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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. Perpendicular to the surface
Explanation: Step 1: If there were a tangential field at the surface, charges would move. Step 2: This cannot happen in electrostatic equilibrium. Step 3: Hence field just outside is perpendicular to the surface.
Correct answer: A. Stronger
Explanation: Step 1: Field near a conductor surface is related to surface charge density. Step 2: Larger surface charge density indicates stronger field. Step 3: Remember the direct relation between surface charge density and field.
Correct answer: A. Because plane symmetry leaves no distance dependence
Explanation: Step 1: An infinite sheet looks the same everywhere. Step 2: Using a pillbox Gaussian surface gives a field independent of distance. Step 3: This result is for the ideal infinite sheet.
Correct answer: A. It becomes half
Explanation: Step 1: For an infinite line charge, field is inversely proportional to distance. Step 2: Doubling distance makes the field half. Step 3: Keep this different from point charge inverse-square relation.
Correct answer: A. It becomes one-fourth
Explanation: Step 1: Field of a point charge varies inversely with square of distance. Step 2: Doubling distance makes square of distance four times. Step 3: Hence field becomes one-fourth.
Correct answer: A. It remains unchanged
Explanation: Step 1: Field due to an infinite charged sheet is independent of distance. Step 2: Doubling distance does not change the field value. Step 3: This is a special result for an ideal infinite sheet.
Correct answer: A. Inversely proportional to square of distance
Explanation: Step 1: For a point charge, the Gaussian surface is a sphere. Step 2: Area of sphere increases with square of distance. Step 3: Hence field varies inversely with square of distance.
Correct answer: A. Because curved area of cylindrical Gaussian surface is proportional to distance
Explanation: Step 1: A cylindrical surface is chosen for a line charge. Step 2: The curved area of the cylinder increases with distance. Step 3: Therefore field varies inversely with distance for a uniform line charge.
Correct answer: A. Same
Explanation: Step 1: An infinite sheet has the same plane symmetry on both sides. Step 2: Gauss's law gives equal magnitude of field on both sides. Step 3: Direction depends on the sign of charge on the sheet.
Correct answer: A. Away from the sheet
Explanation: Step 1: Field lines go outward from positive charge. Step 2: An infinite sheet has symmetry on both sides. Step 3: Therefore field points away from the sheet on both sides.
Correct answer: A. Toward the sheet
Explanation: Step 1: Field lines go toward negative charge. Step 2: Symmetry is the same on both sides of the sheet. Step 3: Hence field points toward the sheet on both sides.
Correct answer: A. Because electric field inside a conductor is zero in electrostatic equilibrium
Explanation: Step 1: In electrostatic equilibrium, electric field inside a conductor is zero. Step 2: If field is zero, flux through every part is zero. Step 3: Therefore total flux is zero.
Correct answer: A. Net enclosed charge is zero
Explanation: Step 1: Electric field inside conductor material is zero. Step 2: Thus total flux through an internal Gaussian surface is zero. Step 3: By Gauss's law, net enclosed charge is zero.
Correct answer: A. Because such a component would move free charges along the surface
Explanation: Step 1: Conductors contain free charges. Step 2: A tangential field would make them move along the surface. Step 3: In electrostatic equilibrium this cannot occur, so field is normal.
Correct answer: A. No, field may exist but net contribution may be zero
Explanation: Step 1: Total flux is the sum over the entire closed surface. Step 2: Positive and negative contributions on different parts may cancel. Step 3: So zero total flux does not require zero field everywhere.
Correct answer: A. Yes, but its net contribution to total flux will be zero
Explanation: Step 1: An external charge creates an electric field around it. Step 2: Its field may exist on the closed surface. Step 3: But lines entering and leaving balance, so net flux contribution is zero.
Correct answer: A. It remains unchanged
Explanation: Step 1: In Gauss's law, total flux depends on enclosed charge. Step 2: Changing radius still encloses the same point charge. Step 3: Therefore total flux remains unchanged.
Correct answer: A. It becomes one-fourth
Explanation: Step 1: Field of a point charge is inversely proportional to square of distance. Step 2: Doubling the sphere radius doubles distance from the charge. Step 3: Hence field on the surface becomes one-fourth.
Correct answer: A. Because field decreases but surface area increases and total result remains same
Explanation: Step 1: As distance increases, field of a point charge decreases. Step 2: At the same time, sphere area increases. Step 3: Since enclosed charge is same, total flux remains same by Gauss's law.
Correct answer: A. Because electric field is perpendicular to the area vector of those ends
Explanation: Step 1: Field of a line charge is radial. Step 2: Area vectors of the flat ends of the cylinder are along the axis. Step 3: These directions are perpendicular, so flux through the ends is zero.
Correct answer: A. Because electric field is parallel to the side surface
Explanation: Step 1: Field of an infinite sheet is perpendicular to the sheet. Step 2: For the side surface of the pillbox, the field does not cross the surface. Step 3: Hence flux through the side surface is zero.
Correct answer: A. Because field passes through the two flat faces of the pillbox
Explanation: Step 1: A pillbox Gaussian surface is taken across the sheet. Step 2: Field is perpendicular to the sheet on both sides. Step 3: Thus both flat faces contribute to flux.
Correct answer: A. Electric field is zero everywhere on the surface
Explanation: Step 1: Zero total flux directly means net enclosed charge is zero. Step 2: Field may still exist at different points on the surface. Step 3: Therefore zero field everywhere is not always correct.
Correct answer: A. Positive
Explanation: Step 1: Lines leaving the surface give positive outward flux. Step 2: If more lines leave than enter, total flux is positive. Step 3: By Gauss's law, net enclosed charge is positive.
Correct answer: A. Negative
Explanation: Step 1: Lines entering the surface give negative contribution to outward flux. Step 2: If this contribution is larger, total flux is negative. Step 3: Hence net enclosed charge is negative.
