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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. They are equal
Explanation: Step 1: Excess charge on a conducting sphere resides on the surface. Step 2: For outside points, spherical symmetry is maintained. Step 3: Hence the outside field is same as if the total charge were concentrated at the centre.
Correct answer: A. Effects from all directions cancel due to symmetry
Explanation: Step 1: At the centre, all parts of the sphere surround the point symmetrically. Step 2: Field contribution in every direction has an equal opposite contribution. Step 3: Therefore the net electric field at the centre is zero by symmetry.
Correct answer: A. Proportional to cube of radius
Explanation: Step 1: For uniform volume charge density, charge is proportional to volume. Step 2: Volume of a smaller spherical Gaussian surface is proportional to cube of radius. Step 3: Hence enclosed charge increases as cube of radius.
Correct answer: A. Because enclosed charge grows as cube of radius while Gaussian area grows as square of radius
Explanation: Step 1: Inside the sphere, enclosed charge increases as cube of radius. Step 2: Spherical area increases as square of radius. Step 3: Their ratio is proportional to radius, so the field increases linearly.
Correct answer: A. In a conductor excess charge moves to the surface, while in an insulator it can remain in volume
Explanation: Step 1: In a conductor, free charges move and settle on the surface. Step 2: In an insulator, charge can remain distributed throughout volume. Step 3: Therefore field is zero inside conductor but can vary inside insulator.
Correct answer: A. It does not change
Explanation: Step 1: For a line charge, enclosed charge depends on the length of line inside the cylinder. Step 2: Changing radius does not change the enclosed length of the line. Step 3: Therefore enclosed charge remains unchanged if length is same.
Correct answer: A. Double
Explanation: Step 1: In Gauss's law, total flux depends on enclosed charge. Step 2: Doubling cylinder length doubles the enclosed line charge. Step 3: Therefore total flux also doubles.
Correct answer: A. Because enclosed charge is decided by the area cut on the sheet, not by height
Explanation: Step 1: Charge on a plane sheet is spread on the surface. Step 2: Enclosed charge in a pillbox depends on the area cut on the sheet. Step 3: Changing height does not change that area, so total flux is unchanged.
Correct answer: A. The two fields are in the same direction and add
Explanation: Step 1: Field is away from a positive sheet and toward a negative sheet. Step 2: Between the two sheets, these directions are the same. Step 3: Hence the fields add in the region between them.
Correct answer: A. Zero
Explanation: Step 1: Field due to a positive sheet is away from the sheet on both sides. Step 2: Between two identical positive sheets, fields due to the two sheets are opposite. Step 3: Equal magnitudes cancel, so the field between them is zero.
Correct answer: A. It is larger because the two fields add
Explanation: Step 1: Positive sheets produce field away from themselves on both sides. Step 2: In an outer region of two sheets, the two fields point in the same direction. Step 3: Therefore they add there.
Correct answer: A. Zero
Explanation: Step 1: A Gaussian surface inside the shell encloses no charge. Step 2: By symmetry, total flux on that surface is zero. Step 3: Therefore electric field inside the shell is zero.
Correct answer: A. Inversely as square of distance
Explanation: Step 1: A Gaussian sphere outside the shell encloses the entire charge. Step 2: Outside, spherical symmetry makes it behave like a point charge at the centre. Step 3: Therefore field falls inversely as square of distance.
Correct answer: A. Electric field will be stronger
Explanation: Step 1: Electric field just outside a conductor surface is related to surface charge density. Step 2: Where surface charge is denser, field lines are also denser. Step 3: Hence the electric field just outside is stronger.
Correct answer: A. Surface charge density can be higher there
Explanation: Step 1: In an electrostatic conductor, excess charge stays on the surface. Step 2: Sharp parts can have higher charge density. Step 3: Higher charge density indicates stronger external electric field.
Correct answer: A. Because tangential component would move charges on the surface
Explanation: Step 1: In an electrostatic conductor, surface charges are at rest. Step 2: An oblique field has a component along the surface. Step 3: That component would move charges, so the field must be perpendicular to the surface.
Correct answer: A. Electric field inside the cavity is zero
Explanation: Step 1: Electric field inside conductor material is zero. Step 2: With no charge inside the empty cavity, field lines cannot persist there. Step 3: In electrostatic condition, the conductor shields the cavity from external fields.
Correct answer: A. Electric field inside conductor is zero
Explanation: Step 1: Inside conductor material in electrostatic condition, electric field is zero. Step 2: Zero field gives zero total flux through a closed surface. Step 3: By Gauss's law, net enclosed charge for such a surface is zero.
Correct answer: A. Net enclosed charge is zero, but local field can exist
Explanation: Step 1: A dipole has equal positive and negative charges. Step 2: Since net charge is zero, total closed flux is zero. Step 3: Still, electric field can exist at different points.
Correct answer: A. Flux due to only the enclosed positive charge
Explanation: Step 1: Total flux through a closed surface depends only on net enclosed charge. Step 2: Here only the positive charge is inside. Step 3: Therefore total flux is positive according to that enclosed positive charge.
Correct answer: A. Total flux will not change
Explanation: Step 1: A complete dipole inside has zero net enclosed charge. Step 2: The external charge is not counted inside the closed surface. Step 3: Hence total flux remains zero even though field on the surface may change.
Correct answer: A. Yes, but finding field simply will be difficult
Explanation: Step 1: Gauss's law is true for total flux through any closed surface. Step 2: If the field is not uniform, it cannot be easily taken out of the flux sum. Step 3: So the law is valid, but calculation may be difficult.
Correct answer: A. Symmetry makes field magnitude and direction simple on the Gaussian surface
Explanation: Step 1: Gauss's law is always true. Step 2: To find field, we need to write flux in a simple form. Step 3: Symmetry makes field uniform or perpendicular on chosen parts, making calculation easier.
Correct answer: A. Because field magnitude is not same everywhere on the cube surface
Explanation: Step 1: A cube is also a closed surface, so total flux can be found. Step 2: But distance from the point charge is not same everywhere on the cube surface. Step 3: Hence field cannot be treated as constant and taken out simply.
Correct answer: A. Because field magnitude is not same on the surface
Explanation: Step 1: Total flux remains fixed by the enclosed charge. Step 2: But an off-centre charge has different distances to different surface points. Step 3: Therefore field is not uniform and the simple multiplication method fails.
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