97 + Interview Questions in Electrostatic Fields in Electromagnetic Theory Page 2 InterviewSolution

51.	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 = 6m.(a) 17/6(b) -17/6(c) 13/6(d) -13/6The question was asked in unit test.This interesting question is from Applications of Gauss Law in portion Electrostatic Fields of Electromagnetic Theory
Answer» The correct ANSWER is (d) -13/6 The best explanation: The radius R = 6m ENCLOSES all the three Gaussian cylinders. By Gauss law,ψ = Q D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 =σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We GET D = -13/6 units.

Discussion

52.	Gauss law can be evaluated in which coordinate system?(a) Cartesian(b) Cylinder(c) Spherical(d) Depends on the Gaussian surfaceI have been asked this question during an internship interview.This intriguing question comes from Applications of Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Right option is (d) Depends on the Gaussian surface The best I can EXPLAIN: The Gauss law exists for all MATERIALS. Depending on the Gaussian surface of the MATERIAL, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate SYSTEM.

Discussion

53.	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 = 4.5m.(a) 4/4.5(b) 3/4.5(c) 2/4.5(d) 1/4.5This question was addressed to me in semester exam.The doubt is from Applications of Gauss Law topic in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct choice is (c) 2/4.5 To explain: The Gaussian cylinder of R = 4.5m encloses sum of CHARGES of two cylinders (R = 2M and R = 4m). By GAUSS LAW,ψ = Q D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 =σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.

Discussion

54.	Gauss law can be used to compute which of the following?(a) Permittivity(b) Permeability(c) Radius of Gaussian surface(d) Electric potentialThis question was posed to me in an online interview.My question is from Applications of Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Correct answer is (c) Radius of Gaussian SURFACE Explanation: Gauss law relates the electric flux DENSITY and the charge density. THUS it can be USED to COMPUTE radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

Discussion

55.	Gauss law cannot be used to find which of the following quantity?(a) Electric field intensity(b) Electric flux density(c) Charge(d) PermittivityThis question was posed to me during a job interview.My question comes from Gauss Law in division Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

56.	Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.(a) 18π(b) 24π(c) 36π(d) 72πThis question was posed to me in homework.Asked question is from Gauss Law topic in division Electrostatic Fields of Electromagnetic Theory
Answer» The correct choice is (d) 72π To explain I would say: The POTENTIAL DUE to a charged ring is GIVEN by λa/2εr, where a = 2m and r = 1m. We GET V = 72π volts.

Discussion

57.	The total charge of a surface with densities 1,2,…,10 is(a) 11(b) 33(c) 55(d) 77The question was posed to me in quiz.My question is from Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer»

Discussion

58.	A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?(a) 3(b) 2(c) 1(d) 0This question was posed to me in an international level competition.My doubt is from Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct choice is (d) 0 The EXPLANATION: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on INTEGRATING with r = 0->5 and φ = 0->2π, we GET Q = ψ = 0.

Discussion

59.	The Gaussian surface for a point charge will be(a) Cube(b) Cylinder(c) Sphere(d) CuboidThe question was posed to me in a national level competition.The above asked question is from Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Right option is (c) Sphere The BEST explanation: A point CHARGE is single dimensional. The three dimensional IMAGINARY enclosed surface of a point charge will be sphere.

Discussion

60.	The Gaussian surface for a line charge will be(a) Sphere(b) Cylinder(c) Cube(d) CuboidI had been asked this question during a job interview.My doubt is from Gauss Law topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» Right CHOICE is (b) CYLINDER The best explanation: A LINE charge can be visualized as a rod of ELECTRIC charges. The three DIMENSIONAL imaginary enclosed surface of a rod can be a cylinder.

Discussion

61.	What is the potential difference between 10sinθcosφ/r^2 at A(1,30,20) and B(4,90,60)?(a) 2.386(b) 3.386(c) 4.386(d) 5.386The question was asked during a job interview.Origin of the question is Electric Potential in chapter Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT answer is (c) 4.386 The explanation is: POTENTIAL at A, Va = 10sin30cos20/1^2 = 4.6985 and Potential at B, VB = 10sin90cos60/4^2 = 0.3125. Potential DIFFERENCE between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts.

Discussion

62.	Divergence theorem is based on(a) Gauss law(b) Stoke’s law(c) Ampere law(d) Lenz lawThe question was posed to me by my school teacher while I was bunking the class.Query is from Gauss Law topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» Right OPTION is (a) GAUSS law The explanation is: The divergence theorem relates SURFACE integral and VOLUME integral. DIV(D) = ρv, which is Gauss’s law.

Discussion

63.	The potential taken between two points across a resistor will be(a) Positive(b) Negative(c) Zero(d) InfinityI had been asked this question in my homework.Origin of the question is Electric Potential in portion Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT option is (b) Negative The best explanation: The resistor will ABSORB power and dissipate it in the form of heat energy. The POTENTIAL between two points across a resistor will be negative.

Discussion

64.	The potential difference in an open circuit is(a) Zero(b) Unity(c) Infinity(d) Circuit does not exist openThe question was posed to me in unit test.My question is taken from Electric Potential topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT choice is (C) INFINITY Best explanation: In an OPEN circuit no current EXISTS due to non-existence of loops. Also voltage/potential will be infinity in an open circuit.

Discussion

65.	Given E = 40xyi + 20x^2j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3).(a) 105(b) 106(c) 107(d) 108The question was asked in unit test.My question is from Electric Potential in section Electrostatic Fields of Electromagnetic Theory
Answer» The correct answer is (b) 106 For explanation: V = -∫ E.dl = -∫ (40xy dx + 20x^2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get Vpq on integrating from Q to P. Vpq = 106 volts.

Discussion

66.	Find the potential of V = 60sin θ/r^2 at P(3,60,25)(a) 5.774(b) 6.774(c) 7.774(d) 8.774I got this question during an interview for a job.This is a very interesting question from Electric Potential in division Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT ANSWER is (a) 5.774 Explanation: V = 60sin θ/R^2, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/3^2 = 5.774 VOLTS.

Discussion

67.	A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.(a) 2.5(b) 2.6(c) 2.7(d) 2.8I had been asked this question by my school teacher while I was bunking the class.This key question is from Electric Potential topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» Right option is (C) 2.7 Easiest explanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are POSITION vectorsrA = 1M and rB = 4m. Thus Vab = 2.7 VOLTS.

Discussion

68.	Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.(a) 140.35(b) 141.35(c) 142.35(d) 143.35This question was posed to me in an online interview.Origin of the question is Electric Potential topic in section Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT choice is (d) 143.35 Best EXPLANATION: V = (1/4πεo) ∑Q/r = (10 X 10^-9/4πεo) (0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 VOLTS.

Discussion

69.	Potential difference is the work done in moving a unit positive charge from one point to another in an electric field. State True/False.(a) True(b) FalseThis question was addressed to me in an internship interview.The doubt is from Electric Potential topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» Right choice is (a) True Explanation: The electric POTENTIAL is the ratio of work done to the CHARGE. Also it is the work done in moving a UNIT positive charge from infinity to a point in an electric field.

Discussion

70.	A point charge 2nC is located at origin. What is the potential at (1,0,0)?(a) 12(b) 14(c) 16(d) 18I have been asked this question by my college professor while I was bunking the class.I'd like to ask this question from Electric Potential topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT answer is (d) 18 The explanation: V = Q/(4πεr), where R = 1m V = (2 X 10^-9)/(4πε x 1) = 18 VOLTS.

Discussion

71.	If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,(a) 2(b) 3(c) 4(d) 5The question was posed to me in unit test.The question is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory
Answer» Correct CHOICE is (c) 4 Explanation: Total flux leaving the ENTIRE surface is, ψ = 4πr^2D from GAUSS law. Ψ = 4π(1/16π^2) X 16π = 4.

Discussion

72.	Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?(a) 1(b) 0.75(c) 0.5(d) 0.25I have been asked this question in a job interview.My doubt is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory
Answer» Correct answer is (d) 0.25 For explanation: The electric FIELD of a line charge is given by, E = λ/(2περ), where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge DENSITY. The density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.

Discussion

73.	A uniform surface charge of σ = 2 μC/m^2, is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?(a) 10^-6(b) -10^-6(c) 10^6(d) -10^6I had been asked this question in homework.My doubt is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory
Answer» Right ANSWER is (B) -10^-6 Best explanation: The FLUX density of any field is independent of the position (point). D = σ/2 = 2 X 10^-6(-az)/2 = -10^-6.

Discussion

74.	Find the flux density of a sheet of charge density 25 units in air.(a) 25(b) 12.5(c) 6.25(d) 3.125I have been asked this question in homework.I'm obligated to ask this question of Electric Field Density in section Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT choice is (b) 12.5 The best I can EXPLAIN: Electric field INTENSITY of infinite SHEET of charge E = σ/2ε. Thus D = εE = σ/2 = 25/2 = 12.5.

Discussion

75.	The Gaussian surface is(a) Real boundary(b) Imaginary surface(c) Tangential(d) NormalI have been asked this question in class test.The query is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT choice is (b) Imaginary surface Easy EXPLANATION: It is any physical or imaginary closed surface around a charge which satisfies the following condition: D is everywhere EITHER normal or tangential to the surface so that D.ds BECOMES either DDS or 0 respectively.

Discussion

76.	Which of the following correctly states Gauss law?(a) Electric flux is equal to charge(b) Electric flux per unit volume is equal to charge(c) Electric field is equal to charge density(d) Electric flux per unit volume is equal to volume charge densityThe question was posed to me in semester exam.This key question is from Electric Field Density topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT option is (d) Electric flux PER unit volume is EQUAL to volume charge density Explanation: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other WORDS, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density.

Discussion

77.	The lines of force are said to be(a) Real(b) Imaginary(c) Drawn to trace the direction(d) Not significantThe question was asked by my school teacher while I was bunking the class.Question is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory
Answer» The correct choice is (c) Drawn to TRACE the direction Explanation: The LINES drawn to trace the direction in which a POSITIVE TEST charge will experience force due to the MAIN charge are called lines of force. They are not real but drawn for our interpretation.

Discussion

78.	The electric flux density is the(a) Product of permittivity and electric field intensity(b) Product of number of flux lines and permittivity(c) Product of permeability and electric field intensity(d) Product of number of flux lines and permeabilityI had been asked this question by my school principal while I was bunking the class.I want to ask this question from Electric Field Density in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct option is (a) Product of permittivity and electric FIELD INTENSITY Easy EXPLANATION: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric FLUX density is the product of permittivity and electric field intensity.

Discussion

79.	Electric flux density in electric field is referred to as(a) Number of flux lines(b) Ratio of flux lines crossing a surface and the surface area(c) Direction of flux at a point(d) Flux lines per unit areaThis question was posed to me in semester exam.My question is from Electric Field Density in division Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT answer is (b) RATIO of flux lines CROSSING a SURFACE and the surface area To explain I would say: Electric flux density is GIVEN by the ratio between number of flux lines crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.

Discussion

80.	For a test charge placed at infinity, the electric field will be(a) Unity(b) +∞(c) Zero(d) -∞I have been asked this question in semester exam.The doubt is from Electric Field Intensity in section Electrostatic Fields of Electromagnetic Theory
Answer» The correct choice is (c) Zero For explanation: E = Q/ (4∏εor^2) When distance d is INFINITY, the electric FIELD will be zero, E= 0.

Discussion

81.	Electric field intensity due to infinite sheet of charge σ is(a) Zero(b) Unity(c) σ/ε(d) σ/2εThis question was addressed to me in an interview.I want to ask this question from Electric Field Intensity in section Electrostatic Fields of Electromagnetic Theory
Answer» Right answer is (d) σ/2ε For explanation I would say: E = σ/2ε.(1- cos α), where α = h/(√(h^2+a^2)) Here, h is the DISTANCE of the SHEET from point P and a is the radius of the sheet. For INFINITE sheet, α = 90. Thus E = σ/2ε.

Discussion

82.	Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.(a) True(b) FalseThis question was posed to me in my homework.The origin of the question is Electric Field Intensity topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT option is (a) True To explain I WOULD say: The electric FIELD intensity of an infinitely long conductor is given by, E = λ/(4πεh).(SIN α2 – sin α1)i + (cos α2 + cos α1)j For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN.

Discussion

83.	Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10^-8C)(a) 4(b) 2(c) 8(d) 6This question was addressed to me in a national level competition.The question is from Electric Field Intensity in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The correct CHOICE is (a) 4 Explanation: E = Q/ (4∏εor^2) Q = (4000 X 0.3^2)/ (9 X 10^9) = 4 X 10^-8 C.

Discussion

84.	The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False(a) True(b) FalseI had been asked this question in homework.I would like to ask this question from Electric Field Intensity topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT ANSWER is (a) True The best I can explain: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed RADIALLY outward from Q. Since electric field is inversely PROPORTIONAL to distance, thus the statement is true.

Discussion

85.	What is the electric field intensity at a distance of 20cm from a charge 2 X 10^-6 C in vacuum?(a) 250,000(b) 350,000(c) 450,000(d) 550,000The question was asked in an interview for internship.This intriguing question comes from Electric Field Intensity in section Electrostatic Fields of Electromagnetic Theory
Answer» Correct option is (C) 450,000 Explanation: E = Q/ (4∏εor2) = (2 X 10^-6)/(4∏ X εo X 0.2^2) = 450,000 V/m.

Discussion

86.	Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.(a) 18 X 10^9(b) 9 X 10^9(c) 36 X 10^9(d) -18 X 10^9The question was posed to me during an interview.My question is from Electric Field Intensity in division Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT OPTION is (b) 9 X 10^9 The BEST explanation: F = q1q2/(4∏εor2) = -2 X 9/(10^-9 X 12) = -18 X 10^9 E = F/q = 18 X 10^9/2 = 9 X 10^9.

Discussion

87.	Find the force on a charge 2C in a field 1V/m.(a) 0(b) 1(c) 2(d) 3The question was asked in unit test.I'd like to ask this question from Electric Field Intensity topic in section Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT answer is (c) 2 Best explanation: FORCE is the product of charge and electric field. F = Q X E = 2 X 1 = 2 N.

Discussion

88.	The electric field intensity is defined as(a) Force per unit charge(b) Force on a test charge(c) Force per unit charge on a test charge(d) Product of force and chargeThe question was asked in exam.My enquiry is from Electric Field Intensity in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct OPTION is (c) Force PER unit charge on a test charge For explanation: The electric field INTENSITY is the force per unit charge on a test charge, i.E, q1 = 1C. E = F/Q = Q/(4∏εr2).

Discussion

89.	The Coulomb law is an implication of which law?(a) Ampere law(b) Gauss law(c) Biot Savart law(d) Lenz lawI got this question during an interview.Origin of the question is Coulomb Law in portion Electrostatic Fields of Electromagnetic Theory
Answer» The correct CHOICE is (b) Gauss LAW Easiest explanation: The COULOMB law can be formulated from the Gauss law, using the DIVERGENCE theorem. THUS it is an implication of Gauss law.

Discussion

90.	Two small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving.(a) 0.5(b) 0.4(c) 0.3(d) 0.2I have been asked this question in unit test.My question comes from Coulomb Law in division Electrostatic Fields of Electromagnetic Theory
Answer» Correct answer is (c) 0.3 Best EXPLANATION: F = mg = 10 X 10^-3 X 9.81 = 9.81 X 10^-2 N. On calculating R by SUBSTITUTING charges, we get r = 0.3m.

Discussion

91.	A charge of 2 X 10^-7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10^-7 C, both the charges are in vacuum.(a) 0.03(b) 0.05(c) 0.07(d) 0.09I have been asked this question during an internship interview.This intriguing question originated from Coulomb Law topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT answer is (d) 0.09 The EXPLANATION: F = q1q2/(4∏εor^2) , substituting Q1, Q2 and F, r2 = q1q2/(4∏εoF) = We get r = 0.09m.

Discussion

92.	Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.(a) 1.44(b) 2.44(c) 1.404(d) 2.404I have been asked this question during an online interview.My question is based upon Coulomb Law topic in division Electrostatic Fields of Electromagnetic Theory
Answer» Right ANSWER is (c) 1.404 Easy explanation: Before the charges are brought into contact, F = 11.234 μN. After charges are brought into contact and then SEPARATED, charge on each sphere is, (q1 + Q2)/2 = 0.5nC On calculating the FORCE with q1 = q2 = 0.5nC, F = 1.404μN.

Discussion

93.	Find the force of interaction between 60 stat coulomb and 37.5 stat coulomb spaced 7.5cm apart in transformer oil(εr=2.2) in 10^-4 N,(a) 8.15(b) 5.18(c) 1.518(d) 1.815The question was asked in my homework.I need to ask this question from Coulomb Law in portion Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT OPTION is (d) 1.815 The explanation: 1 stat COULOMB = 1/(3 X 10^9) C F = (1.998 X 1.2488 X 10^-16)/(4∏ X 8.854 X 10^-12 X 2.2 X (7.5 X 10^-2)^2) = 1.815 X 10^-4 N.

Discussion

94.	Find the force between 2C and -1C separated by a distance 1m in air(in newton).(a) 18 X 10^6(b) -18 X 10^6(c) 18 X 10^-6(d) -18 X 10^-6I have been asked this question during an interview.Enquiry is from Coulomb Law topic in portion Electrostatic Fields of Electromagnetic Theory
Answer» CORRECT choice is (b) -18 X 10^6 The best I can explain: F = q1q2/(4∏εor^2) = -2 X 9/(10^-9 X 12) = -18 X 10^9.

Discussion

95.	Two charges 1C and -4C exists in air. What is the direction of force?(a) Away from 1C(b) Away from -4C(c) From 1C to -4C(d) From -4C to 1CThis question was addressed to me by my college director while I was bunking the class.This key question is from Coulomb Law topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» RIGHT option is (c) From 1C to -4C The EXPLANATION: SINCE the charges are unlike, the force will be ATTRACTIVE. Thus the force directs from 1C to -4C.

Discussion

96.	Coulomb law is employed in(a) Electrostatics(b) Magnetostatics(c) Electromagnetics(d) Maxwell theoryThe question was asked in an internship interview.My question is from Coulomb Law topic in chapter Electrostatic Fields of Electromagnetic Theory
Answer» The CORRECT choice is (a) Electrostatics Best explanation: Coulomb law is applied to STATIC charges. It states that force between any TWO point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. THUS it is EMPLOYED in electrostatics.

Discussion

97.	Coulomb is the unit of which quantity?(a) Field strength(b) Charge(c) Permittivity(d) ForceI have been asked this question in an interview.This is a very interesting question from Coulomb Law topic in section Electrostatic Fields of Electromagnetic Theory
Answer» Correct option is (B) Charge Explanation: The standard unit of charge is COULOMB. One coulomb is defined as the 1 Newton of force APPLIED on 1 unit of electric field.

Discussion

Explore topic-wise InterviewSolutions in .

Gauss law can be used to compute which of the following?(a) Permittivity(b) Permeability(c) Radius of Gaussian surface(d) Electric potentialThis question was posed to me in an online interview.My question is from Applications of Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory

Gauss law cannot be used to find which of the following quantity?(a) Electric field intensity(b) Electric flux density(c) Charge(d) PermittivityThis question was posed to me during a job interview.My question comes from Gauss Law in division Electrostatic Fields of Electromagnetic Theory

The total charge of a surface with densities 1,2,…,10 is(a) 11(b) 33(c) 55(d) 77The question was posed to me in quiz.My question is from Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory

The Gaussian surface for a point charge will be(a) Cube(b) Cylinder(c) Sphere(d) CuboidThe question was posed to me in a national level competition.The above asked question is from Gauss Law in chapter Electrostatic Fields of Electromagnetic Theory

The Gaussian surface for a line charge will be(a) Sphere(b) Cylinder(c) Cube(d) CuboidI had been asked this question during a job interview.My doubt is from Gauss Law topic in portion Electrostatic Fields of Electromagnetic Theory

What is the potential difference between 10sinθcosφ/r^2 at A(1,30,20) and B(4,90,60)?(a) 2.386(b) 3.386(c) 4.386(d) 5.386The question was asked during a job interview.Origin of the question is Electric Potential in chapter Electrostatic Fields of Electromagnetic Theory

Divergence theorem is based on(a) Gauss law(b) Stoke’s law(c) Ampere law(d) Lenz lawThe question was posed to me by my school teacher while I was bunking the class.Query is from Gauss Law topic in portion Electrostatic Fields of Electromagnetic Theory

The potential taken between two points across a resistor will be(a) Positive(b) Negative(c) Zero(d) InfinityI had been asked this question in my homework.Origin of the question is Electric Potential in portion Electrostatic Fields of Electromagnetic Theory

The potential difference in an open circuit is(a) Zero(b) Unity(c) Infinity(d) Circuit does not exist openThe question was posed to me in unit test.My question is taken from Electric Potential topic in portion Electrostatic Fields of Electromagnetic Theory

Given E = 40xyi + 20x^2j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3).(a) 105(b) 106(c) 107(d) 108The question was asked in unit test.My question is from Electric Potential in section Electrostatic Fields of Electromagnetic Theory

Find the potential of V = 60sin θ/r^2 at P(3,60,25)(a) 5.774(b) 6.774(c) 7.774(d) 8.774I got this question during an interview for a job.This is a very interesting question from Electric Potential in division Electrostatic Fields of Electromagnetic Theory

Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.(a) 140.35(b) 141.35(c) 142.35(d) 143.35This question was posed to me in an online interview.Origin of the question is Electric Potential topic in section Electrostatic Fields of Electromagnetic Theory

If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,(a) 2(b) 3(c) 4(d) 5The question was posed to me in unit test.The question is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory

Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?(a) 1(b) 0.75(c) 0.5(d) 0.25I have been asked this question in a job interview.My doubt is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory

A uniform surface charge of σ = 2 μC/m^2, is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?(a) 10^-6(b) -10^-6(c) 10^6(d) -10^6I had been asked this question in homework.My doubt is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory

Find the flux density of a sheet of charge density 25 units in air.(a) 25(b) 12.5(c) 6.25(d) 3.125I have been asked this question in homework.I'm obligated to ask this question of Electric Field Density in section Electrostatic Fields of Electromagnetic Theory

The Gaussian surface is(a) Real boundary(b) Imaginary surface(c) Tangential(d) NormalI have been asked this question in class test.The query is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory

The lines of force are said to be(a) Real(b) Imaginary(c) Drawn to trace the direction(d) Not significantThe question was asked by my school teacher while I was bunking the class.Question is from Electric Field Density in portion Electrostatic Fields of Electromagnetic Theory

For a test charge placed at infinity, the electric field will be(a) Unity(b) +∞(c) Zero(d) -∞I have been asked this question in semester exam.The doubt is from Electric Field Intensity in section Electrostatic Fields of Electromagnetic Theory

Electric field intensity due to infinite sheet of charge σ is(a) Zero(b) Unity(c) σ/ε(d) σ/2εThis question was addressed to me in an interview.I want to ask this question from Electric Field Intensity in section Electrostatic Fields of Electromagnetic Theory

Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.(a) True(b) FalseThis question was posed to me in my homework.The origin of the question is Electric Field Intensity topic in chapter Electrostatic Fields of Electromagnetic Theory

Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.(a) 18 X 10^9(b) 9 X 10^9(c) 36 X 10^9(d) -18 X 10^9The question was posed to me during an interview.My question is from Electric Field Intensity in division Electrostatic Fields of Electromagnetic Theory

Find the force on a charge 2C in a field 1V/m.(a) 0(b) 1(c) 2(d) 3The question was asked in unit test.I'd like to ask this question from Electric Field Intensity topic in section Electrostatic Fields of Electromagnetic Theory

The electric field intensity is defined as(a) Force per unit charge(b) Force on a test charge(c) Force per unit charge on a test charge(d) Product of force and chargeThe question was asked in exam.My enquiry is from Electric Field Intensity in division Electrostatic Fields of Electromagnetic Theory

The Coulomb law is an implication of which law?(a) Ampere law(b) Gauss law(c) Biot Savart law(d) Lenz lawI got this question during an interview.Origin of the question is Coulomb Law in portion Electrostatic Fields of Electromagnetic Theory

Find the force between 2C and -1C separated by a distance 1m in air(in newton).(a) 18 X 10^6(b) -18 X 10^6(c) 18 X 10^-6(d) -18 X 10^-6I have been asked this question during an interview.Enquiry is from Coulomb Law topic in portion Electrostatic Fields of Electromagnetic Theory

Coulomb law is employed in(a) Electrostatics(b) Magnetostatics(c) Electromagnetics(d) Maxwell theoryThe question was asked in an internship interview.My question is from Coulomb Law topic in chapter Electrostatic Fields of Electromagnetic Theory

Coulomb is the unit of which quantity?(a) Field strength(b) Charge(c) Permittivity(d) ForceI have been asked this question in an interview.This is a very interesting question from Coulomb Law topic in section Electrostatic Fields of Electromagnetic Theory