Gauss’s Law Class 12: Complete Notes, Derivation & Important Questions

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Introduction


    Learn about Gauss’s Law Class 12 Physics with derivation, electric flux formula, important concepts, applications, and MCQs for board exams.

Gauss’s Law Class 12 Physics


Statement


    The total electric flux passing through a closed surface kept in an electric field in vacuum or air is equal to the product of the net charge `Sigma q` inside the volume of the closed surface and `frac{1}{epsilon_0}.`


Mathematical Form


Thus, the total flux


`phi = frac{Sigma q}{epsilon_0}`        ……eq.(1)


Here 


`Sigma q ` is the algebraic sum of charges that exist inside the surface.


`epsilon_0` is the permittivity of vacuum


We know that


`frac{1}{4 pi epsilon_0} = K`


`frac{1}{ epsilon_0} = 4 pi K`


Then from Equation (1)


`phi = frac{1}{epsilon_0}times Sigma q`


`phi = 4 pi K times Sigma q`


    The total electric flux is the product of the net charge `Sigma q` inside the volume of the closed surface and `phi = 4 pi K Sigma q`.

Important Points for Exams


  • Gaussian Surface: An imaginary 3D surface used to calculate the electric field.

  • Sign Convention: Electric flux leaving the surface is always taken to be positive, and entering is taken to be negative.

  • Independence: This law is independent of the size and shape of the surface, as long as it is closed.

  • It is valid where Coulomb’s inverse-square law holds.

  • Applicable in both vacuum and medium.

  • Flux does not depend on shape, size of the surface, position of the charge inside, or distribution of charge.

  • Flux `phi` depends on enclosed charge, the nature of the charge, as well as the medium.

Derivation of Gauss’s Law


Step 1: Setup


    Consider a single point charge q located at the origin. We want to find the total electric flux passing through a spherical Gaussian surface of radius r centered on the charge. 

    According to Coulomb’s Law, the electric field E at any point on this sphere is:

    `vec E = frac{1}{4piepsilon_0}frac{q}{r^2}hat r`

where:

`epsilon_0` is the permittivity of free space.

`hat r` is the unit vector pointing radially outward.

Step 2: Definition of Electric Flux


Electric flux through an infinitesimal area element `dvecA` is defined as:

        `dphi_E = vecE.dvecA`

    On our spherical surface, the area vector `dvecA` always points radially outward, just like the electric field. Therefore, the angle `theta` between `vecE` and `dvecA` is `^{circ}`, making cos(`0^{circ})= 1.`

Step 3: Integration Over the Surface


To find the total flux, we integrate over the entire closed surface:

`phi_E=oint vecEcdotdvecA=oint EdA cos(0^{circ}) = oint EdA`

    Since the distance r is constant for every point on the sphere, the magnitude of the electric field `vecE` is also constant and can be moved outside the integral:

    `phi_E = E oint dA`

    The term `oint dA`  is simply the total surface area of a sphere, which is `4 pi r^2.`

Step 4: Final Formula


Now, substitute the expression for E from Step 1

`phi_E = (frac {1}{4 pi epsilon_0}frac{q}{r^2})(4 pi r^2)`

The terms `4pi` and `r^2` cancel out, leaving us with the standard form of Gauss’s Law:

`phi_E = frac{q_(text{enclosed})}{epsilon_0}`

Importance of Gauss’s Law in Electrostatics


  • Uses of Gauss’s Law in Electrostatics are given below –

  • This law is the foundation for electrostatic analysis in physics.

  • It is used to understand the electric flux and electric fields.

  • Gauss’s Law relates the electric field to an enclosed charge.

  • It simplifies calculations of electric fields for symmetrical charge distributions in electrostatics.


Applications of Gauss’s Law 

Gauss’s Law is used for calculating –


  1. Electric field intensity due to a uniformly charged conducting sphere.
  2. Electric field intensity due to a uniformly charged non-conducting sphere.
  3. Electric field intensity due to an infinite line charge.
  4. Electric field due to an infinite uniformly charged non-conducting sheet.
  5. Electric field due to an infinite uniformly charged conducting sheet.


MCQs on Gauss’s Law


1. The total electric flux passing through a closed surface in an electric field is equal to:

   A) Flux = charge / permittivity

  B) The product of net charge and magnetic field strength

  C) The sum of net charge and permittivity

 D) The sum of the net charge and the magnetic field strength


Answer: A) Flux = charge / permittivity


2. Gauss’s law is valid for regions that follow:

   A) Ampere’s law

   B) Coulomb’s inverse square law

   C) Faraday’s law

   D) Ohm’s law


Answer: B) Coulomb’s inverse square law


3. Gauss’s law is applicable in:

   A) Vacuum only

   B) Medium only

   C) Both vacuum and medium

   D) Neither vacuum nor medium


Answer: C) Both vacuum and medium


4. The value of outgoing flux from a closed surface depends on:

   A) Size and shape of the surface

   B) Position of charge inside the surface

   C) Distribution of charge inside the surface

   D) None of the above


Answer: D) None of the above


5. The value of flux does not depend on the:

   A) Position of charge in the closed surface

   B) Quantity and nature of the charge

   C) Permittivity of the medium

   D) Size of the closed surface


Answer: D) Size of the closed surface


6. Which equation represents Gauss’s law?


   A) `phi = frac{1}{epsilon_0}times Sigma q`


   B) `phi = frac{Sigma q}{epsilon_0}`


   C) `phi = 4 pi K times Sigma q`


   D) All of the above


Answer:    D) All of the above


7. What is the permittivity of vacuum denoted by?


   A) `Sigma q`


   B) `epsilon_0`


   C) K


   D) `phi`


Answer: B) `epsilon_0`


8. The value of flux depends on the:

   A) Distribution of charge inside the closed surface

   B) Distance between charges inside the closed surface

   C) Total charge enclosed by the closed surface

   D) Electric field strength inside the closed surface


Answer: C) Total charge enclosed by the closed surface


9. The value of phi in Gauss’s law is equal to:


   A) `1/epsilon_0`


   B) `frac{Sigma q}{4 pi K}`


   C) `frac{4 pi K}{Sigma q}`


   D) `frac{Sigma q}{epsilon_0}`


Answer: D) `frac{Sigma q}{epsilon_0}`


10. Gauss’s law relates electric flux to the:

    A) Magnetic field

    B) Electric field

    C) Electric potential

    D) Electric current


Answer: B) Electric field


Practice Questions


1. What is the equation for the total electric flux passing through a closed surface in an electric field in vacuum or air? (ϕ = (Σq) / ε₀)


2. What is the permittivity of vacuum? (ε₀)


3. Rewrite the equation for total flux (ϕ) in terms of the permittivity of vacuum (ε₀). (ϕ = 4πK `times` Σq)


4. What does the value of outgoing flux from a closed surface depend on? (It does not depend on the size and shape of the closed surface.)


5. Does Gauss’s law hold true in both a vacuum and a medium? (Yes)

This law is the foundation

6. What law does Gauss’s law rely on? (Coulomb’s inverse square law)


7. Does the value of flux (ϕ) depend on the position of charge within the closed surface? (No)


8. Does the value of flux (ϕ) depend on the distribution of charge inside the charge? (No)


9. What does the value of flux (ϕ) depend on? (The quantity and nature of the charge, as well as the medium)


10. Rewrite the equation for the permittivity of vacuum (ε₀) in terms of the constant K. (1 / ε₀ = 4πK)


11. What is the equation for the total electric flux passing through a closed surface in an electric field in vacuum? (Answer: `phi = frac{Sigma q}{epsilon_0}`)


12. If the net charge inside a closed surface is 5 C and the permittivity of vacuum is `8.854 times 10^-12 C^2/(N·m^2)`, what is the total flux passing through the surface? (Answer: `{5 C} / {8.854 times 10^-12} frac{C^2}{N·m^2}`)


13. If the value of K (Coulomb’s constant) is `9 times 10^9 frac{Nm^2}{C^2}` and the net charge inside a closed surface is 8 C, what is the total flux passing through the surface? (Answer: 8 C `times` 4πK)


14. True or False: Gauss’s law is valid only for regions that follow Coulomb’s inverse square law. (Answer: True)


15. True or False: Gauss’s law is followed in both vacuum and medium. (Answer: True)


16. Does the value of outgoing flux from a closed surface depend on the size and shape of the surface? (Answer: No)


17. Does the value of flux (`phi`) depend on the position of the charge inside a closed surface? (Answer: No)


18. Does the value of flux (`phi`) depend on the distribution of charge inside the closed surface? (Answer: No)


19. Does the value of flux (`phi`) depend on the quantity and nature of the charge? (Answer: Yes)


20. Does the value of flux (`phi`) depend on the medium? (Answer: Yes)


21. A point charge of +4 μC is located at the center of a spherical surface with a radius of 0.5 m. What is the electric flux through the spherical surface? `(text{Answer}: 2.88 times 10^4  frac{Nm^2}{C})`


22. A uniform electric field of magnitude 1.2 x 10^4 N/C passes through a cylindrical surface with an area of 0.2 m². What is the total electric flux through the surface? `(text{Answer}: 2.4 times 10^3  frac{Nm^2}{C})`


23. A point charge of -10 μC is located at the center of a cube with an edge length of 0.2 m. What is the electric flux through one face of the cube? `(text{Answer}: 1.8 times 10^4  frac{Nm^2}{C})`


24. An electric dipole consists of two point charges: +5 μC and -5 μC, separated by a distance of 0.3 m. What is the net electric flux through a spherical surface of radius 0.5 m centred on the dipole? (Answer: 0)