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Gauss Law and Its Applications - UNSOLVED PRACTICE SET

Class 12

Chapter: Electric Charges and Fields | Topic: Gauss Law and Its Applications

Study Material.
Class 12

GAUSS LAW AND ITS APPLICATIONS - UNSOLVED PRACTICE SET

Topic: Gauss Law and Its Applications

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Gauss's Law in electrostatics states that the total electric flux through a closed surface is equal to:

  1. The total charge enclosed divided by ε₀
  2. The total charge enclosed multiplied by ε₀
  3. The surface area of the Gaussian surface
  4. The electric field at the surface

Q2. The Gaussian surface is:

  1. Any open surface
  2. A closed imaginary surface
  3. A real physical surface
  4. A surface with zero area

Q3. A point charge +Q is placed at the center of a spherical Gaussian surface of radius R. If the radius is doubled, the electric flux through the surface:

  1. Doubles
  2. Halves
  3. Remains the same
  4. Becomes zero

Q4. Using Gauss's Law, the electric field inside a uniformly charged solid sphere at a distance r from the center (r < R) is:

  1. Zero
  2. Proportional to r
  3. Proportional to 1/r²
  4. Constant

Q5. The electric field just outside a charged conducting surface is:

  1. Zero
  2. σ/2ε₀
  3. σ/ε₀
  4. 2σ/ε₀

Q6. A charge Q is placed at the center of a cube. The electric flux through one face of the cube is:

  1. Q/ε₀
  2. Q/6ε₀
  3. Q/3ε₀
  4. Zero

Short Answer Questions

Q7. State Gauss's Law in mathematical form. What does each term in the equation represent?

Q8. Why is Gauss's Law more useful for calculating electric fields of symmetric charge distributions (spherical, cylindrical, planar) rather than asymmetric ones?

Q9. A charge Q is enclosed by a spherical Gaussian surface. If the charge is moved slightly off-center (but still inside), does the electric flux through the surface change? Explain.

Q10. Using Gauss's Law, derive the expression for the electric field due to a uniformly charged spherical shell at a point
(a) inside the shell, and
(b) outside the shell.

Q11. Explain why the electric field inside a charged conductor is zero in electrostatic equilibrium. Use Gauss's Law in your explanation.

Q12. A closed surface encloses a dipole (two equal and opposite charges). What is the net electric flux through this surface? Justify your answer.

Long Answer Questions

Q13. State and prove Gauss's Law in electrostatics. Use a diagram of a point charge enclosed by an arbitrary closed surface to explain your derivation. Why is this law considered a fundamental law of nature?

Q14. Using Gauss's Law, derive the expression for the electric field due to a uniformly charged infinite plane sheet. Draw the Gaussian surface you use and explain each step clearly.

Q15. A solid sphere of radius 10 cm has a uniform volume charge density ρ = 2 × 10⁻⁶ C/m³.

(a) Calculate the total charge enclosed in the sphere.

(b) Find the electric field at a point 5 cm from the center.

(c) Find the electric field at a point 15 cm from the center.

[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]

Numerical / Application-Based Problems

Q16. A point charge q = +5 μC is placed at the center of a spherical Gaussian surface of radius 20 cm.

(a) Calculate the total electric flux through the Gaussian surface.

(b) If the radius of the Gaussian surface is doubled to 40 cm, what is the new flux? Explain why.

(c) If an additional charge of −3 μC is placed inside the surface, what is the net flux?

(d) Draw a diagram showing the Gaussian surface and the direction of the electric field on the surface.

[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]

Q17. A long straight wire has a uniform linear charge density λ = 4 × 10⁻⁶ C/m.

(a) Using Gauss's Law, derive the expression for the electric field at a perpendicular distance r from the wire. Draw the Gaussian surface.

(b) Calculate the electric field at a distance of 10 cm from the wire.

(c) A proton is placed at this distance. Calculate the force on the proton and its initial acceleration.

[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², e = 1.6 × 10⁻¹⁹ C, mₚ = 1.67 × 10⁻²⁷ kg]

Q18. A solid conducting sphere of radius 8 cm carries a total charge Q = +6 μC. It is surrounded by a concentric hollow conducting spherical shell of inner radius 12 cm and outer radius 15 cm, carrying a total charge of −4 μC.

(a) Find the charge distribution on the inner and outer surfaces of the shell.

(b) Calculate the electric field at r = 6 cm, r = 10 cm, and r = 20 cm from the center.

(c) Draw a rough graph showing the variation of electric field E with distance r from the center.

[Given: k = 9 × 10⁹ N m² C⁻²]


Total: 30 Marks | Time: 40 mins

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