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HIGH VOLTAGE TECHNIQUES Refraction of Electric Fields Assistant Professor Suna BOLAT

Eastern Mediterranean University

Department of Electric & Electronic Engineering

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.

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.

Refraction of electric field lines & equipotential lines

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!

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

Refraction of electric displacement field

Similar to the electric field lines, displacement lines are refracted at the interface so that;

𝐷𝑛1 = 𝐷𝑛2

Displacement vectors

According to 𝐷. 𝑑𝑆 = 0 (one more time)

𝐸𝑛1𝐸𝑛2

=𝜀2𝜀1

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

Because 𝐸𝑡1 = 𝐸𝑡2 and 𝐷𝑛1 = 𝐷𝑛2 ; we can write

𝐸𝑡1 = 𝐸𝑡2𝐷𝑛1 = 𝐷𝑛2

→

𝐸1 sin 𝛼1𝜀1𝐸1cos 𝛼1

=𝐸2 sin 𝛼2𝜀2𝐸2cos 𝛼2

tan𝛼1tan 𝛼2

=𝜀1𝜀2

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

TIME FOR EXAMPLES