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If 6 C charge is placed inside a cube, then the electric flux linked with any one face of the cube is:
1. ϵo
2. \(\frac{1}{ϵ_o}\)
3. \(\frac{6}{ϵ_o}\)
4. 6ϵo

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Correct Answer - Option 2 : \(\frac{1}{ϵ_o}\)


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.


Given Q = 6C

  • ​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 cube is given as,

\(⇒ ϕ=\frac{Q}{ϵ_o}\)

\(⇒ ϕ=\frac{6}{ϵ_o}\)

So the flux linked with anyone face of the cube is given as,

\(⇒ ϕ'=\frac{\phi}{6}\)

\(⇒ ϕ'=\frac{1}{ϵ_o}\)

  • Hence, option 2 is correct.

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