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At what distance along the central perpendicular axis of a uniformly charged plastic disk of radius 0.390 m is the magnitude of the electric field equal to one-half the magnitude of the field at the center of the surface of the disk?

Respuesta :

Kalahira
Radius, the distance from the centre = 0.390
 Electric field is equal to half of the magnitude. E2 = E / 2
 Given
E1 = E2 E1 = k x Q / r^2
  E2 = (k x Q / r2^2) / 2
  Equating the both we get 2 x r^2 = r2^2
 r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
  r2 = 1.414 x 0.390 = 0.551 m

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