Multiple choice question for engineering
1. The divergence of a vector is a scalar. State True/False.
Answer: a [Reason:] Divergence can be computed only for a vector. Since it is the measure of outward flow of flux from a small closed surface as the volume shrinks to zero, the result will be directionless (scalar).
2. The divergence concept can be illustrated using Pascal’s law. State True/False.
Answer: a [Reason:] Consider the illustration of Pascal’s law, wherein a ball is pricked with holes all over its body. After water is filled in it and pressure is applied on it, the water flows out the holes uniformly. This is analogous to the flux flowing outside a closed surface as the volume reduces.
3. Compute the divergence of the vector xi + yj + zk.
Answer: d [Reason:] The vector given is a position vector. The divergence of any position vector is always 3.
4. Find the divergence of the vector yi + zj + xk.
Answer: b [Reason:] Div (yi + zj + xk) = Dx(y) + Dy(z) + Dz(x), which is zero. Here D refers to partial differentiation.
5. Given D = e-xsin y i – e-xcos y j
Find divergence of D.
Answer: d [Reason:] Div (D) = Dx(e-xsin y) + Dy(-e-xcos y ) = -e-xsin y + e-xsin y = 0.
6. Find the divergence of the vector F= xe-x i + y j – xz k
a) (1 – x)(1 + e-x)
b) (x – 1)(1 + e-x)
c) (1 – x)(1 – e)
d) (x – 1)(1 – e)
Answer: a [Reason:] Div(F) = Dx(xe-x) + Dy(y)+Dz(-xz) = -xe-x + e-x + 1 – x =
e-x(1 – x) + (1 – x) = (1 – x)(1 + e-x).
7. Determine the divergence of F = 30 i + 2xy j + 5xz2 k at (1,1,-0.2) and state the nature of the field.
a) 1, solenoidal
b) 0, solenoidal
c) 1, divergent
d) 0, divergent
Answer: b [Reason:] Div(F) = Dx(30) + Dy(2xy) + Dz(5xz2) = 0 + 2x + 10xz = 2x + 10xz
Divergence at (1,1,-0.2) will give zero. As the divergence is zero, field is solenoidal.
Alternate/Shortcut: Without calculation, we can easily choose option b, as by theory when the divergence is zero, the vector is solenoidal. Option b is the only one which is satisfying this condition.
8. Find whether the vector is solenoidal, E = yz i + xz j + xy k
a) Yes, solenoidal
b) No, non-solenoidal
c) Solenoidal with negative divergence
d) Variable divergence
Answer: a [Reason:] Div(E) = Dx(yz) + Dy(xz) + Dz(xy) = 0. The divergence is zero, thus vector is divergentless or solenoidal.
9. Find the divergence of the field, P = x2yz i + xz k
a) xyz + 2x
b) 2xyz + x
c) xyz + 2z
d) 2xyz + z
Answer: b [Reason:] Div(P) = Dx(x2yz) + Dy(0) + Dz(xz) = 2xyz + x, which is option b. For different values of x,y,z the divergence of the field varies.
10. Identify the nature of the field, if the divergence is zero and curl is also zero.
a) Solenoidal, irrotational
b) Divergent, rotational
c) Solenoidal, irrotational
d) Divergent, rotational
Answer: c [Reason:] Since the vector field does not diverge (moves in a straight path), the divergence is zero. Also, the path does not possess any curls, so the field is irrotational.
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 [Reason:] 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 [Reason:] 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.
Answer: b [Reason:] 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.
Answer: d [Reason:] 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.
Answer: a [Reason:] 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
Answer: a [Reason:] 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.
Answer: b [Reason:] 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
Answer: c [Reason:] 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
Answer: d [Reason:] 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
Answer: b [Reason:] 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.
1. The lines of force are said to be
c) Drawn to trace the direction
d) Not significant
Answer: c [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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
Answer: b [Reason:] 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.
Answer: b [Reason:] 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?
Answer: b [Reason:] 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?
Answer: d [Reason:] 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,
Answer: c [Reason:] 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)
Answer: c [Reason:] D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
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.
Answer: a [Reason:] 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)?
Answer: d [Reason:] 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.
Answer: d [Reason:] 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.
Answer: c [Reason:] 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)
Answer: a [Reason:] 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).
Answer: b [Reason:] 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
d) Circuit does not exist open
Answer: c [Reason:] 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
Answer: b [Reason:] 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)?
Answer: c [Reason:] 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 [Reason:] 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.
1. The permittivity is also called
a) Electrostatic energy
b) Dielectric constant
c) Dipole moment
Answer: b [Reason:] 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
Answer: c [Reason:] 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.
Answer: a [Reason:] 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?
c) Provide electrical insulation
d) Allows current leakage at interfaces
Answer: c [Reason:] 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 [Reason:] 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?
Answer: b [Reason:] 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.
Answer: a [Reason:] 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.
Answer: c [Reason:] 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
Answer: b [Reason:] 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 [Reason:] 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.