Gauss's Law and its Applications

The governing principle is Gauss's law, Φ = Q_enclosed/ε₀. It is the algebraic, or net, sum of all charges inside the chosen closed surface that determines the total flux. External charges can alter the electric field at individual points on the surface, but their net contribution to total closed-surface flux is zero. Therefore option A is correct; the other choices ignore charge sign, magnitude, or location improperly.

Gauss's law gives the total flux as Φ = Q_enclosed/ε₀. Substituting Q_enclosed = 0 gives Φ = 0/ε₀ = 0, so option A is correct. This conclusion concerns the algebraic total over the entire closed surface. It does not necessarily mean that the electric field is zero everywhere; fields from charges inside and outside may still exist and cancel in the net flux.

A Gaussian surface is an imaginary, mathematically chosen closed surface used to calculate electric flux and apply Gauss's law. It is selected to fit the symmetry of the charge distribution, such as a sphere for spherical symmetry or a cylinder for cylindrical symmetry. Therefore option A is correct. It need not be a physical metal object, an isolated line, or an instrument for measuring current.

The defining mathematical requirement is that a Gaussian surface be closed, because Gauss's law evaluates the net flux through a complete boundary: ∮E·dA = Q_enclosed/ε₀. It may be spherical, cylindrical, or another shape, and it need not be metallic or planar. Thus option A is correct. An open line or open surface cannot enclose charge and cannot provide the required complete flux integral.

For every small element of a closed surface, the area vector is defined perpendicular to the element and directed outward by convention. This outward normal fixes the sign in the flux expression Φ = ∮E·dA: an outward-pointing field contributes positively, while an inward-pointing field contributes negatively. Hence option A is correct; the vector is not tangential, always inward, or random.

Electric flux through a surface element is dΦ = E·dA = E dA cosθ. For a closed surface, dA points outward. When the electric field leaves the surface, it points in the same direction as the outward area vector, so θ is less than 90° and the contribution is positive. Therefore option A is correct; entering field gives negative flux, while zero occurs only for perpendicular field or cancellation.

The governing relation is dΦ = E·dA. On a closed surface, the area vector dA is directed outward. An electric field entering the surface points opposite to dA, so the angle between them is 180° and cos180° = −1; therefore the local flux contribution is negative. Option A is correct. It is not necessarily maximum or zero, and positive flux corresponds to an outward field.

Gauss’s law states that the net electric flux through a closed surface is Φ = q_enclosed/ε₀. Since the enclosed charge is positive, q_enclosed > 0, so the total flux is positive. Equivalently, more electric field emerges outward than enters inward. The answer is A; a negative charge would give negative flux, while zero flux occurs only when the net enclosed charge is zero.

By Gauss’s law, the net flux through a closed surface is Φ = q_enclosed/ε₀. For a negative enclosed charge, q_enclosed is less than zero, so the total flux is negative. Field lines enter the surface overall rather than leaving it. Option A is correct; positive flux corresponds to a positive net enclosed charge, and zero flux requires zero net enclosed charge.

Gauss’s law gives the net closed-surface flux as Φ = q_enclosed/ε₀, so only the net charge inside the surface determines the total flux. An external charge can produce an electric field at different points on the surface, but its entering and leaving contributions cancel in the integral. Therefore its net contribution is zero, making A correct.

Gauss’s law is universally valid, but it becomes especially useful for finding the field when the charge distribution has high symmetry. Spherical, cylindrical, or planar symmetry allows a Gaussian surface on which the field magnitude is constant or has a simple direction. Then the flux integral is easy to evaluate. Thus A is correct; lack of symmetry usually makes the integral difficult.

The electric field of an isolated point charge has spherical symmetry: at equal distance from the charge, the field magnitude is the same and its direction is radial. A spherical Gaussian surface centered on the charge therefore has constant field magnitude over its area, making Φ = E(4πr²) straightforward. Hence A is correct; the other choices do not match the symmetry or are not closed Gaussian surfaces.

An infinitely long uniformly charged line has cylindrical symmetry. Choose a closed cylinder whose axis coincides with the line charge. The electric field is radial and has the same magnitude at every point on the curved surface at a fixed radius; the flux through the two flat ends is zero because the field is parallel to those faces. Therefore A is correct.

An infinite charged plane has planar symmetry, so the electric field is perpendicular to the sheet and has equal magnitude on both sides. A short cylindrical pillbox crossing the sheet captures this symmetry. The curved side contributes no flux because the field is parallel to it, while flux passes through the two flat faces. Thus A is the appropriate Gaussian surface.

For a point charge, choose a spherical Gaussian surface of radius r. Gauss’s law gives E(4πr²) = q/ε₀, so E = q/(4πε₀r²). Thus, for fixed charge, the field decreases as 1/r². If the distance doubles, the field becomes one-fourth. Therefore A is correct; the other options contradict the spherical-area dependence.

In electrostatic equilibrium, free charges in a conductor move until the electric field within the conducting material becomes zero. Any nonzero internal field would continue to drive charge motion. A Gaussian surface entirely inside the conductor encloses no net charge and has zero flux, consistent with E = 0. Therefore A is correct; the outward field exists outside the sphere, not inside the conductor.

For an ideal infinite plane sheet with surface charge density σ, choose a cylindrical Gaussian pillbox crossing the sheet. By symmetry, the field has equal magnitude on both flat faces and is perpendicular to them. Gauss’s law gives 2EA = σA/ε₀, so E = σ/(2ε₀). The area cancels, and no distance appears; hence the field is independent of distance. Option A is correct, whereas inverse-square behavior belongs to a point charge.

For an infinitely long line charge with linear charge density λ, use a coaxial cylindrical Gaussian surface of radius r and length L. The flux is E(2πrL), while the enclosed charge is λL. Gauss’s law gives E(2πrL) = λL/ε₀, or E = λ/(2πε₀r). Thus the field magnitude varies as 1/r and decreases inversely with distance. Option A is correct; distance independence applies to an ideal infinite plane sheet, not a line charge.

Gauss’s law is written as Φ_E = Q_enclosed/ε₀ in vacuum, where Φ_E is the total electric flux through a closed surface and Q_enclosed is the net charge inside it. The permittivity ε₀ sets the proportionality between enclosed charge and flux and reflects the electrical property of vacuum. Therefore option A is correct. It is not a quantity for mass, temperature, or current direction, so B, C, and D are unrelated.

Gauss’s law states that the total electric flux through any closed Gaussian surface is Φ_E = Q_enclosed/ε₀. If the enclosed net charge changes from Q to 2Q while the medium remains vacuum, the flux changes from Q/ε₀ to 2Q/ε₀. Therefore it doubles, so option A is correct. The result does not depend on the shape or size of the Gaussian surface, provided the enclosed net charge is the quantity being changed.

Electric flux through a surface is Φ_E = ∮ E·dA. If the magnitude of E is constant over the relevant Gaussian surface and its angle with the area element is also fixed by symmetry, E can be taken outside the integral; for a perpendicular field, Φ_E = E∮dA = EA. This converts a difficult surface integral into a simple product. Hence A is correct; uniformity does not make charge zero or remove field direction.

A Gaussian surface is an imaginary closed surface selected according to the symmetry of the charge distribution. The purpose is not to create a physical object but to make the electric field have a simple magnitude or direction on suitable parts of the surface. Then the flux integral becomes manageable and Gauss’s law can yield the field efficiently. Thus option A is correct. The other choices describe actions unrelated to the mathematical construction of a Gaussian surface.

The flux through a small surface element is dΦ = E·dA = EA cos θ, where dA is directed normal to the surface. If the electric field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos 90° = 0. Therefore dΦ = 0 for that part. Option A is correct; maximum magnitude occurs when the field is normal to the surface, not parallel to it.

For a surface element, dΦ = EA cos θ, with θ measured between the electric field and the outward area vector. When the field is perpendicular to the surface, it is parallel or antiparallel to the area vector, so |cos θ| = 1. Consequently, the flux magnitude is maximum, although its sign is positive for outward flow and negative for inward flow. Therefore A is correct; zero flux occurs for a field parallel to the surface.

Spherical symmetry means that the physical situation is unchanged under rotations about a central point. Therefore the electric field cannot prefer a tangential direction; it must lie along the radius at every point. For a positive central charge or positive enclosed charge, the field points radially outward, while for a negative charge it points radially inward. Thus option A is correct. The other directions would break spherical symmetry or are not defined by the charge distribution.