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SubjectsPhysics
गाउस का नियम और इसके अनुप्रयोग
In this Class 12 Physics topic from Chapter 1, Electric Charges and Fields, students learn how electric flux is related to the net charge enclosed by a closed surface through Gauss’s law. The topic develops the idea of Gaussian surfaces, uses symmetry to simplify electric-field calculations, and applies the law to charged spherical shells, uniformly charged spheres, infinite line charges, and plane sheets. It also helps students understand the electric field inside conductors and choose suitable surfaces for solving electrostatic problems.
Correct answer: A. Same total charge placed at the centre
Explanation: Step 1: Excess charge of a conducting sphere resides on the surface. Step 2: For outside points, spherical symmetry exists. Step 3: Thus outside field is like that of total charge placed at the centre.
Correct answer: A. It increases with distance from centre
Explanation: Step 1: In an insulating sphere, charge may be distributed through volume. Step 2: Enclosed charge inside a smaller Gaussian sphere increases with cube of radius. Step 3: As a result, inside field increases with distance from centre.
Correct answer: A. Zero inside conductor, may vary with distance inside insulator
Explanation: Step 1: In a conductor, excess charge moves to the surface. Step 2: Hence inside conductor field is zero in electrostatic condition. Step 3: In an insulator, charge can remain in volume, so inside field may vary.
Correct answer: A. It becomes half
Explanation: Step 1: Field of a long charged line is inversely proportional to distance. Step 2: Doubling distance makes field half. Step 3: Do not confuse it with square dependence of point charge.
Correct answer: A. One fourth
Explanation: Step 1: Field of a point charge is inversely proportional to square of distance. Step 2: When distance is doubled, square factor becomes four. Step 3: Therefore field becomes one fourth.
Correct answer: A. It remains unchanged
Explanation: Step 1: Field of an infinite plane sheet is independent of distance. Step 2: Even if distance doubles, field magnitude does not change in the ideal case. Step 3: This result is based on symmetry of an infinite sheet.
Correct answer: A. Point charge inverse square, line charge inverse distance, infinite sheet distance independent
Explanation: Step 1: For a point charge, spherical area grows as square of distance. Step 2: For a line charge, cylindrical area grows with distance. Step 3: For an infinite sheet, field comes independent of distance.
Correct answer: A. Electric field times total area
Explanation: Step 1: Outward perpendicular field crosses every small part directly. Step 2: Since magnitude is same everywhere, total flux is field multiplied by total area. Step 3: This is simple only when direction and magnitude are uniform.
Correct answer: A. It can be zero
Explanation: Step 1: Outward field gives positive flux. Step 2: Inward field gives negative flux. Step 3: If the two contributions are equal, total flux can be zero.
Correct answer: A. When symmetry makes magnitude same everywhere
Explanation: Step 1: To take field out of flux calculation, its magnitude must be same on the surface. Step 2: This usually comes from symmetry. Step 3: Without symmetry, assuming this can be a mistake.
Correct answer: A. First identify closed surface, enclosed charge, and symmetry
Explanation: Step 1: A closed surface is essential in Gauss's law. Step 2: Total flux is decided by net enclosed charge. Step 3: To find field, use symmetry and choose the correct Gaussian surface.
Correct answer: A. Net charge enclosed inside the surface
Explanation: Step 1: Gauss's law connects total flux through a closed surface with enclosed net charge. Step 2: Shape or colour of the surface does not directly decide net flux. Step 3: In such questions, first check the net charge inside the closed surface.
Correct answer: B. It becomes double
Explanation: Step 1: According to Gauss's law, total flux is proportional to enclosed charge. Step 2: If enclosed charge doubles, flux also doubles. Step 3: In ratio questions, remember the direct relation between enclosed charge and flux.
Correct answer: C. Positive
Explanation: Step 1: First add the charges enclosed by the closed surface. Step 2: Three coulomb positive and one coulomb negative give a net positive charge. Step 3: Positive net charge gives positive net flux.
Correct answer: D. Zero
Explanation: Step 1: In Gauss's law, net flux is decided only by net enclosed charge. Step 2: An outside charge can create field on the surface, but its net flux contribution is zero. Step 3: If no charge is inside, write net flux as zero.
Correct answer: A. Because symmetry allows electric field to be treated simply
Explanation: Step 1: Gauss's law gives total flux. Step 2: To find electric field, the field magnitude or direction should often be simple on the Gaussian surface. Step 3: So choosing a surface using symmetry is very useful in exams.
Correct answer: B. Because field is radial and same at equal distance everywhere
Explanation: Step 1: A point charge has spherical symmetry around it. Step 2: With the charge at the centre, every point of the sphere is at the same distance and field is normal to the surface. Step 3: For point charge, remember spherical Gaussian surface.
Correct answer: A. The wire has cylindrical symmetry
Explanation: Step 1: Field of a long straight wire is radial from the wire. Step 2: On the curved part of a cylindrical surface, flux can be calculated simply. Step 3: Cylindrical symmetry is most useful for line charge.
Correct answer: B. Because equal flux on both sides can be added simply
Explanation: Step 1: For an infinite plane sheet, field on both sides is equal in magnitude and normal to the sheet. Step 2: A small cylindrical pillbox gives flux through the two flat faces. Step 3: For plane sheet, add contributions from both sides carefully.
Correct answer: B. Net charge enclosed by the surface is zero
Explanation: Step 1: In Gauss's law, net flux is directly related to net enclosed charge. Step 2: If net flux is zero, net enclosed charge is zero. Step 3: It is not necessary that electric field is zero at every point on the surface.
Correct answer: C. Irregular and asymmetric charge distribution
Explanation: Step 1: Gauss's law remains true in every case. Step 2: But finding electric field is simple only when symmetry exists. Step 3: In asymmetric distributions, separating field from total flux is difficult.
Correct answer: A. Because electric field inside the conductor is zero
Explanation: Step 1: In electrostatic equilibrium, electric field inside a conductor is zero. Step 2: Hence no net electric flux passes through a Gaussian surface inside. Step 3: By Gauss's law, net charge enclosed by it is also zero.
Correct answer: A. Because zero field inside means no excess enclosed charge can remain inside
Explanation: Step 1: Electric field inside a conductor is zero. Step 2: A small Gaussian surface inside gives zero flux and zero enclosed excess charge. Step 3: Therefore excess charge resides on the surface.
Correct answer: A. Because charge resides on the surface and field inside a conductor is zero
Explanation: Step 1: In a conductor, excess charge stays on the surface in electrostatic equilibrium. Step 2: Inside the conducting material, electric field remains zero. Step 3: Remember zero field inside a conducting sphere.
Correct answer: A. Because an inner Gaussian surface encloses no charge
Explanation: Step 1: Charge of a shell lies on its surface. Step 2: A spherical Gaussian surface inside the shell encloses no charge. Step 3: With symmetry and Gauss's law, field inside is zero.
