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Mark the correct option

A

Gauss law is valid fro unsymmetrical charged distributions

B

Gauss law is valid only for charge palced in vacuum

C

The electric field is calculated by Gauss law is the field due to the charges outside the Gauss law is the field due to the charges outside the Gaussian surface.

D

The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.

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The correct Answer is:
To solve the question regarding Gauss's law and to mark the correct option, we will go through the following steps: ### Step 1: Understand Gauss's Law Gauss's law states that the electric flux (Φ) through a closed surface is proportional to the charge (Q) enclosed within that surface. Mathematically, it is expressed as: \[ \Phi = \frac{Q_{\text{in}}}{\epsilon_0} \] where \( \epsilon_0 \) is the permittivity of free space. ### Step 2: Analyze the Options We need to evaluate the given options based on our understanding of Gauss's law: 1. **Option A**: "Gauss's law is valid only for unsymmetrical charge distribution." - This statement is incorrect. Gauss's law is valid for both symmetrical and unsymmetrical charge distributions. 2. **Option B**: "Gauss's law is valid only in vacuum." - This statement is also incorrect. Gauss's law holds true in any medium, not just in vacuum. 3. **Option C**: "The electric field calculated by Gauss's law is due to the charge outside the Gaussian surface." - This statement is incorrect. The electric field calculated using Gauss's law is due to the charge enclosed within the Gaussian surface, not outside. 4. **Option D**: "The flux of the electric field through a closed surface is equal to the charge enclosed by the surface." - This statement is correct as it directly reflects the essence of Gauss's law. ### Step 3: Conclusion After analyzing all the options, we conclude that the correct option is: **Option D**: "The flux of the electric field through a closed surface is equal to the charge enclosed by the surface."
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NARAYNA-ELECTRIC CHARGES AND FIELDS-C.U.Q
  1. A cubical Gaussian surface encloses electric flux of 30C per unit perm...

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  2. As one penetrates uniformly charged conducting sphere,what happens to ...

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  3. Mark the correct option

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  4. If the flux of the electric field through a closed surface is zero,

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  5. If infinite parallel plane sheet of a metal is charged to charge densi...

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  6. The electric flux from a cube of edge l is phi. If an edge of the cube...

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  7. If the electric flux entering and leaving an enclosed surface respecti...

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  8. Electric flux at a point in an electric field is

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  9. Electric flux over a surface in an electric field

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  10. A charge Qis placed at the mouth of a conical flask. The flux of the e...

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  11. A charge Qis placed at the mouth of a conical flask. The flux of the e...

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  12. Electric field intensity at a point due to an infinite sheet of charge...

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  13. Two thin infinite parallel plates have uniform charge densities + sigm...

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  14. In the above question, if the sheets were thick and conducting , value...

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  15. In the above problem the value of E in the space outside the sheets is...

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  16. The Gaussian surface for calculating the electric field due to a charg...

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  17. The electric flux over a sphere of radius 1m is phi. If radius of the ...

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  18. A charge q is placed at the centre of a cube.What is the electric fl...

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  19. A charge Q is situated at the corner of a cube the electric flux passe...

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  20. A charge +q is placed at the mid point of a cube of side L. The electr...

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