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?

Question

Physics

Ducky

9 October, 05:39

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?

+2

Answers (1)

Know the Answer?

Kane Braun · Answer 1

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.