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If a unit positive charge is placed inside a sphere of radius r, then the electric flux through the sphere will be:
1. ϵor
2. \(\frac{ϵ_o}{r}\)
3. \(ϵ_o^{-1}\)
4. ϵo

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Correct Answer - Option 3 : \(ϵ_o^{-1}\)

CONCEPT:

Gauss's law:

  • According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(⇒ ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

  1. Gauss’s law is true for any closed surface, no matter what its shape or size.
  2. The charges may be located anywhere inside the surface.

EXPLANATION:

Given Q = 1, and r = radius of the sphere

​By the Gauss law, if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,

\(⇒ ϕ=\frac{Q}{ϵ_o}\)     -----(1)

By equation 1 the total flux linked with the sphere is given as,

\(⇒ ϕ=\frac{Q}{ϵ_o}\)

\(⇒ ϕ=\frac{1}{ϵ_o}\)

\(⇒ ϕ = ϵ_o^{-1}\)

  • Hence, option 3 is correct.

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