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If three point charges -3q, +2q and +4q are are placed in a sphere of radius r, then the flux linked with the sphere will be:
1. \(\frac{3q}{\epsilon_o}\)
2. \(\frac{9q}{\epsilon_o}\)
3. Zero
4. None of these

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Correct Answer - Option 1 : \(\frac{3q}{\epsilon_o}\)

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

CALCULATION:

Given Q1 = -3q, Q2 = +2q, and Q3 = +4q

The net charge enclosed in the sphere is given as,

⇒ Q = Q1 + Q2 + Q3

⇒ Q = -3q + 2q + 4q

⇒ Q = 3q

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,

\(⇒ ϕ=\frac{Q}{ϵ_o}\)

\(⇒ ϕ=\frac{3q}{ϵ_o}\)

  • Hence, option 1 is correct.

  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.

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