happens to electric field if the plates are NOT
parallel?
In order to evaluate that, we have to know the
behaviour of field lines at boundary surfaces.
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Refraction
Assume an electric field line coming through a boundary which connects two different dielectric.
In such a case, Electric Field vector will REFRACT to the other dielectric.
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Refraction of electric field lines & equipotential lines
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What is what?
• 1, α2: Refraction angles of electric field lines.
• β1, β2: Refraction angles of equipotential lines
(REMEMBER: always perpendicular to the field lines!!!)
• Et1, Et2: tangent component of the electric fields E1 and E2
• En1, En2: normal (perpendicular) component of the electric fields E1
and E2
• Normal component of the field vector forces dielectric to BREAKDOWN!
• Tangent component of the field vector forces the dielectric to FLASHOVER!
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At the interface field lines are refracted so that;
• According to 𝐷. 𝑑𝑆 = 0 and considering D = E E = D/
𝐸𝑡1 = 𝐸𝑡2
Could you follow this equation?
Considering D = E E = D/
𝐷𝑡1𝐷𝑡2
=𝜀1𝜀2
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Refraction of electric displacement field
Similar to the electric field lines, displacement lines are refracted at the interface so that;
𝐷𝑛1 = 𝐷𝑛2
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Displacement vectors
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According to 𝐷. 𝑑𝑆 = 0 (one more time)
𝐸𝑛1𝐸𝑛2
=𝜀2𝜀1
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Simple trigonometric concepts
For electric field vector components:
𝐸𝑡1 = 𝐸1 sin 𝛼1 𝐸𝑡2 = 𝐸2 sin 𝛼2
For displacement vector components:
𝐷𝑛1 = 𝐷1 cos 𝛼1 = 𝜀1𝐸1cos 𝛼1 𝐷𝑛2 = 𝐷2 cos 𝛼2 = 𝜀2𝐸2cos 𝛼2
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Because 𝐸𝑡1 = 𝐸𝑡2 and 𝐷𝑛1 = 𝐷𝑛2 ; we can write
𝐸𝑡1 = 𝐸𝑡2𝐷𝑛1 = 𝐷𝑛2
→
𝐸1 sin 𝛼1𝜀1𝐸1cos 𝛼1
=𝐸2 sin 𝛼2𝜀2𝐸2cos 𝛼2
tan𝛼1tan 𝛼2
=𝜀1𝜀2
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Refraction of equipotential lines
𝛽1 = 90 − 𝛼1 𝛽2 = 90 − 𝛼2
tan(90 − 𝛽1)
tan(90 − 𝛽2)=𝜀1𝜀2
refraction angles of equipotential lines.
tan𝛽1tan 𝛽2
=𝜀2𝜀1
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TIME FOR EXAMPLES