Up To Date Electromagnetic Theory MCQs – Electrostatic Fields MCQs ( Electromagnetic Theory ) MCQs
Latest Electromagnetic Theory MCQs
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Electromagnetic Theory MCQs – Electrostatic Fields MCQs ( Electromagnetic Theory ) MCQs
The most occurred mcqs of Electrostatic Fields MCQs ( Electromagnetic Theory ) in past papers. Past papers of Electrostatic Fields MCQs ( Electromagnetic Theory ) Mcqs. Past papers of Electrostatic Fields MCQs ( Electromagnetic Theory ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Electrostatic Fields MCQs ( Electromagnetic Theory ) Mcqs. The Important series of Electrostatic Fields MCQs ( Electromagnetic Theory ) Mcqs are given below:
Coulomb Law
1. Coulomb is the unit of which quantity?
a) Field strength
b) Charge
c) Permittivity
d) Force
Answer: b
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.
2. Coulomb law is employed in
a) Electrostatics
b) Magnetostatics
c) Electromagnetics
d) Maxwell theory
Answer: a
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.
3. Find the force between 2C and -1C separated by a distance 1m in air(in newton).
a) 18 X 106
b) -18 X 106
c) 18 X 10-6
d) -18 X 10-6
Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109.
4. 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 1C
Answer: c
Explanation: Since the charges are unlike, the force will be attractive. Thus the force directs from 1C to -4C.
5. 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) 1.815
Answer: d
Explanation: 1 stat coulomb = 1/(3 X 109) 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.
6. 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) 2.404
Answer: c
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.
7. The Coulomb law is an implication of which law?
a) Ampere law
b) Gauss law
c) Biot Savart law
d) Lenz law
Answer: b
Explanation: The Coulomb law can be formulated from the Gauss law, using the divergence theorem. Thus it is an implication of Gauss law.
8. 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) 0.2
Answer: c
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.
9. 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) 0.09
Answer: d
Explanation: F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) =
We get r = 0.09m.
10. For a charge Q1, the effect of charge Q2 on Q1 will be,
a) F1 = F2
b) F1 = -F2
c) F1 = F2 = 0
d) F1 and F2 are not equal
Answer: b
Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.
Electric Field Intensity
1. 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 charge
Answer: c
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).
2. Find the force on a charge 2C in a field 1V/m.
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.
3. Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.
a) 18 X 109
b) 9 X 109
c) 36 X 109
d) -18 X 109
Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109
E = F/q = 18 X 109/2 = 9 X 109.
4. 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) 550,000
Answer: c
Explanation: E = Q/ (4∏εor2)
= (2 X 10-6)/(4∏ X εo X 0.22) = 450,000 V/m.
5. 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) 6
Answer: a
Explanation: E = Q/ (4∏εor2)
Q = (4000 X 0.32)/ (9 X 109) = 4 X 10-8 C.
6. 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) False
Answer: a
Explanation: 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.
7. Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.
a) True
b) False
Answer: a
Explanation: 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.
8. Electric field intensity due to infinite sheet of charge σ is
a) Zero
b) Unity
c) σ/ε
d) σ/2ε
Answer: d
Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h2+a2))
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.
9. For a test charge placed at infinity, the electric field will be
a) Unity
b) +∞
c) Zero
d) -∞
Answer: c
Explanation: E = Q/ (4∏εor2)
When distance d is infinity, the electric field will be zero, E= 0.
10. In electromagnetic waves, the electric field will be perpendicular to which of the following?
a) Magnetic field intensity
b) Wave propagation
c) Both H and wave direction
d) It propagates independently
Answer: c
Explanation: In an electromagnetic wave, the electric field and magnetic field will be perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.
Electric Field Density
1. The lines of force are said to be
a) Real
b) Imaginary
c) Drawn to trace the direction
d) Not significant
Answer: c
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.
2. 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 area
Answer: b
Explanation: 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.
3. 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 permeability
Answer: a
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.
4. 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 density
Answer: d
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.
5. The Gaussian surface is
a) Real boundary
b) Imaginary surface
c) Tangential
d) Normal
Answer: b
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.
6. Find the flux density of a sheet of charge density 25 units in air.
a) 25
b) 12.5
c) 6.25
d) 3.125
Answer: b
Explanation: Electric field intensity of infinite sheet of charge E = σ/2ε.
Thus D = εE = σ/2 = 25/2 = 12.5.
7. A uniform surface charge of σ = 2 μC/m2, 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) 106
d) -106
Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 = 2 X 10-6(-az)/2 = -10-6.
8. 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) 0.25
Answer: d
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.
9. 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) 5
Answer: c
Explanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2) X 16π = 4.
10. Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109 units)
a) 2.5
b) 3.5
c) 4.5
d) 5.5
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
Electric Potential
1. 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) False
Answer: a
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.
2. A point charge 2nC is located at origin. What is the potential at (1,0,0)?
a) 12
b) 14
c) 16
d) 18
Answer: d
Explanation: V = Q/(4πεr), where r = 1m
V = (2 X 10-9)/(4πε x 1) = 18 volts.
3. 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) 143.35
Answer: d
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.
4. 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) 2.8
Answer: c
Explanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA = 1m and rB = 4m. Thus Vab = 2.7 volts.
5. Find the potential of V = 60sin θ/r2 at P(3,60,25)
a) 5.774
b) 6.774
c) 7.774
d) 8.774
Answer: a
Explanation: V = 60sin θ/r2, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 volts.
6. Given E = 40xyi + 20x2j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3).
a) 105
b) 106
c) 107
d) 108
Answer: b
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get Vpq on integrating from Q to P. Vpq = 106 volts.
7. The potential difference in an open circuit is
a) Zero
b) Unity
c) Infinity
d) Circuit does not exist open
Answer: c
Explanation: In an open circuit no current exists due to non-existence of loops. Also voltage/potential will be infinity in an open circuit.
8. The potential taken between two points across a resistor will be
a) Positive
b) Negative
c) Zero
d) Infinity
Answer: b
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.
9. What is the potential difference between 10sinθcosφ/r2 at A(1,30,20) and B(4,90,60)?
a) 2.386
b) 3.386
c) 4.386
d) 5.386
Answer: c
Explanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb = 10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts.
10. The voltage at any point in an ac circuit will be
a) Peak voltage
b) RMS voltage
c) Average voltage
d) Source voltage
Answer: b
Explanation: In any ac circuit, the voltage measured will not be exact maximum. In order to normalise, we assume the instantaneous voltage at any point be 70.7% of the peak value, which is called the root mean square (RMS)voltage.
Gauss Law
1. Divergence theorem is based on
a) Gauss law
b) Stoke’s law
c) Ampere law
d) Lenz law
Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.
2. The Gaussian surface for a line charge will be
a) Sphere
b) Cylinder
c) Cube
d) Cuboid
Answer: b
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.
3. The Gaussian surface for a point charge will be
a) Cube
b) Cylinder
c) Sphere
d) Cuboid
Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.
4. 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) 0
Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.
5. The total charge of a surface with densities 1,2,…,10 is
a) 11
b) 33
c) 55
d) 77
Answer: c
Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.
6. 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) 65.68
Answer: b
Explanation: By Gauss law principles, W = Q X V = 10 X 10-6 X 4.386 = 43.86 X 10-6 joule.
7. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
a) 12.74
b) 13.47
c) 12.47
d) 13.74
Answer: c
Explanation: 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 109 volts.
8. 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) 72π
Answer: d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.
9. Gauss law cannot be used to find which of the following quantity?
a) Electric field intensity
b) Electric flux density
c) Charge
d) Permittivity
Answer: d
Explanation: Permittivity is constant for a particular material(say permittivity of water is 1). It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.
10. Gauss law for magnetic fields is given by
a) Div(E) = 0
b) Div(B) = 0
c) Div(H) = 0
d) Div(D) = 0
Answer: b
Explanation: The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.
Applications of Gauss Law
1. Gauss law can be used to compute which of the following?
a) Permittivity
b) Permeability
c) Radius of Gaussian surface
d) Electric potential
Answer: c
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.
Coordinate Systems And Transforms MCQs
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) 3
Answer: a
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.
3. 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) 4
Answer: b
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.
4. 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) 1/4.5
Answer: c
Explanation: 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.
5. 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) -13/6
Answer: d
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.
6. Gauss law can be evaluated in which coordinate system?
a) Cartesian
b) Cylinder
c) Spherical
d) Depends on the Gaussian surface
Answer: d
Explanation: 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.
7. Gauss law cannot be expressed in which of the following forms?
a) Differential
b) Integral
c) Point
d) Stokes theorem
Answer: d
Explanation: 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.
8. The tangential component of electric field intensity is always continuous at the interface. State True/False.
a) True
b) False
Answer: a
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.
9. The normal component of the electric flux density is always discontinuous at the interface. State True/False.
a) True
b) False
Answer: a
Explanation: 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.
10. With Gauss law as reference which of the following law can be derived?
a) Ampere law
b) Faraday’s law
c) Coulomb’s law
d) Ohm’s law
Answer: c
Explanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity. We know that F = qE. Hence force can be computed. This gives the Coulomb’s law.
Relation of E,D,V
1. The electric flux density and electric field intensity have which of the following relation?
a) Linear
b) Nonlinear
c) Inversely linear
d) Inversely nonlinear
Answer: a
Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.
2. The electric field intensity is the negative gradient of the electric potential. State True/False.
a) True
b) False
Answer: a
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.
3. Find the electric potential for an electric field 3units at a distance of 2m.
a) 9
b) 4
c) 6
d) 3/2
Answer: c
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.
4. Find the potential at a point (4, 3, -6) for the function V = 2x2y + 5z.
a) 96
b) 66
c) 30
d) -66
Answer: b
Explanation: The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.
5. Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is
(in 10-10 units)
a) 2.1
b) 2.33
c) 2.5
d) 2.77
Answer: c
Explanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.
6. If potential V = 20/(x2 + y2). The electric field intensity for V is
40(x i + y j)/(x2 + y2)2. State True/False.
a) True
b) False
Answer: a
Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.
7. Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).
a) 20
b) 10
c) 30
d) 60
Answer: b
Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.
8. Find the work done moving a charge 2C having potential V = 24volts is
a) 96
b) 24
c) 36
d) 48
Answer: d
Explanation: The work done is the product of charge and potential.
W = Q X V = 2 X 24 = 48 units.
9. If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0)
(in 10-12 units)
a) 13.25
b) 22.13
c) 26.31
d) 31.52
Answer: b
Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.
10. If V = 2x2y + 20z – 4/(x2 + y2), find the density at A(6, -2.5, 3) in nC/m2.
a) 0.531i – 0.6373j – 0.177k
b) 0.6373i – 0.177j -0.531k
c) 0.177i – 0.6373j – 0.531k
d) 0.531i – 0.177j – 0.6373k
Answer: a
Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10-12 X
(59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2.
Real Time Applications
1. 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) 3.536nF
Answer: a
Explanation: 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.
2. The resistance of a material with conductivity 2millimho/m2, length 10m and area 50m is
a) 500
b) 200
c) 100
d) 1000
Answer: c
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.
3. 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) 94.131H
Answer: a
Explanation: The inductance is given by L = μ N2A/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 1002 X 6/2 = 131.94mH.
4. Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.
a) 400
b) 300
c) 200
d) 100
Answer: d
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.
5. Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.
a) 1A
b) 10mA
c) 10A
d) 100mA
Answer: a
Explanation: 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.
6. 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 A
Answer: b
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.
7. 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) 200
Answer: b
Explanation: 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.
8. From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
a) True
b) False
Answer: a
Explanation: We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.
Vector Analysis And Applications MCQs
9. 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) 80
Answer: b
Explanation: Power is defined as the product of voltage and current.
P = V X I, where V = E X d. Thus P = E X d X I = 100 X 0.1 X 2 = 20 units.
10. Compute the power consumed by a material with current density 15 units in an area of 100 units. The potential measured across the material is 20V.
a) 100kJ
b) 250kJ
c) 30kJ
d) 15kJ
Answer: c
Explanation: Power is given by, P= V X I, where I = J X A is the current.
Thus power P = V X J X A = 20 X 15 X 100 = 30,000 joule = 30kJ.
Electric Dipole
1. 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 distance
Answer: c
Explanation: 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.
2. 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 charges
Answer: b
Explanation: 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.
3. 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) 0.08
Answer: b
Explanation: 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.
4. 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) 60
Answer: d
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.
5. 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) -∞
Answer: a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr2). When the angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will become zero.
6. 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) 9
Answer: a
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=r2
Now, r2 = 5 x 7 = 35. We get r = 5.91cm.
7. 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) 2.5
Answer: b
Explanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the formula d = M/Q = 6/4 = 1.5units.
8. The potential due to the dipole on the midpoint of the two charges will be
a) 0
b) Unity
c) ∞
d) -∞
Answer: c
Explanation: The potential due a dipole at a point P will be V = m cos θ/(4πεr2).
Now it is given that potential on the midpoint, which means P is on midpoint, then the distance from midpoint and P will be zero. When r = 0 is put in the above equation, we get V = ∞. This shows that the potential of a dipole at its midpoint will be maximum/infinity.
9. Dipoles in any electric field undergo
a) Magnetism
b) Electromagnetism
c) Magnetisation
d) Polarisation
Answer: d
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.
10. Dipole moments are used to calculate the
a) Electric field intensity
b) Polarisation patterns
c) Strength of the dipole in the field
d) Susceptibility
Answer: b
Explanation: Dipole moment implicates the strength of the dipole in the electric field. They are then used to compute the polarisation patterns based on the applied field. Once the polarisation is determined we can find its susceptibility. Though all options seem to be correct, the apt answer is to calculate polarisation, provided applied field is known.
Electrostatic Energy
1. 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 lines
Answer: c
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.
2. 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) 48
Answer: a
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.
3. 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) 9.16
Answer: a
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 E2. On substituting the data, we get 3.16 microjoule.
4. Equipotential surface is a
a) Real surface
b) Complex surface
c) Imaginary surface
d) Not existing surface
Answer: c
Explanation: 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.
5. 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) False
Answer: a
Explanation: Since the electric potential in the equipotential surface is the same, the work done will be zero.
6. When curl of a path is zero, the field is said to be conservative. State True/False.
a) True
b) False
Answer: a
Explanation: 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.
7. 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) Permittivity
Answer: a
Explanation: 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.
8. Superconductors exhibit which of the following properties?
a) Ferromagnetism
b) Polarisation
c) Diamagnetism
d) Ferrimagnetism
Answer: c
Explanation: Since superconductors have very good conductivity at low temperatures (σ->∞), they have nearly zero resistivity and exhibit perfect diamagnetism.
9. Debye is the unit used to measure
a) Permittivity
b) Electric dipole moment
c) Magnetic dipole moment
d) Susceptibility
Answer: b
Explanation: Debye is the standard unit for measurement of electric dipole moment. 1 Debye = 3.336 x 10-30 Coulomb-meter.
10. Ceramic materials possess which of the following properties?
a) Brittle and low dielectric constant
b) Rigid and low dielectric constant
c) Brittle and high dielectric constant
d) Rigid and high dielectric constant
Answer: c
Explanation: Ceramic materials are generally brittle. Since these materials are used in capacitors, they have higher dielectric constant than polymer. With respect to energy, they possess high electrostatic energy due to very high dielectric constant (W α ε).
Electrostatic Properties
1. The permittivity is also called
a) Electrostatic energy
b) Dielectric constant
c) Dipole moment
d) Susceptibility
Answer: b
Explanation: The term permittivity or dielectric constant is the measurement of electrostatic energy stored within it and therefore depends on the material.
2. Dielectric constant will be high in
a) Conductors
b) Semiconductors
c) Insulators
d) Superconductors
Answer: c
Explanation: 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.
3. Under the influence of electric field, the dielectric materials will get charged instantaneously. State True/False.
a) True
b) False
Answer: a
Explanation: 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.
4. Insulators perform which of the following functions?
a) Conduction
b) Convection
c) Provide electrical insulation
d) Allows current leakage at interfaces
Answer: c
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.
5. 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 density
Answer: c
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.
6. Semiconductors possess which type of bonding?
a) Metallic
b) Covalent
c) Ionic
d) Magnetic
Answer: b
Explanation: Conductors exhibit metallic bonding. Insulators exhibit ionic bonding and semiconductors exhibit covalent bonding due to sharing of atoms.
7. Find the susceptibility of a material whose dielectric constant is 2.26.
a) 1.26
b) 3.26
c) 5.1
d) 1
Answer: a
Explanation: 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.
8. The bound charge density and free charge density are 12 and 6 units respectively. Calculate the susceptibility.
a) 1
b) 0
c) 2
d) 72
Answer: c
Explanation: The electric susceptibility is given by, χe = Bound free density/Free charge density. χe = 12/6 = 2. It has no unit.
9. The susceptibility of free space is
a) 1
b) 0
c) 2
d) ∞
Answer: b
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.
10. When the electric field becomes zero, which of the following relations hold good?
a) E = P
b) D = P
c) B = P
d) H = P
Answer: b
Explanation: The electric flux density of a field is the sum of εE and polarisation P. It gives D = εE + P. When electric field becomes zero, it is clear that D = P.