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Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.
  • a)
    True
  • b)
    False
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Electric field of an infinitely long conductor of charge density &lamb...
Answer: a
Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/(4πεh).(sin α2 – sin α1)i + (cos α2 + cos α1)j
For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN.
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Most Upvoted Answer
Electric field of an infinitely long conductor of charge density &lamb...
The electric field of an infinitely long conductor can be found using Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space (ε0).

In the case of an infinitely long conductor, we can consider a cylindrical surface of radius r and length L, with the conductor passing through the center of the cylinder. Since the conductor is infinitely long, the electric field will be constant along the surface of the cylinder.

The electric flux through the cylindrical surface can be calculated as the product of the electric field (E) and the surface area of the cylinder (2πrL). The total charge enclosed by the cylinder is given by the charge density (ρ) multiplied by the volume of the cylinder (πr^2L).

Using Gauss's law, we have:

E * 2πrL = (ρ * πr^2L) / ε0

Simplifying, we find:

E = (ρ * r) / (2ε0)

Therefore, the electric field of an infinitely long conductor is directly proportional to the charge density and the distance from the conductor.
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Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.a)Trueb)FalseCorrect answer is option 'A'. Can you explain this answer?
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