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By Karan Singh Bisht
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Updated on 13 Apr 2026, 12:49 IST
Gauss’s Law is one of the most important topics in electrostatics and plays a major role in understanding electric fields and charge distribution. For students preparing for CBSE Class 12 exams, a strong command of this chapter is important for both board exam success and concept clarity. This section is designed to help you revise the topic in a simple and effective way with key theory, conceptual questions, and practice-based learning.
Here, you will find Gauss law numericals for Class 12 that help improve calculation skills and strengthen your understanding of formulas and applications. You will also come across important questions on Gauss Law for Class 12 that focus on core concepts, derivations, and exam-oriented practice. All the questions are based on the CBSE Class 12 syllabus and aim to support better preparation.
To make your study process even more helpful, NCERT Solutions for Class 12 Physics are also included so you can quickly refer to accurate explanations and strengthen your revision.
Gauss’s Law is a fundamental principle in electrostatics that relates the electric flux passing through a closed surface to the total electric charge enclosed within that surface.
It states that:
Φ = Q / ε0
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Explanation:
In simple terms, Gauss’s Law tells us that the electric field flowing out of a closed surface depends only on the charge inside it, not on the shape or size of the surface.
Q. State Gauss’s Law.
Answer: Electric flux through a closed surface equals charge enclosed divided by ε₀.
Q. What is electric flux?
Answer: Measure of electric field passing through a surface.
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Q. SI unit of electric flux?
Answer: Nm²/C
Q. What is Gaussian surface?
Answer: Imaginary closed surface used to apply Gauss’s Law.
Q. Does external charge affect flux?
Answer: No
Q. Formula of Gauss’s Law?
Answer: Φ = Q/ε₀
Q. What is ε₀?
Answer: Permittivity of free space
Q. Flux through closed surface with no charge?
Answer: Zero
Q. Nature of flux for positive charge?
Answer: Positive
Q. Nature of flux for negative charge?
Answer: Negative
Q. Why is Gauss’s Law useful?
Answer: Simplifies electric field calculations for symmetry.
Q. Define electric field.
Answer: Force per unit charge.
Q. Flux formula?
Answer: Φ = EA cosθ
Q. What is symmetry?
Answer: Uniform distribution of charge.
Q. Gaussian surface for sphere?
Answer: Concentric sphere
Q. Electric field inside conductor?
Answer: Zero
Q. Flux depends on?
Answer: Enclosed charge
Q. Field lines direction?
Answer: From positive to negative
Q. Unit of electric field?
Answer: N/C
Q. Charge density formula?
Answer: ρ = Q/V
Q. Surface charge density?
Answer: σ = Q/A
Q. Line charge density?
Answer: λ = Q/L
Q. Field due to plane sheet?
Answer: σ/2ε₀
Q. Field inside hollow sphere?
Answer: Zero
Q. Field outside sphere?
Answer: Same as point charge
Q. Derive Gauss’s Law.
Answer: Based on Coulomb’s law and flux definition.
Q. Explain flux with diagram.
Answer: Lines passing through surface represent flux.
Q. Why Gaussian surface chosen?
Answer: To exploit symmetry.
Q. Electric field of infinite line charge?
Answer: E = λ/2πε₀r
Q. Electric field of sphere?
Answer: E = kQ/r²
Q. Why field inside conductor zero?
Answer: Charges redistribute.
Q. Explain shielding.
Answer: Charges stay on surface.
Q. Flux independence from shape?
Answer: Depends only on charge.
What is electrostatic equilibrium?
Answer: No net movement of charges.
Q. What happens if charge doubled?
Answer: Flux doubles
Q. Define closed surface.
Answer: Surface enclosing volume
Q. What is dipole?
Answer: Two equal opposite charges
Q. A charge of 4 × 10-6 C is enclosed inside a closed surface. Find the electric flux through the surface.
Solution:
Using Gauss’s Law,
Φ = Q / ε0
Φ = (4 × 10-6) / (8.85 × 10-12)
Φ ≈ 4.52 × 105 Nm2/C
Answer: 4.52 × 105 Nm2/C
Q. A closed surface encloses no charge. Find the electric flux through it.
Solution:
By Gauss’s Law,
Φ = Qenclosed / ε0
Since enclosed charge is zero,
Φ = 0
Answer: 0
Q. A long straight wire has linear charge density 3 × 10-6 C/m. Find the electric field at a distance of 0.2 m.
Solution:
E = λ / (2πε0r)
E = (3 × 10-6) / (2 × 3.14 × 8.85 × 10-12 × 0.2)
E ≈ 2.7 × 105 N/C
Answer: 2.7 × 105 N/C
Q. Surface charge density is 6 × 10-6 C/m2. Find the electric field near the sheet.
Solution:
E = σ / (2ε0)
E = (6 × 10-6) / (2 × 8.85 × 10-12)
E ≈ 3.39 × 105 N/C
Answer: 3.39 × 105 N/C
Q. A charge of 9 × 10-9 C is placed inside a cube. Find the total flux through the cube.
Solution:
Φ = Q / ε0
Φ = (9 × 10-9) / (8.85 × 10-12)
Φ ≈ 1.02 × 103 Nm2/C
Answer: 1.02 × 103 Nm2/C
Q. A charge q is placed at the center of a cube. Find the flux through one face.
Solution:
Total flux through cube:
Φtotal = q / ε0
Since cube has 6 equal faces, flux through one face:
Φone face = q / 6ε0
Answer: q / 6ε0
Q. A spherical shell carries charge 5 × 10-8 C. Find the electric field at a point 0.5 m from its center.
Solution:
Outside the shell, it behaves like a point charge.
E = (1 / 4πε0) × (Q / r2)
E = 9 × 109 × (5 × 10-8 / (0.5)2)
E = 9 × 109 × (5 × 10-8 / 0.25)
E = 1.8 × 103 N/C
Answer: 1.8 × 103 N/C
Q. A charged conducting shell encloses a cavity. Find the electric field inside the conductor.
Solution:
For a conductor in electrostatic equilibrium,
E = 0
Answer: 0
Q. A closed surface has electric flux 2.26 × 105 Nm2/C. Find the charge enclosed.
Solution:
Q = Φε0
Q = 2.26 × 105 × 8.85 × 10-12
Q ≈ 2 × 10-6 C
Answer: 2 × 10-6 C
Q. A sphere of radius 0.1 m carries charge 8 × 10-9 C. Find the electric field at its surface.
Solution:
E = (1 / 4πε0) × (Q / R2)
E = 9 × 109 × (8 × 10-9 / (0.1)2)
E = 9 × 109 × (8 × 10-9 / 0.01)
E = 7.2 × 103 N/C
Answer: 7.2 × 103 N/C
Q. A hollow spherical shell has charge Q. Find electric field at a point inside it.
Solution:
By Gauss’s Law, charge enclosed inside a Gaussian surface within the shell is zero.
E = 0
Answer: 0
Q. A wire has λ = 4 × 10-6 C/m. Find electric field at r = 0.5 m.
Solution:
E = λ / (2πε0r)
E = (4 × 10-6) / (2 × 3.14 × 8.85 × 10-12 × 0.5)
E ≈ 1.44 × 105 N/C
Answer: 1.44 × 105 N/C
Q. The electric field near a large plane sheet is 5 × 104 N/C. Find surface charge density.
Solution:
E = σ / (2ε0)
σ = 2ε0E
σ = 2 × 8.85 × 10-12 × 5 × 104
σ = 8.85 × 10-7 C/m2
Answer: 8.85 × 10-7 C/m2
Q. A sphere encloses a charge of -3 × 10-6 C. Find flux.
Solution:
Φ = Q / ε0
Φ = (-3 × 10-6) / (8.85 × 10-12)
Φ ≈ -3.39 × 105 Nm2/C
Answer: -3.39 × 105 Nm2/C
Q. The electric field at 0.3 m from the center of a spherical shell is 2 × 103 N/C. Find the charge on shell.
Solution:
E = (1 / 4πε0) × (Q / r2)
Q = Er2 / (9 × 109)
Q = (2 × 103 × (0.3)2) / (9 × 109)
Q = (2 × 103 × 0.09) / (9 × 109)
Q = 180 / (9 × 109)
Q = 2 × 10-8 C
Answer: 2 × 10-8 C
Q. A sheet carries charge density 1 × 10-5 C/m2. Find field near it.
Solution:
E = σ / (2ε0)
E = (1 × 10-5) / (2 × 8.85 × 10-12)
E ≈ 5.65 × 105 N/C
Answer: 5.65 × 105 N/C
Q. A charge 6 × 10-6 C is kept inside a cube. Find total flux.
Solution:
Φ = Q / ε0
Φ = (6 × 10-6) / (8.85 × 10-12)
Φ ≈ 6.78 × 105 Nm2/C
Answer: 6.78 × 105 Nm2/C
Q. A charge of 12 × 10-6 C is placed at the center of a cube. Find flux through one face.
Solution:
Φtotal = Q / ε0
Φone face = Q / 6ε0
Φone face = (12 × 10-6) / (6 × 8.85 × 10-12)
Φone face ≈ 2.26 × 105 Nm2/C
Answer: 2.26 × 105 Nm2/C
Q. Linear charge density is 1.5 × 10-6 C/m. Find field at 0.1 m.
Solution:
E = λ / (2πε0r)
E = (1.5 × 10-6) / (2 × 3.14 × 8.85 × 10-12 × 0.1)
E ≈ 2.7 × 105 N/C
Answer: 2.7 × 105 N/C
Q. A point charge q is placed at the center of a sphere. Find flux through one hemisphere.
Solution:
Total flux through full sphere:
Φ = q / ε0
Since hemisphere is half the sphere,
Φhemisphere = q / 2ε0
Answer: q / 2ε0
Q. A charge 9 × 10-9 C produces electric field 900 N/C at distance r. Find r.
Solution:
E = (1 / 4πε0) × (Q / r2)
900 = 9 × 109 × (9 × 10-9 / r2)
900 = 81 / r2
r2 = 81 / 900 = 0.09
r = 0.3 m
Answer: 0.3 m
Flux through a closed surface is 8.85 × 103 Nm2/C. Find enclosed charge.
Solution:
Q = Φε0
Q = 8.85 × 103 × 8.85 × 10-12
Q ≈ 7.83 × 10-8 C
Answer: 7.83 × 10-8 C
Q. A spherical shell has radius R. Find electric field at r < R.
Solution:
Inside a spherical shell, enclosed charge is zero.
E = 0
Answer: 0
Q. If surface charge density becomes 2σ, what is the new field?
Solution:
E = σ / (2ε0)
For 2σ,
E' = 2σ / (2ε0) = σ / ε0
So the field becomes double.
Answer: 2E
Q. Charges +2µC, -1µC, and +3µC are enclosed in a surface. Find net flux.
Solution:
Net enclosed charge:
Q = 2 - 1 + 3 = 4µC = 4 × 10-6 C
Φ = Q / ε0
Φ = (4 × 10-6) / (8.85 × 10-12)
Φ ≈ 4.52 × 105 Nm2/C
Answer: 4.52 × 105 Nm2/C
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Gauss's Law is crucial for understanding electric fields and flux, and it forms the basis for solving complex electrostatic problems in the CBSE Class 12 Physics syllabus. Mastery of this topic helps students tackle important questions effectively.
Gauss's Law is applied by selecting an appropriate Gaussian surface where the symmetry of the problem simplifies the calculation of electric fields. It’s especially useful for spherical, cylindrical, and planar symmetries.
You can find important questions on Gauss’s Law, along with detailed solutions and explanations, on the Infinity Learn page dedicated to CBSE Class 12 Physics study material, which covers all key concepts and practice problems.
Gauss’s Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. It helps in calculating electric fields for symmetrical charge distributions.
The SI unit associated with Gauss’s Law (electric flux) is: Newton meter² per coulomb (Nm²/C)
Gauss’s principle refers to the idea that the electric flux passing through any closed surface depends only on the total charge enclosed inside that surface, regardless of the shape or size of the surface.