Ask Question
19 November, 10:53

At what distance along the central axis of a uniformly charged plastic disk of radius R = 0.525 m is the magnitude of the electric field equal to 1/5 times the magnitude of the field at the center of the surface of the disk?

+3
Answers (1)
  1. 19 November, 11:03
    0
    The general formula for the strength of magnetic field is written below:

    E = kQ/d²,

    where k is the Coulomb's constant

    Q is the charge of the object

    d is the distance

    The ratio of the magnetic fields should be 1/5:

    E₁/E₂ = 1/5

    kQ/d₁² : kQ/d₂² = 1/5

    The k and Q will be cancelled out because only the distance is varied, the charge does not change. The equation becomes

    1/d₁² : 1/d₂² = 1/5

    Substitute d₁ = 0.525 m

    1 / (0.525 m) ² : 1/d₂² = 1/5

    Solving for d₂,

    d₂ = 0.235 m

    The distance is 0.235 m along the central axis.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “At what distance along the central axis of a uniformly charged plastic disk of radius R = 0.525 m is the magnitude of the electric field ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers