Gauss's Law and its Applications

Gauss’s law gives Φ = Qenc/ε₀. With a negative enclosed charge and a positive constant ε₀, the quotient is negative, so option A is correct. Equivalently, electric field lines point toward a negative charge and enter the closed surface overall. Positive or zero flux would require different net enclosed charge conditions.

Gauss’s law counts only net charge enclosed by the closed surface: Φ = Qenc/ε₀. An external charge can create an electric field on the surface, but field lines entering the surface are balanced by lines leaving it, so its net flux contribution is zero. Therefore option A is correct; the other choices confuse local field with net flux.

Gauss’s law states that the net electric flux through any closed surface is Φ = Q_enclosed/ε₀. Therefore, changing the shape of the Gaussian surface does not alter the total flux as long as the net charge enclosed by it remains unchanged. The colour, temperature, or mass outside the surface does not determine the net electric flux. Hence option A is correct.

Gauss’s law always gives the total electric flux through a closed surface, but it gives the electric-field magnitude directly only when the charge distribution has sufficient symmetry. Spherical, cylindrical, or planar symmetry can make the field constant or its direction simple on a chosen Gaussian surface. Without symmetry, the surface integral is usually difficult. Thus option A is correct.

A point charge produces an electric field with spherical symmetry: at equal distances from the charge, the field has the same magnitude and is directed radially. A spherical Gaussian surface centred on the charge therefore has constant field magnitude and a simple angle with the area vector, making the flux calculation easy. A cube or open plane does not match this symmetry. Hence option A is correct.

An infinitely long, uniformly charged straight wire has cylindrical symmetry. A cylindrical Gaussian surface is chosen with the wire along its axis. The electric field is radial and has the same magnitude at every point on the curved surface, while the flux through the end caps is zero because the field is parallel to those caps. Therefore option A is the most convenient choice.

An infinite uniformly charged plane sheet has planar symmetry, and the electric field is perpendicular to the sheet on both sides. A short cylindrical Gaussian surface, called a pillbox, is placed with its flat faces parallel to the sheet and its axis normal to it. Flux passes through the two flat faces, while the curved side contributes zero. Thus option A is correct.

Consider a spherical Gaussian surface of radius smaller than the radius of the uniformly charged shell. This inner Gaussian surface encloses no charge, so Gauss’s law gives ∮E·dA = Q_enclosed/ε₀ = 0. By spherical symmetry, the field would have the same magnitude over the surface; therefore its magnitude must be zero. Thus the field is zero everywhere inside the shell, making option A correct.

For a uniformly charged spherical shell, take a spherical Gaussian surface outside the shell. It encloses the entire charge Q, and spherical symmetry makes the electric field constant on that surface. Gauss’s law gives E(4πr²) = Q/ε₀, or E = (1/4πε₀)Q/r². This is exactly the field of a point charge Q placed at the centre, so option A is correct.

For an infinite uniformly charged plane sheet, translational symmetry makes the electric field have the same magnitude at every point at a given side of the sheet. Applying Gauss’s law to a pillbox Gaussian surface gives 2EA = sigma A divided by epsilon_0, so E = sigma/(2epsilon_0). Distance does not appear in this expression; therefore option A is correct. Options B and C describe distance dependence for other charge geometries, not an infinite sheet.

A long uniformly charged straight wire has cylindrical symmetry. Choose a coaxial cylindrical Gaussian surface: the electric field must point perpendicular to the wire along the radial direction, because no radial direction is preferred around the axis. For positive linear charge density, field lines point outward; for negative density, they point inward. Thus option A is correct. The field is not along the wire, circular, or necessarily zero.

Gauss’s law, integral E dot dA = Q_enclosed/epsilon_0, is valid for every charge distribution, but it gives the electric field easily only when the distribution has sufficient symmetry. Spherical, cylindrical, or planar symmetry can make the field constant on a suitable Gaussian surface and simplify the flux integral. Therefore option A is correct; irregular or unknown distributions usually require direct integration or another method.

Electric flux is defined as the surface integral of electric field: Phi_E = integral E dot dA. The SI unit of electric field is N/C and the unit of area is m^2. Multiplying them gives N m^2/C, which is also equivalent to V m. Hence option A is correct. N/m is not the flux unit, C/m represents a different quantity, and J/s is the unit of power.

In vacuum, Gauss’s law is written as Phi_E = Q_enclosed/epsilon_0, where epsilon_0 is the permittivity of free space. This constant describes how electric fields and electric flux are related to charge in vacuum. It is an electrical property, not a mechanical, thermal, or optical property. Therefore option A is correct; the density of water, air temperature, and metal colour are unrelated to this law’s constant.

Gauss’s law states that the net electric flux through a closed surface is Phi_E = Q_enclosed/epsilon_0. If the enclosing surface and the medium remain unchanged, flux is directly proportional to the net enclosed charge. Replacing Q_enclosed by 2Q_enclosed gives Phi'_E = 2Phi_E. Therefore option A is correct. The result concerns net enclosed charge, not merely the number of individual charges.

The integral form of Gauss’s law is Phi_E = Q_enclosed/epsilon_0. With the same closed surface and the same medium, epsilon_0 is fixed, so total flux changes in the same ratio as net enclosed charge. If Q_enclosed becomes Q_enclosed/2, then Phi_E becomes Phi_E/2. Thus option A is correct. Doubling or quadrupling would require increasing the charge, while an unchanged flux would contradict the law.

Let the enclosed charges be +q and -q. Their algebraic sum is Q_net = +q - q = 0. Gauss’s law then gives Phi_E = Q_net/epsilon_0 = 0, regardless of the individual electric fields inside or outside the surface. Therefore option A is correct. The fields at particular points need not vanish, but their total outward flux cancels. Positive or negative flux would require a nonzero net enclosed charge.

A Gaussian surface is an imaginary closed surface selected to make the flux integral simple. Its shape should match the symmetry of the charge distribution: spherical for spherical symmetry, cylindrical for a long line charge, and pillbox-shaped for an infinite plane. Then the field may be constant or have a simple direction on the surface. Hence option A is correct; colour, naming, and paper size have no physical role.

Gauss’s law is universally valid for any closed surface and any charge distribution. However, it determines the field conveniently only when symmetry lets us know the field’s direction and magnitude over parts of a Gaussian surface. Without useful symmetry, the field varies over the surface, so the flux integral cannot be simplified to E times area; direct integration may be needed. Thus option A is correct.

For an infinitely long line charge with linear charge density λ, Gauss’s law gives E(2πrL) = λL/ε₀, so E = λ/(2πε₀r). The field is therefore inversely proportional to the distance r from the line. As r increases, the field magnitude decreases. Hence option A is correct; it does not increase, remain constant, or become zero at every distance.

The governing concept is symmetry. An infinite uniformly charged plane looks identical in every direction along its surface, so no tangential direction can be preferred. Consequently, the electric field must be normal, or perpendicular, to the sheet on both sides. A cylindrical Gaussian pillbox confirms this result because flux passes through its two flat faces. Therefore option A is correct; the field is not parallel, circular, or necessarily zero.

Electric field lines originate on positive charge and point away from it. Because an infinite sheet has identical surroundings on both sides, its field is perpendicular to the sheet and directed outward on each side. A Gaussian pillbox gives equal outward flux through both faces, consistent with this direction. Thus option A is correct; inward directions apply to a negative sheet, while one-sided choices violate the sheet’s symmetry.

Electric field lines terminate on negative charge, so the field direction is toward a negatively charged source. For an infinite plane sheet, symmetry requires the field to be perpendicular to the sheet, and therefore it points inward toward the sheet on both sides. Option A is correct. Outward directions describe a positive sheet, while upward-only or downward-only directions ignore the equal geometry on the two sides.

In electrostatic equilibrium, free charges in a conductor redistribute until the electric field inside the conducting material is zero. A Gaussian surface wholly inside then has zero electric flux, and Gauss’s law, Φ = Qenc/ε₀, implies zero net enclosed charge. This explains option A. Temperature, colour, and mass are not properties established by Gauss’s law, so the remaining choices are irrelevant.

Gauss’s law states that the total electric flux through a closed surface is Φ = Qenc/ε₀. Since the electric field is zero everywhere on the small Gaussian surface inside the conductor, the flux integral is zero. Therefore Qenc must also be zero. Option A is correct. A positive, negative, or infinite enclosed charge would require nonzero flux and would contradict the stated zero field.