Electromagnetic Theory MCQs – Electrostatic Boundary Value Problem MCQs ( Electromagnetic Theory ) MCQs
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Electromagnetic Theory MCQs – Electrostatic Boundary Value Problem MCQs ( Electromagnetic Theory ) MCQs
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Poisson and Laplace Equation
1. The given equation satisfies the Laplace equation.
V = x2 + y2 – z2. State True/False.
a) True
b) False
Answer: a
Explanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del2(V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.
2. In free space, the Poisson equation becomes
a) Maxwell equation
b) Ampere equation
c) Laplace equation
d) Steady state equation
Answer: c
Explanation: The Poisson equation is given by Del2(V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del2(V) = 0, which is the Laplace equation.
3. If Laplace equation satisfies, then which of the following statements will be true?
a) Potential will be zero
b) Current will be infinite
c) Resistance will be infinite
d) Voltage will be same
Answer: b
Explanation: Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.
4. Suppose the potential function is a step function. The equation that gets satisfied is
a) Laplace equation
b) Poisson equation
c) Maxwell equation
d) Ampere equation
Answer: a
Explanation: Step is a constant function. The Laplace equation Div(Grad(step)) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.
5. Calculate the charge density when a potential function x2 + y2 + z2 is in air(in 10-9 order)
a) 1/6π
b) 6/2π
c) 12/6π
d) 10/8π
Answer: a
Explanation: The Poisson equation is given by Del2(V) = -ρ/ε. To find ρ, put ε = 8.854 x 10-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10-9/36π = 1/6π units.
6. The function V = exsin y + z does not satisfy Laplace equation. State True/False.
a) True
b) False
Answer: b
Explanation: Grad (V) = exsin y i + ex cos y j + k. Div(Grad(V)) = exsin y – exsin y + 0= 0.Thus Laplacian equation Div(Grad(V)) = 0 is true.
7. Poisson equation can be derived from which of the following equations?
a) Point form of Gauss law
b) Integral form of Gauss law
c) Point form of Ampere law
d) Integral form of Ampere law
Answer: a
Explanation: The point of Gauss law is given by, Div (D)= ρv. On putting
D= ε E and E=- Grad (V) in Gauss law, we get Del2 (V)= -ρ/ε, which is the Poisson equation.
8. Find the charge density from the function of flux density given by 12x – 7z.
a) 19
b) -5
c) 5
d) -19
Answer: c
Explanation: From point form of Gauss law, we get Div (D) = ρv
Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.
9. Find the electric field of a potential function given by 20 log x + y at the point (1,1,0).
a) -20 i – j
b) -i -20 j
c) i + j
d) (i + j)/20
Answer: a
Explanation: The electric field is given by E = -Grad(V). The gradient of the given function is 20i/x + j. At the point (1,1,0), we get 20i + j. The electric field E = -(20i + j) = -20i – j.
10. When a material has zero permittivity, the maximum potential that it can possess is
a) ∞
b) -∞
c) Unity
d) Zero
Answer: d
Explanation: Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.
Resistances and Capacitances
1. Find the resistivity of a material having resistance 20kohm, area 2 units and length of 12m.
a) 6666.6
b) 3333.3
c) 1200
d) 2000
Answer: b
Explanation: The resistance of a material is given by R = ρL/A. To get ρ, put R = 20 x 103, A = 2 and L = 12. We get ρ = 3333.3 units.
2. A resistor value of colour code orange violet orange will be
a) 37 kohm
b) 37 Mohm
c) 48 kohm
d) 48 Mohm
Answer: a
Explanation: Orange refers to number 3. Violet refers to number 7. The third colour code orange refers to 103. Thus the resistor value will be 37 kilo ohm.
3. A infinite resistance is considered as a/an
a) Closed path(short circuit)
b) Open path
c) Not defined
d) Ammeter with zero reading
Answer: b
Explanation: When there exists infinite resistance in a path, the current flowing will ideally be zero. This is possible only for an open path/circuit.
4. Find the time constant in a series R-L circuit when the resistance is 4 ohm and the inductance is 2 H.
a) 0.25
b) 0.2
c) 2
d) 0.5
Answer: d
Explanation: The time constant for an R-L series circuit will be τ = L/R. Put R = 4 and L = 2. We get τ = 2/4 = 0.5 second.
Electric Fields In Material Space MCQs
5. Find the time constant for a R-C circuit for resistance R = 24 kohm and C = 16 microfarad.
a) 1.5 millisecond
b) 0.6 nanosecond
c) 384 millisecond
d) 384 microsecond
Answer: c
Explanation: The time constant for R-C circuit is τ = RC. Put R = 24 kilo ohm and C = 16 micro farad. We get τ = 24 x 103 x 16 x 10-6 = 0.384 = 384 millisecond.
6. Find the capacitance when charge is 20 C has a voltage of 1.2V.
a) 32.67
b) 16.67
c) 6.67
d) 12.33
Answer: b
Explanation: Capacitance is related to Q and V as C = Q/V. Put C = 20C and V = 1.2V, we get Q = 20/1.2 = 16.67 farad.
7. Calculate the capacitance of two parallel plates of area 2 units separated by a distance of 0.2m in air(in picofarad)
a) 8.84
b) 88.4
c) 884.1
d) 0.884
Answer: b
Explanation: Capacitance is given by, C = εo A/d. Put A = 2, d = 0.2, εo = 8.854 x 10-12, we get C = 8.841 x 10-11 = 88. 41 pF.
8. Compute the capacitance between two concentric shells of inner radius 2m and the outer radius is infinitely large.
a) 0.111 nF
b) 0.222 nF
c) 4.5 nF
d) 5.4 nF
Answer: b
Explanation: The concentric shell with infinite outer radius is considered to be an isolated sphere. The capacitance C = 4πε/(1/a – 1/b). If b->∞, then C = 4πεa. Put a = 2m, we get C = 4π x 8.854 x 10-12 x 2 = 0.222 nF.
9. The capacitance of a material refers to
a) Ability of the material to store magnetic field
b) Ability of the material to store electromagnetic field
c) Ability of the material to store electric field
d) Potential between two charged plates
Answer: c
Explanation: The capacitance of a material is a measure of the ability of the material to store electric field. It is the ratio of charge stored to the voltage across the parallel plates.
10. A cable of core radius 1.25cm and impregnated paper insulation of thickness 2.13cm and relative permittivity 3.5. Compute the capacitance of the cable/km(in nF)
a) 195.7
b) 179.5
c) 157.9
d) 197.5
Answer: a
Explanation: Capacitance between coaxial cylinders of inner radius 1.25cm and outer radius 1.25 + 2.13 = 3.38cm will be C = 2πεL/ ln(b/a). Put b = 3.38, a = 1.25 and L = 1000m, we get C = 1.957 x 10-7 = 195.7 nF.
Method of Images
1. Identify the advantage of using method of images.
a) Easy approach
b) Boundaries are replaced by charges
c) Boundaries are replaced by images
d) Calculation using Poisson and Laplace equation
Answer: a
Explanation: Electrostatic boundary value problems are difficult if Poisson and Laplace equation is solved directly. But method of images helps us to solve problems without the equations. This is done by replacing boundary surfaces with appropriate image charges.
2. Calculate the electric field intensity of a line charge of length 2m and potential 24V.
a) 24
b) 12
c) 0.083
d) 12.67
Answer: b
Explanation: The electric field intensity is given by the ratio of potential to distance or length. E = V/d = 24/2 = 12 V/m.
3. Calculate potential of a metal plate of charge 28C and capacitance 12 mF.
a) 3.33 kohm
b) 2.33 kohm
c) 3.33 Mohm
d) 2.33 Mohm
Answer: b
Explanation: Potential is given by V = Q/C. Put Q = 28C and C = 12 mF. We get V = 28/12 x 10-3 = 2.333 x 103 ohm.
4. Find the dissipation factor when series resistance is 5 ohm and capacitive resistance is 10 unit.
a) 2
b) 0.5
c) 1
d) 0
Answer: b
Explanation: The dissipation factor is nothing but the tangent of loss angle of loss tangent. Tan δ = Series resistance/Capacitive resistance = 5/10 = 0.5.
5. A material with zero resistivity is said to have
a) Zero conductance
b) Infinite conductance
c) Zero resistance
d) Infinite resistance
Answer: c
Explanation: Since resistivity is directly proportional to the resistance, when the resistivity is zero, resistance is also zero. Thus we get zero resistance. The option infinite conductance is also possible ideally, but it is not possible practically. As there is always some loss in the form of heat, thus infinite conductance is impossible, but a short circuit (zero resistance) is practically possible.
6. Find the energy stored by the capacitor 3F having a potential of 12V across it.
a) 432
b) 108
c) 216
d) 54
Answer: c
Explanation: The energy stored in a capacitor is given by, E = 0.5 CV2.
E = 0.5 x 3 x 122 = 0.5 x 432 = 216 units.
7. By method of images, the problem can be easily calculated by replacing the boundary with which polygon?
a) Rectangle
b) Trapezoid
c) Square
d) Triangle
Answer: d
Explanation: When any field or potential needs to be calculated for either line charge or coaxial cable or concentric cylinder, the method of images uses a triangle which converts the three dimensional problem to one dimensional analysis. From this, the result can be calculated.
8. Calculate the electric field due to a surface charge of 20 units on a plate in air(in 1012 order)
a) 2.19
b) 1.12
c) 9.21
d) 2.91
Answer: b
Explanation: The electric field due to plate of charge will be E = ρs/2εo. Put ρs = 20, we get E = 20/(2 x 8.854 x 10-12) = 1.129 x 1012 units.
9. Find the electric field due to charge density of 1/18 and distance from a point P is 0.5 in air(in 109 order)
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: The electric field for this case is given by, E = ρl/2πεd. Put ρl = 1/18 and d = 0.5. We get E = 2 x 109 units.
10. Find the total capacitances when two capacitors 2F and 5F are in series.
a) 5/7
b) 12/7
c) 2/5
d) 10/7
Answer: d
Explanation: Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7 farad.