Conductors - Electromagnetic Theory

Questions and Answers - Sanfoundry

by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Conductors”.

1. Which of the following are conductors?

a) Ceramics
b) Plastics
c) Mercury
d) Rubber
View Answer

Answer: c
Explanation: Normally, metals are said to be good conductors. Here mercury is the only metal
(which is in liquid form). The other options are insulators.

2. Find the range of band gap energy for conductors.

a) >6 eV
b) 0.2-0.4 eV
c) 0.4-2 eV
d) 2-6 eV
View Answer

Answer: b
Explanation: Conductors are materials with least band gap energy. The smallest range in this
group is 0.2-0.4 eV.

3. Conduction in metals is due to

a) Electrons only
b) Electrons and holes
c) Holes only
d) Applied electric field
View Answer

Answer: a
Explanation: Conduction in metals is only due to majority carriers, which are electrons.
Electrons and holes are responsible for conduction in a semiconductor.

4. Find the band gap energy when a light of wavelength 1240nm is incident on it.
a) 1eV
b) 2eV
c) 3eV
d) 4eV
View Answer

Answer: a
Explanation: The band gap energy in electron volt when wavelength is given is, Eg = 1.24(μm)/λ
= 1.24 x 10-6/1240 x 10-9 = 1eV.

5. Alternating current measured in a transmission line will be

a) Peak value
b) Average value
c) RMS value
d) Zero
View Answer

Answer: c
Explanation: The instantaneous current flowing in a transmission line, when measured using an
ammeter, will give RMS current value. This value is 70.7% of the peak value. This is because,
due to oscillations in AC, it is not possible to measure peak value. Hence to normalise, we
consider current at any time in a line will be the RMS current.

6. The current in a metal at any frequency is due to

a) Conduction current
b) Displacement current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
View Answer

Answer: a
Explanation: At any frequency, the current through the metal will be due to conduction current.
Only at high frequencies and when medium is air, the conduction is due to displacement current.
Thus in general the current in metal is due to conduction current, which depends on the mobility
of the carriers.

7. For conductors, the free electrons will exist at

a) Valence band
b) Middle of valence and conduction band
c) Will not exist
d) Conduction band
View Answer

Answer: d
Explanation: In conductors, the free electrons exist in the conduction band. Since the band gap
energy is very low, less energy is required to transport the free electrons to the conduction band,
as they are readily available to conduct.
8. The current flowing through an insulating medium is called
a) Conduction
b) Convection
c) Radiation
d) Susceptibility
View Answer

Answer: b
Explanation: A beam of electrons in a vacuum tube is called convection current. It occurs when
current flows through an insulating medium like liquid, vacuum etc.

9. Find the conduction current density when conductivity of a material is 500 units and
corresponding electric field is 2 units.
a) 500
b) 250
c) 1000
d) 2000
View Answer

Answer: c
Explanation: The conduction current density is given by, J = σE
J = 500 X 2 = 1000 units.

10. Calculate the convection current when electron density of 200 units is travelling at a speed of
12m/s.
a) 16.67
b) 2400
c) 2880
d) 0.06
View Answer

Answer: b
Explanation: The convection current density is given by, J = ρeV
J = 200 X 12= 2400 units.

Dielectrics - Electromagnetic Theory

Questions and Answers - Sanfoundry
by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Dielectrics”.
1. A dielectric is always an insulator. But an insulator is not necessarily a dielectric. State
True/False.
a) True
b) False
View Answer

Answer: a
Explanation: For a material to be dielectric, its permittivity should be very high. This is seen in
insulators. For a material to be insulator, the condition is to have large band gap energy.
However, this is not necessary for a dielectric.

2. Identify a good dielectric.

a) Iron
b) Ceramics
c) Plastic
d) Magnesium
View Answer

Answer: b
Explanation: Iron and magnesium are metals. Hence they need not be considered. Both ceramics
and plastic are insulators. But dielectric constant is more for ceramics always. Hence ceramics is
the best dielectric.

3. A dielectric can be made a conductor by

a) Compression
b) Heating
c) Doping
d) Freezing
View Answer

Answer: b
Explanation: On increasing the temperature, the free electrons in an insulator can be promoted
from valence to conduction band. Gradually, it can act as a conductor through heating process.
This condition is called dielectric breakdown, wherein the insulator loses its dielectric property
and starts to conduct.

4. Find the dielectric constant for a material with electric susceptibility of 4.

a) 3
b) 5
c) 8
d) 16
View Answer

Answer: b
Explanation: The electric susceptibility is given by χe = εr – 1. For a susceptibility of 4, the
dielectric constant will be 5. It has no unit.
5. For a dielectric which of the following properties hold good?
a) They are superconductors at high temperatures
b) They are superconductors at low temperatures
c) They can never become a superconductor
d) They have very less dielectric breakdown voltage
View Answer

Answer: b
Explanation: Superconductors are characterised by diamagnetism behaviour and zero resistivity,
which true for a dielectric. They occur only at low temperature. Thus a dielectric can become a
superconductor at low temperatures with very high dielectric breakdown voltage.

6. The magnetic field which destroys the superconductivity is called

a) Diamagnetic field
b) Ferromagnetic field
c) Ferrimagnetic field
d) Critical field
View Answer

Answer: d
Explanation: Critical field is that strong magnetic field which can destroy the superconductivity
of a material. The temperature at which this occurs is called transition temperature.

7. The magnetic susceptibility in a superconductor will be

a) Positive
b) Negative
c) Zero
d) Infinity
View Answer

Answer: b
Explanation: Due to perfect diamagnetism in a superconductor, its magnetic susceptibility will
be negative. This phenomenon is called Meissner effect.

8. The superconducting materials will be independent of which of the following?

a) Magnetic field
b) Electric field
c) Magnetization
d) Temperature
View Answer

Answer: b
Explanation: Superconducting materials depends only on the applied magnetic field, resultant
magnetization at the temperature considered. It is independent of the applied electric field and
the corresponding polarization.
9. Find the mean free path of an electron travelling at a speed of 18m/s in 2 seconds.
a) 9
b) 36
c) 0.11
d) 4.5
View Answer

Answer: b
Explanation: The mean free path is defined as the average distance travelled by an electron
before collision takes place. It is given by, d = v x τc, where v is the velocity and τc is the
collision time. Thus d = 18 x 2 = 36m.

10. Find the velocity of an electron when its kinetic energy is equal to one electron volt (in
105m/s).
Given charge of an electron e = 1.6 x 10-19 and mass of an electron m = 9.1 x 10-31.
a) 3.9
b) 4.9
c) 5.9
d) 6.9
View Answer

Answer: c
Explanation: When the kinetic energy and one electron volt are equal, we can equate mv2/2 = eV.
Put e and m in the equation to get velocity v = 5.9 x 105 m/s.

Electromagnetic Theory Interview Questions

and Answers for Experienced - Sanfoundry
by staff10

This set of Electromagnetic Theory Interview Questions and Answers for Experienced people
focuses on “Displacement and Conduction Current”.

1. Find the conductivity of a material with conduction current density 100 units and electric field
of 4 units.
a) 25
b) 400
c) 0.04
d) 1600
View Answer

Answer: a
Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E =
100/4 = 25 units.
2. Calculate the displacement current density when the electric flux density is 20sin 0.5t.
a) 10sin 0.5t
b) 10cos 0.5t
c) 20sin 2t
d) 20cos 2t
View Answer

Answer: b
Explanation: The displacement current density is given by, Jd = dD/dt.
Jd = d(20sin 0.5t)/dt = 20cos 0.5t (0.5) = 10cos 0.5t.

3. Find the magnitude of the displacement current density in air at a frequency of 18GHz in
frequency domain. Take electric field E as 4 units.
a) 18
b) 72
c) 36
d) 4
View Answer

Answer: d
Explanation: Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier
transform, Jd = εjωE. The magnitude of
Jd = εωE = ε(2πf)E. On substituting, we get 4 ampere.

4. Calculate the frequency at which the conduction and displacement currents become equal with
unity conductivity in a material of permittivity 2.
a) 18 GHz
b) 9 GHz
c) 36 GHz
d) 24 GHz
View Answer

Answer: b
Explanation: When Jd = Jc , we get εωE = σE. Thus εo(2∏f) = σ. On substituting conductivity as
one and permittivity as 2, we get f = 9GHz.

5. The ratio of conduction to displacement current density is referred to as

a) Attenuation constant
b) Propagation constant
c) Loss tangent
d) Dielectric constant
View Answer

Answer: c
Explanation: Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The
loss tangent is used to determine if the material is a conductor or dielectric.
6. If the loss tangent is very less, then the material will be a
a) Conductor
b) Lossless dielectric
c) Lossy dielectric
d) Insulator
View Answer

Answer: b
Explanation: If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor
and the material should be a dielectric. Since it is specifically mentioned very less, assuming the
conductivity to be zero, the dielectric will be lossless (ideal).

7. In good conductors, the electric and magnetic fields will be

a) 45 in phase
b) 45 out of phase
c) 90 in phase
d) 90 out of phase
View Answer

Answer: b
Explanation: The electric and magnetic fields will be out of phase by 45 in good conductors.
This is because their intrinsic impedance is given by η = √(ωμ/σ) X (1+j). In polar form we get
45 out of phase.

8. In free space, which of the following will be zero?

a) Permittivity
b) Permeability
c) Conductivity
d) Resistivity
View Answer

Answer: c
Explanation: In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be
unity. Since the free space will contain no charges in it, the conductivity will be zero.

9. If the intrinsic angle is 20, then find the loss tangent.

a) tan 20
b) tan 40
c) tan 60
d) tan 80
View Answer

Answer: b
Explanation: The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be
tan 40.
10. The intrinsic impedance of free space is given by
a) 272 ohm
b) 412 ohm
c) 740 ohm
d) 377 ohm
View Answer

Answer: d
Explanation: The intrinsic impedance is given by η = √(μo/εo) ohm. Here εo = 8.854 x 10-12 and
μo = 4π x 10-7.
On substituting the values, we get η = 377 ohm.

Polarization - Electromagnetic Theory

Questions and Answers - Sanfoundry
by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Polarization”.

1. The best definition of polarisation is

a) Orientation of dipoles in random direction
b) Electric dipole moment per unit volume
c) Orientation of dipole moments
d) Change in polarity of every dipole
View Answer

Answer: b
Explanation: The polarisation is defined mathematically as the electric dipole moment per unit
volume. It is also referred to as the orientation of the dipoles in the direction of applied electric
field.

2. Calculate the polarisation vector of the material which has 100 dipoles per unit volume in a
volume of 2 units.
a) 200
b) 50
c) 400
d) 0.02
View Answer

Answer: a
Explanation: Polarisation vector P = N x p, where N = 100 and p = 2. On substituting we get P =
200 units.
3. Polarizability is defined as the
a) Product of dipole moment and electric field
b) Ratio of dipole moment to electric field
c) Ratio of electric field to dipole moment
d) Product of dielectric constant and dipole moment
View Answer

Answer: b
Explanation: Polarizability is a constant that is defined as the ratio of elemental dipole moment
to the electric field strength.

4. Calculate the energy stored per unit volume in a dielectric medium due to polarisation when P
= 9 units and E = 8 units.
a) 1.77
b) 2.25
c) 36
d) 144
View Answer

Answer: c
Explanation: The energy stored per unit volume in a dielectric medium is given by, W = 0.5 X
PE = 0.5 X 9 X 8 = 36 units.

5. Identify which type of polarisation depends on temperature.

a) Electronic
b) Ionic
c) Orientational
d) Interfacial
View Answer

Answer: c
Explanation: The electronic, ionic and interfacial polarisation depends on the atoms which are
independent with respect to temperature. Only the orientational polarisation is dependent on the
temperature and is inversely proportional to it.

6. Calculate the polarisation vector in air when the susceptibility is 5 and electric field is 12
units.
a) 3
b) 2
c) 60
d) 2.4
View Answer

Answer: c
Explanation: The polarisation vector is given by, P = ε0 x χe x E, where χe = 5 and ε0 = 12. On
substituting, we get P = 1 x 5 x 12 = 60 units.
7. In isotropic materials, which of the following quantities will be independent of the direction?
a) Permittivity
b) Permeability
c) Polarisation
d) Polarizability
View Answer

Answer: a
Explanation: Isotropic materials are those with radiate or absorb energy uniformly in all
directions (eg. Isotropic antenna). Thus it is independent of the direction.

8. The total polarisation of a material is the

a) Product of all types of polarisation
b) Sum of all types of polarisation
c) Orientation directions of the dipoles
d) Total dipole moments in the material
View Answer

Answer: b
Explanation: The total polarisation of a material is given by the sum of electronic, ionic,
orientational and interfacial polarisation of the material.

9. In the given types of polarisation, which type exists in the semiconductor?

a) Electronic
b) Ionic
c) Orientational
d) Interfacial or space charge
View Answer

Answer: d
Explanation: The interfacial type of polarisation occurs due to accumulation of charges at the
interface in a multiphase material. This interface or junction is found in a semiconductor
material.

10. Solids do not have which type of polarisation?

a) Ionic
b) Orientational
c) Interfacial
d) Electronic
View Answer

Answer: c
Explanation: Solids possess permanent dipole moments. Moreover they do not have junction like
semiconductors. Thus, solids neglect the interfacial and space charge polarisation. They possess
only electronic, ionic and orientational polarisations.
Electromagnetic Theory Test - Sanfoundry
by staff10

This set of Electromagnetic Theory test focuses on “Dielectric Strength and Constant”.

1. Which of the following is not an example of elemental solid dielectric?

a) Diamond
b) Sulphur
c) Silicon
d) Germanium
View Answer

Answer: c
Explanation: Elemental solid dielectrics are the materials consisting of single type of atoms.
Such materials have neither ions nor permanent dipoles and possess only electronic polarisation.
Its examples are diamond, sulphur and germanium.

2. Ionic non polar solid dielectrics contain more than one type of atoms but no permanent
dipoles. State True/False
a) True
b) False
View Answer

Answer: a
Explanation: In ionic crystals, the total polarisation is electronic and ionic in nature. Thus, it
implies that it contains more than one type of atom and no permanent dipoles.

3. Compute the refractive index when the dielectric constant is 256 in air.
a) 2562
b) 16
c) 256
d) 64
View Answer

Answer: b
Explanation: By Maxwell relation, εr = n2, where εro is the dielectric constant at optical
frequencies and n is the refractive index.For the given dielectric constant we get n = 16.

4. Dielectric property impacts the behaviour of a material in the presence of electric field. State
True/False.
a) True
b) False
View Answer
Answer: a
Explanation: Based on the dielectric property, a material can be classified as piezoelectric,
ferroelectric, pyroelectric and anti-ferroelectric materials under the influence of electric field.

5. Curie-Weiss law is applicable to which of the following materials?

a) Piezoelectric
b) Ferroelectric
c) Pyroelectric
d) Anti-ferroelectric
View Answer

Answer: b
Explanation: Curie-Weiss law is given by χe = εr -1 = C/(T-θ), where C is the curie constant and
θ is the characteristic temperature which is usually a few degrees higher than the curie
temperature for ferromagnetic materials.

6. Curie-Weiss law is used to calculate which one of the following?

a) Permittivity
b) Permeability
c) Electric susceptibility
d) Magnetic susceptibility
View Answer

Answer: c
Explanation: Curie-Weiss law is given by χe = εr -1. Thus it is used to calculate the electric
susceptibility of a material.

7. Calculate the loss tangent when the dielectric constant in AC field is given by 3 + 2j.
a) (2/3)
b) (3/2)
c) (-3/2)
d) (-2/3)
View Answer

Answer: d
Explanation: The AC dielectric constant is given by εr = ε` – jε“, where ε` is the real part of AC
dielectric and ε“ is the imaginary part of AC dielectric. The loss tangent is given by tan δ = ε“/ε`
= -2/3.

8. When a dielectric loses its dielectric property, the phenomenon is called

a) Dielectric loss
b) Dielectric breakdown
c) Polarisation
d) Magnetization
View Answer
Answer: b
Explanation: Due to various treatments performed on the dielectric, in order to make it conduct,
the dielectric reaches a state, where it loses its dielectric property and starts to conduct. This
phenomenon is called as dielectric breakdown.

9. Choose the best definition of dielectric loss.

a) Absorption of electric energy by dielectric in an AC field
b) Dissipation of electric energy by dielectric in a static field
c) Dissipation of heat by dielectric
d) Product of loss tangent and relative permittivity
View Answer

Answer: a
Explanation: In the scenario of an AC field, the absorption of electrical energy by a dielectric
material is called as dielectric loss. This will result in dissipation of energy in the form of heat.

10. Compute the loss factor when the loss tangent is 0.88 and the real part of dielectric is 24.
a) 12.12
b) 12.21
c) 21.21
d) 21.12
View Answer

Answer: d
Explanation: The loss factor is nothing but the imaginary part of AC dielectric. It is given by, ε“
= ε` tan δ. We get loss factor as 24 x 0.88 = 21.12.

Continuity Equation - Electromagnetic

Theory Questions and Answers - Sanfoundry
by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Continuity Equation”.

1. Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.
a) 3
b) 5
c) 7
d) 9
View Answer
Answer: c
Explanation: The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On
differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.

2. The continuity equation is a combination of which of the two laws?

a) Ohm’s law and Gauss law
b) Ampere law and Gauss law
c) Ohm’s law and Ampere law
d) Maxwell law and Ampere law
View Answer

Answer: b
Explanation: I = ∫ J.ds is the integral form of Ohm’s law and Div (J) = dq/dt is the Gauss law
analogous to D. Through these two equations, we get Div(J) = -dρ/dt. This is the continuity
equation.

3. Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
a) 20t
b) 21t
c) 19t
d) -20t
View Answer

Answer: b
Explanation: Using continuity equation, the problem can be solved. Div(J) =
– dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating
the Div(J) with respect to t, the charge density will be 21t.

4. Compute the conductivity when the current density is 12 units and the electric field is 20 units.
Also identify the nature of the material.
a) 1.67, dielectric
b) 1.67, conductor
c) 0.6, dielectric
d) 0.6, conductor
View Answer

Answer: c
Explanation: The current density is the product of conductivity and electric field intensity. J =
σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the
material is a dielectric.

5. Find the electron density when convection current density is 120 units and the velocity is
5m/s.
a) 12
b) 600
c) 24
d) 720
View Answer

Answer: c
Explanation: The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v
= 5. ρe = 120/5 = 24 units.

6. Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and
the velocity is 10m/s. Assume the conduction and convection current densities are same.
a) 2.4
b) 4.8
c) 3.6
d) 1.2
View Answer

Answer: d
Explanation: The conduction current density is given by J = σE and the convection current
density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E
= 2.4 x 10/20 = 1.2 units.

7. Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11
units.
a) 1.1
b) 2.2
c) 3.2
d) 0.9
View Answer

Answer: b
Explanation: The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E =
23/11 = 2.1 units.

8. Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12
units.
a) 1/24
b) 1/48
c) 1/12
d) 1/96
View Answer

Answer: a
Explanation: The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we
get R = 100/(12 x 200) = 1/24 units.

9. Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
a) 26.4
b) 42.6
c) 64.2
d) 24.6
View Answer

Answer: d
Explanation: The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we
get V = E x L = 12.3 x 2 = 24.6 units.

10. On equating the generic form of current density equation and the point form of Ohm’s law,
we can obtain V=IR. State True/False.
a) True
b) False
View Answer

Answer: a
Explanation: The generic current density equation is J = I/A and the point form of Ohm’s law is J
= σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s
law.

Boundary Conditions - Electromagnetic

Theory Questions and Answers - Sanfoundry
by staff10

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on
“Boundary Conditions”.

1. The charge within a conductor will be

a) 1
b) -1
c) 0
d) ∞
View Answer

Answer: c
Explanation: No charges exist in a conductor. An illustration for this statement is that, it is safer
to stay inside a car rather than standing under a tree during lightning. Since the car has a metal
body, no charges will be possessed by it to get ionized by the lightning.

2. For a conservative field which of the following equations holds good?

a) ∫ E.dl = 0
b) ∫ H.dl = 0
c) ∫ B.dl = 0
d) ∫ D.dl = 0
View Answer

Answer: a
Explanation: A conservative field implies the work done in a closed path will be zero. This is
given by ∫ E.dl = 0.

3. Find the electric field if the surface density at the boundary of air is 10-9.
a) 12π
b) 24π
c) 36π
d) 48π
View Answer

Answer: c
Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10-9 and
εo = 10-9/36π. We get E = 36π units.

4. Find the flux density at the boundary when the charge density is given by 24 units.
a) 12
b) 24
c) 48
d) 96
View Answer

Answer: b
Explanation: At the boundary of a conductor- free space interface, the flux density is equal to the
charge density. Thus D = ρv = 24 units.

5. Which component of the electric field intensity is always continuous at the boundary?
a) Tangential
b) Normal
c) Horizontal
d) Vertical
View Answer

Answer: a
Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric
field intensity is always continuous. We get Et1 = Et2.

6. The normal component of which quantity is always discontinuous at the boundary?

a) E
b) D
c) H
d) B
View Answer
Answer: b
Explanation: The normal component of an electric flux density is always discontinuous at the
boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free
surface charge exists at the interface.

7. The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux
density of the surface in air?
a) 24
b) 6
c) 1/6
d) 0
View Answer

Answer: b
Explanation: The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put
Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.

8. The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the
field intensity of the surface in air?
a) 5.14
b) 0.194
c) 63
d) 29
View Answer

Answer: c
Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ ε1.
Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.

9. A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The
reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
a) 9.4
b) 8.4
c) 10.4
d) 7.4
View Answer

Answer: c
Explanation: By Snell’s law, the relation between incident and reflected waves is given by, E1
sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.

10. Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the
surface at 45 in air.
a) 1.41
b) 3.5
c) 2.2
d) 1.73
View Answer

Answer: d
Explanation: From the relations of the boundary conditions of a dielectric-dielectric interface, we
get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 = 1.73.