97 + Interview Questions in Electrostatic Fields in Electromagnetic Theory Page 1 InterviewSolution

1.	Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m.(a) 3(b) 10/3(c) 11/3(d) 4I had been asked this question in class test.I need to ask this question from Applications of Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct answer is (B) 10/3 The EXPLANATION: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.

Discussion

2.	Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.(a) 0(b) 1(c) 2(d) 3I had been asked this question in an international level competition.My question is taken from Applications of Gauss Law topic in division Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT choice is (a) 0 For explanation: Since 1m does not enclose any cylinder (THREE GAUSSIAN surfaces of RADIUS 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss LAW. Thus flux density is also zero.

Discussion

3.	The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 10^9)(a) 12.74(b) 13.47(c) 12.47(d) 13.74This question was posed to me in final exam.I need to ask this question from Gauss Law topic in section Electrostatic Fields of Electromagnetic Theory
Answer» Correct option is (c) 12.47 To ELABORATE: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b = 2M and a = 1m. We GET V = 12.47 X 10^9 VOLTS.

Discussion

4.	The work done by a charge of 10μC with a potential 4.386 is (in μJ)(a) 32.86(b) 43.86(c) 54.68(d) 65.68I had been asked this question by my school principal while I was bunking the class.This is a very interesting question from Gauss Law in portion Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT ANSWER is (b) 43.86 The explanation is: By Gauss law principles, W = Q X V = 10 X 10^-6 X 4.386 = 43.86 X 10^-6 joule.

Discussion

5.	Find the susceptibility of a material whose dielectric constant is 2.26.(a) 1.26(b) 3.26(c) 5.1(d) 1The question was posed to me in class test.The origin of the question is Electrostatic Properties in portion Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT choice is (a) 1.26 To explain: ELECTRIC susceptibility is the measure of ability of the material to get POLARISED. It is given by, χe = εr – 1.Thus we get 1.26.

Discussion

6.	The bound charge density and free charge density are 12 and 6 units respectively. Calculate the susceptibility.(a) 1(b) 0(c) 2(d) 72This question was addressed to me in an international level competition.This intriguing question originated from Electrostatic Properties topic in section Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT CHOICE is (C) 2 The explanation: The electric susceptibility is given by, χe = BOUND free density/Free charge density. χe = 12/6 = 2. It has no unit.

Discussion

7.	The susceptibility of free space is(a) 1(b) 0(c) 2(d) ∞I got this question during an interview.Question is taken from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (b) 0 Easiest explanation: For FREE space/air, the relative permittivity is UNITY i.e, εr = 1. Thus χe = εr – 1 = 0. The SUSCEPTIBILITY will become ZERO in air.

Discussion

8.	Semiconductors possess which type of bonding?(a) Metallic(b) Covalent(c) Ionic(d) MagneticI had been asked this question in class test.This key question is from Electrostatic Properties topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» Correct choice is (b) Covalent Easiest EXPLANATION: Conductors exhibit metallic bonding. Insulators exhibit IONIC bonding and semiconductors exhibit covalent bonding due to sharing of ATOMS.

Discussion

9.	Which of the following properties distinguish a material as conductor, insulator and semiconductor?(a) Free electron charges(b) Fermi level after doping(c) Energy band gap(d) Electron densityThis question was posed to me in examination.This question is from Electrostatic Properties in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (c) ENERGY band GAP Easiest explanation: The only parameter that classifies the material as conductor or insulator or semiconductor is the band gap energy. It is the energy required to make the electrons conduct. This is low of conductors, AVERAGE for semiconductors and very high for INSULATORS. This means it REQUIRES very high energy to make an insulator conduct.

Discussion

10.	Dielectric constant will be high in(a) Conductors(b) Semiconductors(c) Insulators(d) SuperconductorsThe question was asked during an interview for a job.My question comes from Electrostatic Properties topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT option is (c) INSULATORS To elaborate: Materials that have very less CONDUCTIVITY like ceramics, plastics have HIGHER dielectric constants. Due to their low conductivity, the dielectric materials are said to be good insulators.

Discussion

11.	Under the influence of electric field, the dielectric materials will get charged instantaneously. State True/False.(a) True(b) FalseI had been asked this question in an online quiz.Asked question is from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory
Answer» Right option is (a) True To explain I would say: The dielectrics have the ABILITY of storing energy easily when an ELECTRIC field is applied as their permittivity is RELATIVELY HIGHER than any other materials.

Discussion

12.	Insulators perform which of the following functions?(a) Conduction(b) Convection(c) Provide electrical insulation(d) Allows current leakage at interfacesThe question was posed to me during an interview for a job.I need to ask this question from Electrostatic Properties topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (C) PROVIDE electrical insulation Easy explanation: INSULATORS is a non-conducting MATERIAL which prevents the leakage of electric current in unwanted directions. Thus it is used to provide electrical insulation.

Discussion

13.	The permittivity is also called(a) Electrostatic energy(b) Dielectric constant(c) Dipole moment(d) SusceptibilityI had been asked this question in an interview.Enquiry is from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT OPTION is (B) Dielectric CONSTANT To elaborate: The term permittivity or dielectric constant is the measurement of electrostatic energy stored within it and therefore depends on the MATERIAL.

Discussion

14.	Debye is the unit used to measure(a) Permittivity(b) Electric dipole moment(c) Magnetic dipole moment(d) SusceptibilityThis question was addressed to me in a national level competition.My enquiry is from Electrostatic Energy topic in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct choice is (B) ELECTRIC dipole MOMENT The BEST explanation: Debye is the standard UNIT for measurement of electric dipole moment. 1 Debye = 3.336 x 10^-30 Coulomb-meter.

Discussion

15.	Superconductors exhibit which of the following properties?(a) Ferromagnetism(b) Polarisation(c) Diamagnetism(d) FerrimagnetismI got this question at a job interview.This key question is from Electrostatic Energy in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT ANSWER is (C) Diamagnetism To explain I would say: SINCE SUPERCONDUCTORS have very good conductivity at low temperatures (σ->∞), they have nearly zero resistivity and exhibit perfect diamagnetism.

Discussion

16.	If the electric potential is given, which of the following cannot be calculated?(a) Electrostatic energy(b) Electric field intensity(c) Electric flux density(d) PermittivityI had been asked this question during an online exam.Query is from Electrostatic Energy topic in division Electrostatic Fields of Electromagnetic Theory
Answer» The correct answer is (a) ELECTROSTATIC ENERGY The EXPLANATION is: Using potential, we can calculate electric field DIRECTLY by gradient operation. From E, the flux density D can also be calculated. Thus it is not possible to calculate energy directly from potential.

Discussion

17.	The work done in moving a test charge from one point to another in an equipotential surface is zero. State True/False.(a) True(b) FalseThis question was addressed to me in a national level competition.I need to ask this question from Electrostatic Energy topic in section Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (a) True For EXPLANATION: Since the electric potential in the EQUIPOTENTIAL surface is the same, the WORK DONE will be zero.

Discussion

18.	Equipotential surface is a(a) Real surface(b) Complex surface(c) Imaginary surface(d) Not existing surfaceI got this question in an interview for job.This is a very interesting question from Electrostatic Energy in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct choice is (c) Imaginary surface For explanation I would say: Equipotential surface is an imaginary surface in an electric field of a GIVEN charge DISTRIBUTION in which all the points on the surface are at the same electric POTENTIAL.

Discussion

19.	When curl of a path is zero, the field is said to be conservative. State True/False.(a) True(b) FalseThe question was asked in an international level competition.This question is from Electrostatic Energy topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Correct option is (a) True To elaborate: By Stoke’s theorem, when curl of a PATH becomes ZERO, then ∫ E.dl = 0. In other words the work DONE in a closed path will ALWAYS be zero. Fields having this property is called CONSERVATIVE or lamellar fields.

Discussion

20.	Calculate the energy in an electric field with permittivity of 56 and field intensity of 36π(in μJ)(a) 3.16(b) 5.16(c) 7.16(d) 9.16I got this question in an international level competition.The doubt is from Electrostatic Energy topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT option is (a) 3.16 Easy explanation: The energy in an electric field is given by, W = 0.5 x D x E. SINCE D = εE, we get W = 0.5 x ε x E^2. On SUBSTITUTING the DATA, we get 3.16 microjoule.

Discussion

21.	Calculate the energy in an electric field with flux density 6 units and field intensity of 4 units.(a) 12(b) 24(c) 36(d) 48This question was addressed to me in my homework.Origin of the question is Electrostatic Energy topic in division Electrostatic Fields of Electromagnetic Theory
Answer» The correct OPTION is (a) 12 Explanation: The energy in an electric field is given by, W = 0.5 x D x E, where D = 6 and E = 4. We GET W = 0.5 x 6 x 4 = 12 units.

Discussion

22.	The electrostatic energy in an electric field does not depend on which of the following?(a) Magnitude of charges(b) Permittivity(c) Applied electric field(d) Flux linesI have been asked this question in homework.Enquiry is from Electrostatic Energy in division Electrostatic Fields of Electromagnetic Theory
Answer» Right answer is (c) Applied electric FIELD Explanation: The energy in an electric field directly MAGNITUDE of charges. Thus electric field and FLUX DENSITY are also dependent. But the applied field affects only the POLARISATION and it is independent of the energy in the field.

Discussion

23.	Dipoles in any electric field undergo(a) Magnetism(b) Electromagnetism(c) Magnetisation(d) PolarisationThe question was posed to me during an internship interview.Origin of the question is Electric Dipole in portion Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT option is (d) Polarisation Explanation: DIPOLES in any pure electric FIELD will undergo polarisation. It is the process of alignment of dipole moments in ACCORDANCE with the electric field applied.

Discussion

24.	The potential due to the dipole on the midpoint of the two charges will be(a) 0(b) Unity(c) ∞(d) -∞The question was asked during an interview.Asked question is from Electric Dipole topic in section Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

25.	Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units.(a) 1(b) 1.5(c) 2(d) 2.5The question was posed to me in quiz.The origin of the question is Electric Dipole in division Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT choice is (b) 1.5 To explain I WOULD say: The DIPOLE moment is given by, M = Q x d. To GET d, we rearrange the formula d = M/Q = 6/4 = 1.5units.

Discussion

26.	For two charges 3C and -3C separated by 1cm and are located at distances 5cm and 7cm respectively from the point P, then the distance between their midpoint and the point P will be(a) 5.91(b) 12.6(c) 2(d) 9This question was addressed to me during an internship interview.I'd like to ask this question from Electric Dipole in section Electrostatic Fields of Electromagnetic Theory
Answer» Right answer is (a) 5.91 Explanation: For a DISTANT POINT P, the R1 and R2 will approximately be equal. R1 = R2 = R, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r^2 Now, r2 = 5 X 7 = 35. We get r = 5.91cm.

Discussion

27.	Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.(a) 0(b) Unity(c) ∞(d) -∞I have been asked this question during an internship interview.Origin of the question is Electric Dipole topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT answer is (a) 0 The best explanation: The potential due the dipole is GIVEN by, V = m cos θ/(4πεr^2). When the ANGLE becomes perpendicular (θ = 90). The potential becomes ZERO since cos 90 will become zero.

Discussion

28.	Find the angle at which the potential due a dipole is measured, when the distance from one charge is 12cm and that due to other is 11cm, separated to each other by a distance of 2cm.(a) 15(b) 30(c) 45(d) 60I had been asked this question in a national level competition.Enquiry is from Electric Dipole in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Right answer is (d) 60 The explanation: Here, the two CHARGES are separated by d = 2cm. The distance from one charge (say Q1) will be R1 = 11cm. The distance from another charge (say Q2) will be R2 = 12CM. If R1 and R2 is assumed to be parallel, then R2 – R1 = d COS θ. We GET 1 = 2cos θ and cos θ = 0.5. Then θ = cos^-1(0.5) = 60.

Discussion

29.	Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.(a) 0.02(b) 0.04(c) 0.06(d) 0.08This question was posed to me at a job interview.This interesting question is from Electric Dipole topic in section Electrostatic Fields of Electromagnetic Theory
Answer» The correct ANSWER is (b) 0.04 To elaborate: The dipole moment of charge 2C and DISTANCE 2cm will be, M = Q x d. THUS, M = 2 x 0.02 = 0.04 C-m.

Discussion

30.	The potential due to a dipole at a point P from it is the(a) Sum of potentials at the charges(b) Difference of potentials at the charges(c) Multiplication of potentials at the charges(d) Ratio of potentials at the chargesI have been asked this question during an interview.Question is from Electric Dipole topic in section Electrostatic Fields of Electromagnetic Theory
Answer» Correct choice is (b) Difference of potentials at the charges The explanation is: The total potential at the point P due to the dipole is given by the difference of the potentials of the individual charges. V = V1 + (-V2), SINCE both the charges are UNLIKE. Thus V = V1 – V2.

Discussion

31.	Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.(a) 10(b) 20(c) 40(d) 80I have been asked this question during an online exam.The origin of the question is Real Time Applications topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

32.	Choose the best definition of a dipole.(a) A pair of equal and like charges located at the origin(b) A pair of unequal and like charges located at the origin(c) A pair of equal and unlike charges separated by a small distance(d) A pair of unequal and unlike charges separated by a small distanceThe question was asked in exam.The query is from Electric Dipole in division Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT option is (c) A pair of equal and unlike charges SEPARATED by a small distance To explain I would say: An electric DIPOLE generally refers to two equal and unlike (opposite signs) charges separated by a small distance. It can be anywhere, not necessarily at ORIGIN.

Discussion

33.	From the formula F = qE, can prove that work done is a product of force and displacement. State True/False(a) True(b) FalseI got this question in homework.I'm obligated to ask this question of Real Time Applications topic in portion Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

34.	In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be(a) H = μ B(b) B = μ H(c) A = μ B(d) H = μ AThis question was addressed to me in class test.I would like to ask this question from Real Time Applications topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Right OPTION is (B) B = μ H For EXPLANATION: In electric fields, the flux density is a PRODUCT of permittivity and FIELD intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.

Discussion

35.	Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.(a) 60(b) 120(c) 180(d) 200The question was posed to me in an interview for internship.I need to ask this question from Real Time Applications topic in section Electrostatic Fields of Electromagnetic Theory
Answer» Right option is (b) 120 The EXPLANATION is: The force on a CONDUCTOR is GIVEN by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.

Discussion

36.	Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.(a) 1A(b) 10mA(c) 10A(d) 100mAThe question was asked in final exam.I need to ask this question from Real Time Applications topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» Correct OPTION is (a) 1A The explanation is: We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s LAW, V = IR and current I = V/R = 2/2 = 1A.

Discussion

37.	Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.(a) 400(b) 300(c) 200(d) 100I have been asked this question in class test.This intriguing question originated from Real Time Applications in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct answer is (d) 100 The best explanation: The CURRENT density is given by J = σ E, where σ is the CONDUCTIVITY. THUS resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 UNITS.

Discussion

38.	Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.(a) 131.94mH(b) 94.131mH(c) 131.94H(d) 94.131HThis question was addressed to me in an international level competition.This key question is from Real Time Applications in section Electrostatic Fields of Electromagnetic Theory
Answer» The correct CHOICE is (a) 131.94mH The explanation is: The INDUCTANCE is GIVEN by L = μ N^2A/l, where μ= μoμr is the permeability of AIR and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10^-7 X 100^2 X 6/2 = 131.94mH.

Discussion

39.	The resistance of a material with conductivity 2millimho/m^2, length 10m and area 50m is(a) 500(b) 200(c) 100(d) 1000This question was posed to me during an interview.I want to ask this question from Real Time Applications in section Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT answer is (c) 100 The explanation: The resistance is given by, R = ρL/A, where ρ is the RESISTIVITY, the inverse of CONDUCTIVITY. R = 10/(0.002 X 50) = 100 OHM.

Discussion

40.	Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.(a) 35.36pF(b) 3.536pF(c) 35.36nF(d) 3.536nFThe question was posed to me during an internship interview.Question is from Real Time Applications in division Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT OPTION is (a) 35.36pF To ELABORATE: The capacitance of any MATERIAL is GIVEN by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10^-12 X 20/5 = 35.36pF.

Discussion

41.	Find the electric potential for an electric field 3units at a distance of 2m.(a) 9(b) 4(c) 6(d) 3/2I had been asked this question in homework.Origin of the question is Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT choice is (c) 6 The EXPLANATION: The electric field intensity is the RATIO of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.

Discussion

42.	Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).(a) 20(b) 10(c) 30(d) 60I got this question during an interview for a job.This interesting question is from Relation of E,D,V in section Electrostatic Fields of Electromagnetic Theory
Answer» Correct CHOICE is (b) 10 Explanation: GIVEN V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.

Discussion

43.	If potential V = 20/(x^2 + y^2). The electric field intensity for V is(a) /(x^2 + y^2)^2. State True/False.(b) True(c) FalseI got this question during an interview.My question is based upon Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct OPTION is (a) /(X^2 + y^2)^2. State TRUE/False. The EXPLANATION is: E = -Grad (V) = -Grad(20/(x^2 + y^2)) = -(-40x i /(x^2 + y^2)^2 – 40(y j)/(x^2 + y^2)^2) = 40(x i + y j)/(x^2 + y^2)^2. Thus the statement is true.

Discussion

44.	Find the potential at a point (4, 3, -6) for the function V = 2x^2y + 5z.(a) 96(b) 66(c) 30(d) -66I have been asked this question during an online interview.Query is from Relation of E,D,V in section Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT answer is (b) 66 Easiest explanation: The electric POTENTIAL for the function V = 2x^2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.

Discussion

45.	Find the work done moving a charge 2C having potential V = 24volts is(a) 96(b) 24(c) 36(d) 48I have been asked this question in homework.Asked question is from Relation of E,D,V in division Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (d) 48 For explanation: The work DONE is the PRODUCT of charge and potential. W = Q X V = 2 X 24 = 48 units.

Discussion

46.	The electric field intensity is the negative gradient of the electric potential. State True/False.(a) True(b) FalseThe question was posed to me in class test.My enquiry is from Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT CHOICE is (a) True Best explanation: V = -∫E.dl is the integral form. On DIFFERENTIATING both sides, we get E = -Grad (V). Thus the electric field intensity is the negative GRADIENT of the electric potential.

Discussion

47.	The electric flux density and electric field intensity have which of the following relation?(a) Linear(b) Nonlinear(c) Inversely linear(d) Inversely nonlinearThe question was posed to me during an interview.I'm obligated to ask this question of Relation of E,D,V in division Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

48.	The tangential component of electric field intensity is always continuous at the interface. State True/False.(a) True(b) FalseI have been asked this question by my college director while I was bunking the class.This is a very interesting question from Applications of Gauss Law topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT choice is (a) True For explanation: Consider a dielectric-dielectric boundary, the ELECTRIC field intensity in both the surfaces will be Et1 = Et2, which IMPLIES that the TANGENTIAL component of electric field intensity is always continuous at the boundary.

Discussion

49.	The normal component of the electric flux density is always discontinuous at the interface. State True/False.(a) True(b) FalseI had been asked this question in final exam.Origin of the question is Applications of Gauss Law topic in section Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT choice is (a) True The explanation is: In a dielectric-dielectric boundary, if a free surface CHARGE DENSITY exists at the interface, then the NORMAL components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.

Discussion

50.	Gauss law cannot be expressed in which of the following forms?(a) Differential(b) Integral(c) Point(d) Stokes theoremI got this question in my homework.My enquiry is from Applications of Gauss Law topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct option is (d) Stokes theorem For explanation I would say: GAUSS law can be expressed in differential or point FORM as, DIV (D)= ρv and in INTEGRAL form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.

Discussion

Explore topic-wise InterviewSolutions in .

The work done by a charge of 10μC with a potential 4.386 is (in μJ)(a) 32.86(b) 43.86(c) 54.68(d) 65.68I had been asked this question by my school principal while I was bunking the class.This is a very interesting question from Gauss Law in portion Electrostatic Fields of Electromagnetic Theory

Find the susceptibility of a material whose dielectric constant is 2.26.(a) 1.26(b) 3.26(c) 5.1(d) 1The question was posed to me in class test.The origin of the question is Electrostatic Properties in portion Electrostatic Fields of Electromagnetic Theory

The susceptibility of free space is(a) 1(b) 0(c) 2(d) ∞I got this question during an interview.Question is taken from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory

Semiconductors possess which type of bonding?(a) Metallic(b) Covalent(c) Ionic(d) MagneticI had been asked this question in class test.This key question is from Electrostatic Properties topic in portion Electrostatic Fields of Electromagnetic Theory

Dielectric constant will be high in(a) Conductors(b) Semiconductors(c) Insulators(d) SuperconductorsThe question was asked during an interview for a job.My question comes from Electrostatic Properties topic in portion Electrostatic Fields of Electromagnetic Theory

Under the influence of electric field, the dielectric materials will get charged instantaneously. State True/False.(a) True(b) FalseI had been asked this question in an online quiz.Asked question is from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory

The permittivity is also called(a) Electrostatic energy(b) Dielectric constant(c) Dipole moment(d) SusceptibilityI had been asked this question in an interview.Enquiry is from Electrostatic Properties in division Electrostatic Fields of Electromagnetic Theory

Debye is the unit used to measure(a) Permittivity(b) Electric dipole moment(c) Magnetic dipole moment(d) SusceptibilityThis question was addressed to me in a national level competition.My enquiry is from Electrostatic Energy topic in division Electrostatic Fields of Electromagnetic Theory

Superconductors exhibit which of the following properties?(a) Ferromagnetism(b) Polarisation(c) Diamagnetism(d) FerrimagnetismI got this question at a job interview.This key question is from Electrostatic Energy in portion Electrostatic Fields of Electromagnetic Theory

Equipotential surface is a(a) Real surface(b) Complex surface(c) Imaginary surface(d) Not existing surfaceI got this question in an interview for job.This is a very interesting question from Electrostatic Energy in division Electrostatic Fields of Electromagnetic Theory

When curl of a path is zero, the field is said to be conservative. State True/False.(a) True(b) FalseThe question was asked in an international level competition.This question is from Electrostatic Energy topic in chapter Electrostatic Fields of Electromagnetic Theory

Calculate the energy in an electric field with permittivity of 56 and field intensity of 36π(in μJ)(a) 3.16(b) 5.16(c) 7.16(d) 9.16I got this question in an international level competition.The doubt is from Electrostatic Energy topic in portion Electrostatic Fields of Electromagnetic Theory

Calculate the energy in an electric field with flux density 6 units and field intensity of 4 units.(a) 12(b) 24(c) 36(d) 48This question was addressed to me in my homework.Origin of the question is Electrostatic Energy topic in division Electrostatic Fields of Electromagnetic Theory

Dipoles in any electric field undergo(a) Magnetism(b) Electromagnetism(c) Magnetisation(d) PolarisationThe question was posed to me during an internship interview.Origin of the question is Electric Dipole in portion Electrostatic Fields of Electromagnetic Theory

The potential due to the dipole on the midpoint of the two charges will be(a) 0(b) Unity(c) ∞(d) -∞The question was asked during an interview.Asked question is from Electric Dipole topic in section Electrostatic Fields of Electromagnetic Theory

Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units.(a) 1(b) 1.5(c) 2(d) 2.5The question was posed to me in quiz.The origin of the question is Electric Dipole in division Electrostatic Fields of Electromagnetic Theory

Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.(a) 0.02(b) 0.04(c) 0.06(d) 0.08This question was posed to me at a job interview.This interesting question is from Electric Dipole topic in section Electrostatic Fields of Electromagnetic Theory

From the formula F = qE, can prove that work done is a product of force and displacement. State True/False(a) True(b) FalseI got this question in homework.I'm obligated to ask this question of Real Time Applications topic in portion Electrostatic Fields of Electromagnetic Theory

Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.(a) 1A(b) 10mA(c) 10A(d) 100mAThe question was asked in final exam.I need to ask this question from Real Time Applications topic in portion Electrostatic Fields of Electromagnetic Theory

The resistance of a material with conductivity 2millimho/m^2, length 10m and area 50m is(a) 500(b) 200(c) 100(d) 1000This question was posed to me during an interview.I want to ask this question from Real Time Applications in section Electrostatic Fields of Electromagnetic Theory

Find the electric potential for an electric field 3units at a distance of 2m.(a) 9(b) 4(c) 6(d) 3/2I had been asked this question in homework.Origin of the question is Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory

Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).(a) 20(b) 10(c) 30(d) 60I got this question during an interview for a job.This interesting question is from Relation of E,D,V in section Electrostatic Fields of Electromagnetic Theory

If potential V = 20/(x^2 + y^2). The electric field intensity for V is(a) /(x^2 + y^2)^2. State True/False.(b) True(c) FalseI got this question during an interview.My question is based upon Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory

Find the potential at a point (4, 3, -6) for the function V = 2x^2y + 5z.(a) 96(b) 66(c) 30(d) -66I have been asked this question during an online interview.Query is from Relation of E,D,V in section Electrostatic Fields of Electromagnetic Theory

Find the work done moving a charge 2C having potential V = 24volts is(a) 96(b) 24(c) 36(d) 48I have been asked this question in homework.Asked question is from Relation of E,D,V in division Electrostatic Fields of Electromagnetic Theory

The electric field intensity is the negative gradient of the electric potential. State True/False.(a) True(b) FalseThe question was posed to me in class test.My enquiry is from Relation of E,D,V in chapter Electrostatic Fields of Electromagnetic Theory

The normal component of the electric flux density is always discontinuous at the interface. State True/False.(a) True(b) FalseI had been asked this question in final exam.Origin of the question is Applications of Gauss Law topic in section Electrostatic Fields of Electromagnetic Theory

Gauss law cannot be expressed in which of the following forms?(a) Differential(b) Integral(c) Point(d) Stokes theoremI got this question in my homework.My enquiry is from Applications of Gauss Law topic in chapter Electrostatic Fields of Electromagnetic Theory