Solved Exercise_Numericals - Electric Field Intensity (E) - Line, Surface & Mixed Charge Configuration - 2

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Solved Exercise/Numericals Electric Field Intensity (E) - Line, Surface & Mixed Charge Conguration - 2.

Ans: The electric eld intensity (E) due to an innite line charge is given as:

In this case, ρ = (6 – 3) ay + ρ(1 – 5)az = 3ay – 4az Hence aρ = (3ay - 4az) / 5 Therefore E is,

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Ans:

Electric eld due to a line charge density is given as:

In this case

Hence we have,

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Ans: As the both charges are on the y axis, the point at which the elds due to the two charges can cancel has to lie on the y-axis also.

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the point at which the net electric eld will be zero is midway between them, i.e. at point (0, 6, 0)

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Ans: Electric eld at point (1, 1, 1) is given as:

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= (- 0.0216, 0, 0.0288) + (0, 18, 18) – 264.7 (0, 0, 1) = - 0.0216 ax + 18ay – 264.7 az V/m

Ans:

ElectroMagnetic Theory Concepts & Solved Numericals Let f (x, y) = x + 2y – 5 SUBSCRIBE ∇ f

= ax + 2ay

A unit normal vector to any surface or plane is given as:

Now since (-1, 0, 1) lies below the plane, therefore Electric eld is given as:


sir plz hlp me in solving this que innite line charge of p=5nc/m lie along +ve n -ve x and y axes 1)(0,0,4) 2)(0,3,4)

plz hlp me in solving this que innite line charge is located on both +ve n -ve x and y axes a)(0,0,4) b)(0,3,4)

innite long charge is located on +ve and -ve xy axes 1)(0,0,4) 2)(0,3,4)


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Solved Exercise_Numericals - Electric Field Intensity (E) - Line, Surface & Mixed Charge Configuration - 2

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