Kindly share detailed explanation for each step and each options. ALLENFigure shows an isosceles right triangular thin sheet of uniform surface charge densityenergy of the sheet is U. What will be self energy of the sheet if it is folded along the dotted line.No = I Go E2 dv.3.(A) 2Ua+xxxxx+xtX-(B) 2√2U,StxaX++++++(C) √2U.+a+++a(D) U₁Paper-1o. The self

for case-1, from the given figure, the length of unkown side can be calculated by, L = a2+a2 = 2×a…

Answered: G. The self Figure shows an isosceles…

G. The self Figure shows an isosceles right triangular thi energy of the sheet is U. What will be self energy of the sheet if it is folded along the dotted line. 2 dv.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q = 2.00 μc, and L = 7.00 με 60.0° L 0.650 m.) q -4.00 μC x (a) Calculate the electric field at the position of charge q due to the 7.00-μC and -4.00-μC charges. -10636.7 Once you calculate the magnitude of the field contribution from each charge you need to add these as vectors. KN/CI + 128963 Think carefully about the direction of the field due to the 7.00-μC charge. KN/C ĵ (b) Use your answer to part (a) to determine the force on charge q. -21.3 If you know the electric field at a particular point, how do you find the force that acts on a charge at that point? mN I + 257.9 Check units carefully. Note that the field in part (a) is given in KN/J and the force in this part is given in mN. mN ĵ ×
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Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q 7.00 με 60.0° L = 2.00 μc, and L = 0.650 m.) q -4.00 μC x (a) Calculate the electric field at the position of charge q due to the 7.00-μC and -4.00-μC charges. KN/CI + KN/C ĵ (b) Use your answer to part (a) to determine the force on charge q. mNI + mN ĵ
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