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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

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) 29The question was posed to me in an internship interview.My doubt stems from Boundary Conditions in section Electric Fields in Material Space of Electromagnetic Theory

Answer»

The correct answer is (c) 63

To EXPLAIN I would say: 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.

2.

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) 7.4I had been asked this question in exam.My doubt stems from Boundary Conditions topic in division Electric Fields in Material Space of Electromagnetic Theory

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Right choice is (c) 10.4

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.

3.

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) 0This question was posed to me in an online quiz.My enquiry is from Boundary Conditions in portion Electric Fields in Material Space of Electromagnetic Theory

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The correct OPTION is (B) 6

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.

4.

The normal component of which quantity is always discontinuous at the boundary?(a) E(b) D(c) H(d) BI got this question in class test.This interesting question is from Boundary Conditions topic in division Electric Fields in Material Space of Electromagnetic Theory

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5.

Which component of the electric field intensity is always continuous at the boundary?(a) Tangential(b) Normal(c) Horizontal(d) VerticalThe question was posed to me in an internship interview.My enquiry is from Boundary Conditions in portion Electric Fields in Material Space of Electromagnetic Theory

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6.

Find the flux density at the boundary when the charge density is given by 24 units.(a) 12(b) 24(c) 48(d) 96This question was addressed to me by my college professor while I was bunking the class.The origin of the question is Boundary Conditions in section Electric Fields in Material Space of Electromagnetic Theory

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Correct choice is (B) 24

To explain: At the boundary of a conductor- free space INTERFACE, the flux DENSITY is EQUAL to the charge density. Thus D = ρv = 24 units.

7.

Find the electric field if the surface density at the boundary of air is 10^-9.(a) 12π(b) 24π(c) 36π(d) 48πI had been asked this question in an online interview.My question is taken from Boundary Conditions topic in division Electric Fields in Material Space of Electromagnetic Theory

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Right OPTION is (C) 36π

The best I can EXPLAIN: 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.

8.

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 = 0I got this question in an interview.I need to ask this question from Boundary Conditions topic in section Electric Fields in Material Space of Electromagnetic Theory

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Correct choice is (a) ∫ E.dl = 0

The BEST I can explain: A CONSERVATIVE field implies the WORK DONE in a closed path will be zero. This is GIVEN by ∫ E.dl = 0.

9.

The charge within a conductor will be(a) 1(b) -1(c) 0(d) ∞I have been asked this question in a job interview.Origin of the question is Boundary Conditions in portion Electric Fields in Material Space of Electromagnetic Theory

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10.

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) 24.6This question was addressed to me during an online interview.The origin of the question is Continuity Equation topic in section Electric Fields in Material Space of Electromagnetic Theory

Answer»

The correct OPTION is (d) 24.6

Easy 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.

11.

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) 1.2I have been asked this question by my school teacher while I was bunking the class.This is a very interesting question from Continuity Equation in chapter Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (d) 1.2

For EXPLANATION I would say: 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.

12.

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) 1/96I got this question in an internship interview.Query is from Continuity Equation in section Electric Fields in Material Space of Electromagnetic Theory

Answer» RIGHT option is (a) 1/24

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.
13.

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) 0.9The question was asked in semester exam.Question is from Continuity Equation in chapter Electric Fields in Material Space of Electromagnetic Theory

Answer»

The correct CHOICE is (B) 2.2

For explanation: The mobility is DEFINED as the drift velocity per unit electric FIELD. Thus μe = vd/E = 23/11 = 2.1 UNITS.

14.

Find the electron density when convection current density is 120 units and the velocity is 5m/s.(a) 12(b) 600(c) 24(d) 720I got this question during an online exam.This is a very interesting question from Continuity Equation in section Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (c) 24

For EXPLANATION I WOULD say: 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.

15.

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) -20tThe question was posed to me in a job interview.Enquiry is from Continuity Equation in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (B) 21t

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.

16.

Find the current when the charge is a time function given by q(t) = 3t + t^2 at 2 seconds.(a) 3(b) 5(c) 7(d) 9The question was asked in an interview for internship.My doubt stems from Continuity Equation topic in division Electric Fields in Material Space of Electromagnetic Theory

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Right choice is (C) 7

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.

17.

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, conductorThis question was addressed to me in an interview.My question is taken from Continuity Equation topic in chapter Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (c) 0.6, DIELECTRIC

For explanation I would say: 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.

18.

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 lawThe question was asked during an online exam.My question is taken from Continuity Equation in section Electric Fields in Material Space of Electromagnetic Theory

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19.

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 permittivityThe question was asked in homework.This key question is from Electric Fields in Material Space topic in section Electric Fields in Material Space of Electromagnetic Theory

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The correct option is (a) Absorption of electric energy by DIELECTRIC in an AC field

Best 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.

20.

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)The question was asked by my college director while I was bunking the class.Question is from Electric Fields in Material Space in section Electric Fields in Material Space of Electromagnetic Theory

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The correct ANSWER is (d) (-2/3)

EASIEST 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.

21.

Curie-Weiss law is used to calculate which one of the following?(a) Permittivity(b) Permeability(c) Electric susceptibility(d) Magnetic susceptibilityI have been asked this question during an interview.I would like to ask this question from Electric Fields in Material Space in section Electric Fields in Material Space of Electromagnetic Theory

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Correct choice is (C) ELECTRIC SUSCEPTIBILITY

Explanation: Curie-Weiss law is GIVEN by χe = εr -1. Thus it is used to CALCULATE the electric susceptibility of a material.

22.

When a dielectric loses its dielectric property, the phenomenon is called(a) Dielectric loss(b) Dielectric breakdown(c) Polarisation(d) MagnetizationThe question was asked during a job interview.This key question is from Electric Fields in Material Space in section Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (b) Dielectric breakdown

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.

23.

Curie-Weiss law is applicable to which of the following materials?(a) Piezoelectric(b) Ferroelectric(c) Pyroelectric(d) Anti-ferroelectricI had been asked this question in an interview for internship.My query is from Electric Fields in Material Space in chapter Electric Fields in Material Space of Electromagnetic Theory

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Correct CHOICE is (B) Ferroelectric

The 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.

24.

Compute the refractive index when the dielectric constant is 256 in air.(a) 2562(b) 16(c) 256(d) 64This question was posed to me during an internship interview.My doubt stems from Electric Fields in Material Space in chapter Electric Fields in Material Space of Electromagnetic Theory

Answer» CORRECT choice is (b) 16

Easy explanation: By Maxwell RELATION, εr = n^2, where εro is the dielectric CONSTANT at OPTICAL frequencies and n is the refractive index.For the given dielectric constant we GET n = 16.
25.

Dielectric property impacts the behaviour of a material in the presence of electric field. State True/False.(a) True(b) FalseI got this question at a job interview.This intriguing question originated from Electric Fields in Material Space topic in section Electric Fields in Material Space of Electromagnetic Theory

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26.

Ionic non polar solid dielectrics contain more than one type of atoms but no permanent dipoles. State True/False(a) True(b) FalseThis question was posed to me during an online interview.This question is from Electric Fields in Material Space in chapter Electric Fields in Material Space of Electromagnetic Theory

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The correct ANSWER is (a) True

The EXPLANATION is: 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.

27.

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 materialThis question was addressed to me in final exam.Enquiry is from Polarization in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (B) Sum of all TYPES of polarisation

The EXPLANATION: The total polarisation of a material is given by the sum of ELECTRONIC, ionic, orientational and interfacial polarisation of the material.

28.

Which of the following is not an example of elemental solid dielectric?(a) Diamond(b) Sulphur(c) Silicon(d) GermaniumThis question was addressed to me by my school teacher while I was bunking the class.This interesting question is from Electric Fields in Material Space in section Electric Fields in Material Space of Electromagnetic Theory

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The correct OPTION is (c) Silicon

Best 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.

29.

In the given types of polarisation, which type exists in the semiconductor?(a) Electronic(b) Ionic(c) Orientational(d) Interfacial or space chargeI got this question in an interview for internship.My question is taken from Polarization in division Electric Fields in Material Space of Electromagnetic Theory

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30.

In isotropic materials, which of the following quantities will be independent of the direction?(a) Permittivity(b) Permeability(c) Polarisation(d) PolarizabilityI have been asked this question in homework.My enquiry is from Polarization in section Electric Fields in Material Space of Electromagnetic Theory

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31.

Calculate the polarisation vector in air when the susceptibility is 5 and electric field is 12 units.(a) 3(b) 2(c) 60(d) 2.4I have been asked this question in examination.This interesting question is from Polarization topic in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (c) 60

To EXPLAIN I would say: 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.

32.

Identify which type of polarisation depends on temperature.(a) Electronic(b) Ionic(c) Orientational(d) InterfacialI had been asked this question during an online interview.My doubt stems from Polarization topic in division Electric Fields in Material Space of Electromagnetic Theory

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33.

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) 144I got this question in unit test.This intriguing question originated from Polarization in division Electric Fields in Material Space of Electromagnetic Theory

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Right answer is (c) 36

The 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.

34.

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 momentThe question was posed to me during an online interview.My question is based upon Polarization in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct OPTION is (B) RATIO of DIPOLE moment to electric field

Easy explanation: Polarizability is a constant that is defined as the ratio of elemental dipole moment to the electric field strength.

35.

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) 0.02I got this question in class test.The origin of the question is Polarization in division Electric Fields in Material Space of Electromagnetic Theory

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Correct OPTION is (a) 200

For explanation I WOULD say: POLARISATION vector P = N x p, where N = 100 and p = 2. On substituting we GET P = 200 units.

36.

In free space, which of the following will be zero?(a) Permittivity(b) Permeability(c) Conductivity(d) ResistivityThe question was posed to me in unit test.I'd like to ask this question from Displacement and Conduction Current in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (C) Conductivity

The explanation is: 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.

37.

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 dipoleThis question was addressed to me in an internship interview.I'd like to ask this question from Polarization topic in chapter Electric Fields in Material Space of Electromagnetic Theory

Answer» CORRECT option is (b) Electric dipole moment per UNIT volume

The best 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.
38.

If the intrinsic angle is 20, then find the loss tangent.(a) tan 20(b) tan 40(c) tan 60(d) tan 80I got this question during an online interview.Question is taken from Displacement and Conduction Current topic in section Electric Fields in Material Space of Electromagnetic Theory

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Right ANSWER is (B) tan 40

Easy EXPLANATION: The LOSS tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.

39.

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 phaseThis question was posed to me in homework.This is a very interesting question from Displacement and Conduction Current in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (b) 45 out of phase

To EXPLAIN: 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.

40.

If the loss tangent is very less, then the material will be a(a) Conductor(b) Lossless dielectric(c) Lossy dielectric(d) InsulatorThis question was addressed to me in a national level competition.Origin of the question is Displacement and Conduction Current topic in division Electric Fields in Material Space of Electromagnetic Theory

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Correct OPTION is (B) Lossless DIELECTRIC

Best 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).

41.

The ratio of conduction to displacement current density is referred to as(a) Attenuation constant(b) Propagation constant(c) Loss tangent(d) Dielectric constantThe question was asked in an online interview.This intriguing question comes from Displacement and Conduction Current in portion Electric Fields in Material Space of Electromagnetic Theory

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The correct answer is (c) Loss tangent

Easiest 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.

42.

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 GHzThis question was posed to me in an online interview.This interesting question is from Displacement and Conduction Current topic in portion Electric Fields in Material Space of Electromagnetic Theory

Answer» RIGHT choice is (b) 9 GHz

The explanation is: 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.
43.

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) 4The question was asked in unit test.This interesting question is from Displacement and Conduction Current in chapter Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (d) 4

The 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.

44.

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 2tI had been asked this question in class test.Question is taken from Displacement and Conduction Current topic in section Electric Fields in Material Space of Electromagnetic Theory

Answer» CORRECT OPTION is (b) 10cos 0.5t

For 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.
45.

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) 4.5This question was posed to me by my school teacher while I was bunking the class.This key question is from Dielectrics topic in portion Electric Fields in Material Space of Electromagnetic Theory

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Right option is (b) 36

Best 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.

46.

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) 1600The question was posed to me during an interview for a job.The doubt is from Displacement and Conduction Current in section Electric Fields in Material Space of Electromagnetic Theory

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The correct option is (a) 25

For explanation I WOULD say: The CONDUCTION CURRENT density is given by, Jc = σE. To get CONDUCTIVITY, σ = J/E = 100/4 = 25 units.

47.

The superconducting materials will be independent of which of the following?(a) Magnetic field(b) Electric field(c) Magnetization(d) TemperatureThe question was posed to me by my college director while I was bunking the class.My question is taken from Dielectrics in portion Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (b) ELECTRIC field

Easy 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.

48.

The magnetic susceptibility in a superconductor will be(a) Positive(b) Negative(c) Zero(d) InfinityThis question was posed to me during an interview for a job.My enquiry is from Dielectrics in section Electric Fields in Material Space of Electromagnetic Theory

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Correct answer is (b) Negative

Explanation: Due to PERFECT DIAMAGNETISM in a superconductor, its MAGNETIC SUSCEPTIBILITY will be negative. This phenomenon is called Meissner effect.

49.

The magnetic field which destroys the superconductivity is called(a) Diamagnetic field(b) Ferromagnetic field(c) Ferrimagnetic field(d) Critical fieldI have been asked this question in my homework.The origin of the question is Dielectrics topic in chapter Electric Fields in Material Space of Electromagnetic Theory

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50.

Find the dielectric constant for a material with electric susceptibility of 4.(a) 3(b) 5(c) 8(d) 16I got this question in an interview for job.Origin of the question is Dielectrics in division Electric Fields in Material Space of Electromagnetic Theory

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Correct option is (b) 5

To explain I WOULD say: The electric SUSCEPTIBILITY is given by χe = εr – 1. For a susceptibility of 4, the dielectric constant will be 5. It has no UNIT.