Gauss's Law and its Applications

The governing concept is cylindrical symmetry. For an infinitely long uniformly charged line, the physical situation is unchanged by rotation around the line or translation along it. Therefore no preferred tangential or axial direction exists, and the electric field must be radial, perpendicular to the line. It points outward for a positive line charge and inward for a negative one, so option A is correct; the other directions violate the symmetry.

The governing idea is planar symmetry. An infinite uniformly charged sheet looks identical at every position within its plane, so no sideways direction is preferred. Any electric-field components parallel to the sheet cancel by symmetry, leaving only the normal component. The field is perpendicular to the sheet, directed away from a positive sheet and toward a negative sheet. Thus option A is correct; the other choices contradict this symmetry.

Gauss's law is expressed as the closed-surface integral ∮E·dA = Q_enclosed/ε₀. It therefore directly relates the net electric flux through a complete closed surface to the algebraic charge enclosed by that surface. An open surface can have flux, but Gauss's law in this form does not determine it from enclosed charge alone. Hence option A is correct, while the other choices are not the required surface type.

Gauss's law states that the net flux through a closed surface is Φ = Q_net/ε₀. If the enclosed charges are +q and −q, their algebraic sum is Q_net = +q − q = 0. Consequently Φ = 0, regardless of the individual fields at different points on the surface. Option A is correct. A positive or negative answer would ignore charge cancellation, and “double” has no basis.

The governing relation is Φ = Q_enclosed/ε₀, where ε₀ is positive. Therefore the sign of total electric flux is exactly the sign of the net charge enclosed by the closed surface. A negative enclosed charge gives negative flux, meaning that the net field is inward relative to the chosen outward normal. Surface shape does not determine the sign. Thus option A is correct; positive or zero charge gives positive or zero flux.

For a closed surface, Gauss's law gives Φ = Q_enclosed/ε₀. Since ε₀ is positive, the total flux is positive exactly when the algebraic net charge inside the surface is positive. Charges outside may alter the local electric field on the surface, but their net contribution to closed-surface flux is zero. A zero enclosed charge gives zero flux, and an open surface is outside the stated law. Therefore option A is correct.

Planar symmetry determines the direction of the field. Because an infinite sheet has no preferred direction within its plane, all components parallel to the sheet cancel. The remaining field is normal to the sheet on both sides. For a positively charged sheet it points away on both sides; for a negatively charged sheet it points toward the sheet on both sides. Thus option A is correct, while the other choices violate symmetry or describe the wrong geometry.

The governing concept is electric flux, Φ = ∮E·dA, which requires integration over the complete closed Gaussian surface. Each small area contributes E·dA, with the sign determined by the angle between the electric field and outward area vector. Adding all these signed contributions gives total flux; therefore option A is correct. A single point, external charge count, or surface colour cannot determine it.

Gauss's law states that the net flux through a closed surface is Φ = Q_enclosed/ε₀. External charges may produce field lines that enter and leave the Gaussian surface, so individual parts can have nonzero flux. However, when no charge is enclosed, the inward and outward contributions cancel algebraically, giving net flux zero. Thus option A is correct, while the other choices incorrectly claim a fixed nonzero or infinite value.

For any closed Gaussian surface, Gauss's law gives Φ = Q_enclosed/ε₀. Therefore, if the net charge enclosed remains unchanged, deforming or reshaping the surface does not alter the total flux, although the field and flux density at individual points may change. Option A is correct. An open surface is not covered by the closed-surface form of the law, and changing external charge does not generally preserve the same situation.

Gauss's law is universally valid, but it becomes a practical method for finding E only when the charge distribution has sufficient symmetry. Spherical, cylindrical, and planar symmetry can make the field magnitude constant on a suitable Gaussian surface, allowing Φ = EA or an equivalent simple expression. Thus option A is correct. Irregular distributions generally require direct integration; magnetic and gravitational fields are not the requested electric-field application.

For a sufficiently long line charge, the electric field is radial outward from the line. The area vectors of the two flat end caps of a coaxial cylinder point along the line's axis, so the field is tangent to those caps and E·dA = 0 there. Flux through the curved surface is nonzero because its outward area vectors are radial. Hence option A is correct.

An infinite uniformly charged plane produces an electric field perpendicular to the plane. Choose a pillbox with its axis perpendicular to the sheet. On the cylindrical curved wall, the electric field is along the wall, while the outward area vector is radial and perpendicular to it. Thus E·dA = 0 at every point of the curved surface, so its flux is zero. Option A is correct; the field and area are not zero.

This conclusion requires the point charge to be at the centre of the spherical Gaussian surface. Every point on the sphere is then at the same distance r from the charge, so Coulomb's law gives the same magnitude, E = (1/4πε₀)|q|/r², everywhere. The field points radially outward for positive q and inward for negative q. Therefore option A is correct; it is not tangential or necessarily zero.

Electric flux is defined by Φ = ∫E·dA. The SI unit of electric field is N/C and the unit of area is m², so the flux unit is (N/C)m² = N m²/C. Equivalently, it can be written as V m. Therefore option A is correct. N/m is not the flux unit, C/s is current, and J/m has the dimensions of force rather than electric flux.

Gauss's law connects the net electric flux through a closed surface with the enclosed charge: ∮E·dA = Q_enclosed/ε₀. For highly symmetric distributions, such as a spherical charge, an infinite line, or an infinite plane sheet, the field has a simple magnitude and direction, so the law allows E to be found efficiently. Hence option A is correct; the other choices belong to unrelated physical measurements or processes.

Gauss's law gives Φ_net = Q_enclosed/ε₀. If the enclosed net charge is negative, the total flux is negative, meaning the inward contribution is greater than the outward contribution when counted with the outward normal convention. Field lines are conventionally directed toward negative charge, so the surface has a net entering tendency. Thus option A is correct; equal entry and exit would imply zero net flux.

Gauss’s law states that the net electric flux through a closed surface is proportional to the net charge enclosed: Φ = Qenclosed/ε₀. A positive enclosed charge gives positive flux, meaning the outward component of the electric field is greater overall. Option A is correct. Option B corresponds to net negative charge, while option C is not implied; option D is true only when the net enclosed charge is zero.

The governing idea is Gauss’s law, Φ = Qenclosed/ε₀. Before calculating flux or field, identify the closed Gaussian surface and determine the algebraic sum of all charges inside it. This information controls the net flux. Option A is therefore correct. Surface colour, total mass, and temperature do not determine electric flux in this law, so options B, C, and D are irrelevant distractors.

Gauss’s law governs electric flux through a closed surface. Mathematically, Φ = Q enclosed / ε₀, so the total flux is related to the net charge enclosed by that surface. Therefore option A is correct. Current belongs to circuit electricity, temperature to thermal physics, and mass is unrelated to this electrostatic law.

Gauss’s law states that the net electric flux through a closed surface equals the net charge enclosed divided by ε₀: Φ = Qenc/ε₀. The surface must be closed so that “inside” charge is well defined. Hence option A is correct. An open surface can have flux, but it cannot directly be used for the enclosed-charge form of Gauss’s law.

A Gaussian surface is an imaginary, closed mathematical surface selected to calculate electric flux conveniently. Its shape is chosen according to the symmetry of the charge distribution, such as a sphere, cylinder, or pillbox. Thus option A is correct. It need not be metallic, a mere line, or always spherical like Earth.

Gauss’s law gives the relation Φ = Qenc/ε₀. Therefore the total electric flux through a closed surface depends on the net charge enclosed by it, not on its colour, thickness, or the outside temperature. Option A is correct. External charges may affect the field at individual points, but their net contribution to closed-surface flux is zero.

Gauss’s law is Φ = Qenc/ε₀. Substituting Qenc = 0 gives Φ = 0/ε₀ = 0, so option A is correct. This conclusion concerns net flux: positive and negative contributions may cancel. It does not necessarily mean the electric field is zero at every point on or inside the surface.

Using Gauss’s law, Φ = Qenc/ε₀. Since ε₀ is positive and the enclosed charge Qenc is positive, the total flux must also be positive. Thus option A is correct. Field lines from a positive charge leave the closed surface overall. Negative flux would correspond to net negative enclosed charge, while zero requires zero net enclosed charge.