Gauss's Law and its Applications

Gauss’s law states that the net electric flux through any closed surface is Φ = Q_enclosed/ε₀. Therefore, it depends on the algebraic sum of charges enclosed by the surface, not on its colour, material, thickness, or detailed shape. Charges outside may produce electric field on the surface, but their total contribution to closed-surface flux is zero. Hence option A is correct.

By Gauss’s law, the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Substituting Q_enclosed = 0 gives Φ = 0. This conclusion concerns the net flux, not necessarily the electric field at every point; fields from internal opposite charges or external charges may still exist and cancel in total flux. Therefore option A is correct.

For a positive enclosed charge q, Gauss’s law gives Φ = q/ε₀. Since q and ε₀ are positive, the flux is positive. Equivalently, electric field lines emerge from a positive charge and leave the closed surface, producing outward flux. The surface’s shape does not change this sign. Thus option A is correct; negative or zero flux would require a negative or zero net enclosed charge.

For a negative enclosed charge q, Gauss’s law gives Φ = q/ε₀. Because q is negative while ε₀ is positive, the net flux is negative. In terms of field lines, they enter the closed surface toward the negative charge, so inward flux is counted as negative. The result is finite for a finite charge; it is not automatically zero or infinite. Hence option A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Field lines produced by an external charge may enter the surface and must leave it again, so the inward and outward contributions cancel in the net total. Since the charge is not enclosed, Q_enclosed from it is zero and its net flux contribution is zero. Thus option A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. The positive and negative charges have equal magnitudes, so their algebraic sum is zero: Q_enclosed = +q − q = 0. Therefore Φ = 0, making option A correct. Electric fields may still exist at points inside the surface, but their total outward flux cancels.

Gauss’s law gives Φ = Q_enclosed/ε₀ for any closed surface. Here, only the positive charge +q lies inside, while the negative charge is outside and is not included in Q_enclosed. Therefore Φ = +q/ε₀, which is positive. Option A is correct; the external charge can affect the field on the surface but cannot change the net flux determined by enclosed charge.

For a closed surface, Gauss’s law states Φ = Q_enclosed/ε₀. In this case the enclosed charge is only −q, because the positive member of the dipole lies outside. Hence Φ = −q/ε₀, so the net flux is negative and option A is correct. The outside positive charge may contribute to the local electric field, but it contributes no net enclosed charge.

Electric flux measures the electric field passing through a surface, mathematically represented by the surface integral of the electric field. Gauss’s law directly relates the net electric flux through a closed surface to the charge enclosed: Φ = Qₑₙc/ε₀. Therefore, Gauss’s law is the correct answer. Ohm’s law concerns voltage and current, Hooke’s law concerns elasticity, and Newton’s law of cooling concerns heat transfer.

Gauss’s law states that the total electric flux through any closed surface is Φ_E = Q_enclosed/ε₀. Thus the determining quantity is the algebraic, or net, charge enclosed by the surface; external charges may affect the local field but contribute zero net flux through the closed surface. Therefore option A is correct, and colour, thickness, and temperature are irrelevant in this ideal electrostatic relation.

By Gauss’s law, the total electric flux through a closed surface is Φ_E = Q_net/ε₀. Substituting Q_net = 0 gives Φ_E = 0. Therefore option A is correct. This does not require the electric field to be zero at every point on the surface; fields from internal charges can cancel in net flux, and external charges can produce local fields without changing the total closed-surface flux.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. A positive charge makes Q_enclosed positive, so the outward flux is positive. Field lines emerge from a positive charge, which also gives the same sign interpretation. Negative flux would correspond to net inward field lines, while zero flux would require zero net enclosed charge.

By Gauss’s law, the net flux through a closed surface is Φ = Q_enclosed/ε₀. For a negative enclosed charge, Q_enclosed is negative, so the outward flux is negative. Physically, field lines point toward a negative charge and enter the surface, giving negative outward flux. Positive flux would require a net positive enclosed charge; zero would require zero net charge.

For a closed surface, outward electric flux is taken as positive and inward flux as negative. If equal numbers of field lines enter and leave, their signed contributions cancel: Φnet = Φout − Φin = 0. Gauss’s law gives the same result when the net enclosed charge is zero. Therefore option A is correct; the sign cannot be positive or negative after exact cancellation.

Gauss’s law states that the net electric flux through a closed surface is Φnet = Qenclosed/ε0. If the algebraic net charge enclosed is zero, then Φnet = 0, regardless of the surface’s shape or area. External charges may produce electric field on the surface, but their entering and leaving contributions cancel. Thus option A is correct; the flux is not determined by area alone.

Gauss’s law gives Φnet = Qenclosed/ε0. A positive charge means Qenclosed > 0, and ε0 is positive, so the total outward flux is positive. The exact value depends on the charge magnitude, but its sign is fixed. Option A is therefore correct. Negative flux would correspond to net negative enclosed charge, while zero would require zero net enclosed charge.

By Gauss’s law, the total flux through a closed surface is Φnet = Qenclosed/ε0. For an enclosed negative charge, Qenclosed is negative while ε0 is positive, so the net outward flux is negative. Therefore option A is correct. A positive sign would require a net positive enclosed charge, zero would require zero net charge, and “always maximum” is not a valid consequence of charge sign.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Thus, for the same medium and unchanged ε₀, flux is directly proportional to the net enclosed charge. If Q_enclosed changes to 2Q, then Φ changes to 2Φ. Therefore option A is correct. The surface’s shape or the charge outside it does not alter this total-flux relation.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. If the surface encloses only the positive charge +q of the dipole, then Q_enclosed = +q, so Φ = +q/ε₀ and its sign is positive. The negative charge outside the surface contributes no net enclosed charge, although its field may pass through parts of the surface. Therefore, option A is correct; zero applies only when the total enclosed charge is zero.

By Gauss’s law, the net flux through a closed surface is Φ = Q_enclosed/ε₀. When the surface contains only the negative dipole charge −q, the enclosed charge is negative, giving Φ = −q/ε₀. Thus the net flux has a negative sign. The positive charge outside the surface can affect the local electric field and the flux through individual portions, but it does not change the net enclosed charge. Hence option A is correct, while zero would require equal positive and negative charge inside.

Gauss’s law gives the total outward flux through a closed surface as Phi = Q_enclosed/epsilon-naught. Since the enclosed net charge is positive, the quotient is positive. Equivalently, more electric field emerges outward from positive charge than enters the surface. Thus option A is correct. The magnitude would require the numerical charge, but its sign is determined immediately; zero would require zero net enclosed charge.

For a closed surface, Gauss’s law states Phi_out = Q_enclosed/epsilon-naught. A negative enclosed charge therefore produces a negative total outward flux. Physically, field lines tend to enter the surface toward the negative charge, opposite to the outward area direction. Hence option A is correct. A positive or zero flux would require positive or zero net enclosed charge, respectively; the magnitude cannot be called maximum without numerical information.

Gauss’s law gives the total electric flux through a closed surface as Φ = Q_enclosed/ε₀. Since the algebraic sum of the two positive charges and the negative charge is stated to be positive, Q_enclosed > 0 and the net outward flux is positive. The detailed shape of the surface can change the field distribution, but not the total flux. Thus A is correct; B and C contradict the sign of enclosed charge, and D confuses local distribution with total flux.

Gauss’s law gives Φ = Q_enclosed/ε₀ for a closed surface. With the same physical constant ε₀, flux is directly proportional to the net enclosed charge. If Q_enclosed changes to 3Q_enclosed, then Φ changes to 3Φ. Therefore option A is correct. Nine times would require both charge-related factors to be doubled, and the other choices contradict direct proportionality.

Gauss's law states that the net electric flux through a closed surface equals the net charge enclosed divided by the permittivity of free space: Φ = Q_enclosed/ε₀. Hence option A correctly names the two related quantities. The law is not a relation between an open-surface length and current, mechanical quantities, or thermal variables. The word “closed” is essential because the theorem concerns a complete closed boundary.