Electromagnetic Theory Questions and Answers – Method of Images

This set of Electromagnetic Theory Quiz focuses on “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
View Answer

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
View Answer

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
View Answer

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
View Answer

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
View Answer

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.
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6. Find the energy stored by the capacitor 3F having a potential of 12V across it.
a) 432
b) 108
c) 216
d) 54
View Answer

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
View Answer

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
View Answer

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
View Answer

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
View Answer

Answer: d
Explanation: Two capacitances in series gives C = C1C2/C1 + C2 = 2 x 5/2 + 5 = 10/7 farad.

Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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