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EE371 Electromagnetic Field TheorySpring 20 20Mid-Term Exam (SOLUTION)Maximum Time: ( 60 min) Max Marks: 30.....................................................................................................................Question 1: [2]State and explain Coulomb’s Law in vector form.Solution:Coulo...

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EE371 Electromagnetic Field Theory Spring 2020 Mid-Term Exam (SOLUTION) Maximum Time: (60 min) Max Marks: 30 ……………………………………………………………………………………………………… Question 1: [2] State and explain Coulomb’s Law in vector form. Solution: Coulombs law states that the force (F) between any two point charges (Q1 and Q2) is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance R between them. It is directed along the line joining the two charges. ฀฀2 =

฀฀1 ฀฀2 ฀฀� � 12 � 12 4฀฀฀฀0 �฀฀

If Q1 and Q2 have like signs, the vector force ฀฀2 on Q2 is in � 12 . the same direction as the vector ฀฀

Question 2: (CLO_1)(PLO_2) (Cognitive Level 3, i.e., Applying) [2+2+4 = 8] at the origin, if the following charge distributions are present in free space:  Find the electric field ฀฀� i. Point charge of 12 nC at ฀฀(2, 0, 6). ii. Uniform charge of surface charge density ฀฀฀ ฀ = 0.2 nC/m2 at ฀ ฀ = 2. iii.

Solution:

Find the total charge within the given volume having volume charge density ฀฀฀ ฀ = ฀฀2 ฀฀ 2 sin(0.6฀฀);

0 ≤ ฀฀ ≤ 0.1, 0 ≤ ฀฀ ≤ ฀฀, 2 ≤ ฀฀ ≤ 4

Question 3: (CLO_1)(PLO_2) (Cognitive Level 3, i.e., Applying) [10] The figure below shows a closed Gaussian cylindrical surface of radius R immersed in a uniform � , the cylinder axis being parallel to the field. Compute the flux, ψ for the given closed electric field ฀฀

surface.

Solution: The flux ψ can be written as the sum of three terms, an integral over (a) the left cylinder cap, (b) the cylindrical surface, and (c) the right cap. Thus �  ∙ ฀฀฀฀ Ψ = ∮ ฀฀  ฀฀ �  ∙ ฀฀฀฀+ ∫ ฀฀ �  ∙ ฀฀฀฀� ∙ ฀฀฀฀+ ∫ ฀฀ Ψ = ฀฀0 �∫ � (1) (฀฀)

(฀฀)

(฀฀)

For the left cap the unit vectors of the surface and the electric field intensity are antiparallel so; �  ∙ ฀฀฀฀ = −฀฀฀฀฀฀ ฀฀  for all elements ฀฀฀฀ of surface S. Thus �  ∙ ฀฀฀฀= −฀฀฀฀ � ฀฀

(฀฀)

Similarly, for the right cap the unit vectors of the surface and the electric field intensity are parallel so; �  ∙ ฀฀฀฀= ฀฀฀฀ � ฀฀

(฀฀)

Finally, for the cylindrical surface the unit vectors of the surface and the electric field intensity are perpendicular so; �  ∙ ฀฀฀฀= 0 � ฀฀

(฀฀)

Finally the eq (1) gives us the total flux as:  ฀฀ �  ∙ ฀฀฀฀+ � ฀฀ �  ∙ ฀฀฀฀� = ฀฀0 [−฀฀฀฀ + 0 + ฀฀฀฀ ] = 0 ∙ ฀฀฀฀+ � ฀฀ Ψ = ฀฀0 � �� (฀฀)

(฀฀)

(฀฀)

Question 4: (CLO_1)(PLO_2) (Cognitive Level 3, i.e., Applying) [10] In a certain region, the electric field intensity is given as:

Find: i. ii. Solution:

3 �  = (2฀ ฀ cos ฀฀) ⁄ ฀฀฀฀� ฀฀ � ฀฀฀฀3 ฀ ฀ + (฀ ฀ sin⁄฀฀) ฀฀

equation of the stream lines charge contained in a sphere of radius b centered at the origin...