Gauss's Law and its Applications

Gauss’s law gives the total flux through any closed surface as Φ = Q_enclosed/ε₀. The result depends on the algebraic net charge enclosed, not on the shape, size, or detailed distribution of the surface, provided the same charge remains enclosed. If Q_enclosed is changed to 2Q_enclosed, then Φ changes to 2Φ. Therefore option A is correct. Shape may alter the local field distribution, but it does not alter the total flux under these conditions.

Gauss’s law gives the net flux through any closed surface as Φ = Q_enclosed/ε₀. If the enclosed net charge changes from Q to 3Q, the flux changes from Q/ε₀ to 3Q/ε₀, so it becomes three times. Changing the shape of the closed surface does not affect total flux, provided the enclosed net charge is the stated value; shape only changes the local field distribution.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Only the algebraic sum of charges enclosed by the surface appears in this expression. Charges outside can produce electric field lines through different parts of the surface, changing local flux density, but their total contribution to closed-surface flux is zero. Therefore a negative enclosed charge gives negative net flux, regardless of external positive charges.

A dipole consists of equal positive and negative charges, so its net charge is zero. Gauss’s law therefore gives Φ = Q_enclosed/ε₀ = 0 whenever both charges remain inside the closed surface. Changing the surface size may alter the electric field at individual points and the local flux density, but the integrated total flux is unchanged. Hence A is correct; B, C, and D incorrectly relate total flux to size or dipole moment.

The governing result is Gauss’s law: the net flux through a closed surface equals Q_enclosed/ε₀. With no enclosed charge, the net value is zero, even if external charges create electric field over parts of the surface. Positive contributions in some regions cancel negative contributions elsewhere. Therefore A is correct. B and C ignore cancellation, while D focuses on one local direction rather than the complete closed-surface integral.

For a closed surface, Gauss’s law states Φ_total = Q_enclosed/ε₀. The positive charge inside contributes to the net flux, whereas the equal negative charge outside is not enclosed. It can change the electric field at surface points, but its total flux contribution through the closed surface is zero. Therefore the total flux is positive and determined only by the inside charge, so A is correct; B and C incorrectly include the external charge.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Since ε₀ is positive, the sign of flux is the sign of the net enclosed charge. Negative total flux therefore means Q_enclosed is negative. The result does not require any particular external charge, so A is correct and D is unsupported; B and C have the wrong sign or magnitude conclusion.

For any closed surface, Gauss’s law states Φ = Q_enclosed/ε₀. The total flux depends only on the net charge enclosed, not directly on the area, shape, or size of the Gaussian surface. Doubling the area can redistribute the field over the surface and change local flux density, but the integral remains the same because Q_enclosed is unchanged. Hence A is correct.

Electric flux is defined using the outward area vector. Field lines leaving a closed surface contribute positive flux, whereas lines entering it contribute negative flux because the field is opposite to the outward normal. If more lines enter than leave, the algebraic total flux is negative. By Gauss’s law, Φ = Qenclosed/ε₀, so a negative flux implies a negative net enclosed charge. Therefore option A is correct.

Gauss’s law gives the relation ∮E·dA = Qenclosed/ε₀ for any closed surface. If the total flux is zero, then the algebraic sum of all enclosed charge must be zero. This does not require the electric field to vanish at every point; fields from positive and negative enclosed charges can cancel only in the integral. External charges may also be present, since they do not alter the net flux through the closed surface.

Gauss’s law states that the net electric flux through a closed surface is Φ = Qenclosed/ε0. Let each charge magnitude be q. The enclosed charge is 3q − 2q = +q, which is positive. Therefore the net outward flux is positive, regardless of the exact positions of the charges or the shape of the closed surface. Option B would require more negative charge, while option C would require zero net charge. Hence option A is correct.

For any closed surface, Gauss’s law gives Φ = Qenclosed/ε0. If the magnitude of each charge is q, the enclosed net charge is 2q − 4q = −2q. Since ε0 is positive, the total flux must be negative. The negative sign indicates that the algebraic inward contribution dominates the outward contribution. Flux is not determined by surface shape, and it would be zero only if the positive and negative enclosed charges cancelled exactly. Thus option A is correct.

Gauss’s law determines the total flux through a closed surface solely from the algebraic charge enclosed: Φ = Qenclosed/ε0. Moving the same charge from the centre toward the boundary changes the electric field’s distribution and may make the field stronger on some portions and weaker on others. However, the integrated flux over the entire closed surface remains unchanged as long as the charge does not cross the boundary. Therefore option A is correct.

Gauss’s law states that the net electric flux through any closed surface is Φ = Qenclosed/ε0. The result depends only on the algebraic net charge enclosed, not on the shape, size, or orientation of the Gaussian surface. If the enclosed charge changes from Q to 2Q, the flux changes from Q/ε0 to 2Q/ε0. Therefore the total flux doubles. Surface geometry can change the local field distribution, but it cannot change the total flux for the stated enclosed charge.

By Gauss’s law, net flux through a closed surface is Φ = Qnet/ε0. For net negative enclosed charge, Φ is negative. With the outward area-vector convention, outward crossings contribute positive flux and inward crossings contribute negative flux. A negative total therefore means the inward contribution is greater than the outward contribution, or equivalently the surface has a net inward tendency of field lines. The field need not vanish, so D is incorrect and A is correct.

A complete electric dipole contains charges +q and −q, so its net charge is zero. Gauss’s law gives the total flux through any closed surface as Φ = Qenclosed/ε0. Enlarging the surface changes the field and flux density at different parts of the surface, but it still encloses the same total charge, +q − q = 0. Therefore the net flux remains zero. The field itself does not disappear, and the surface area certainly does not become zero, so A is correct.

For a closed surface, the outward area vector defines positive flux. If the electric field has an outward component everywhere and is non-zero, then E · dA is positive throughout, so the total flux is positive. Gauss’s law, Φ = Qenclosed/ε0, then requires Qenclosed > 0. Thus the net enclosed charge is positive. This does not prove that only one positive charge or a particular arrangement is present; it identifies only the sign of the total charge.

The area vector of a closed surface is defined outward. If the electric field points inward everywhere, it is opposite to dA, so E · dA is negative at every point and the total electric flux is negative. By Gauss’s law, Φ = Qenclosed/ε0, a negative flux means Qenclosed is negative. Therefore the net enclosed charge is negative. This conclusion concerns the algebraic total charge, not necessarily a single negative charge.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. A complete dipole contains +q and −q, so its enclosed net charge is zero and the total flux is zero. The external positive charge can create electric field through the surface but contributes no net closed flux because it is not enclosed. Therefore option A is correct; dipole moment and surface shape do not determine the net value.

Gauss’s law gives the total flux through a closed surface as Φ = Q_net/ε₀, where Q_net is the algebraic sum of all enclosed charges. Thus positive charges are added and negative charges are subtracted. Since the magnitudes are unequal, the net charge is generally nonzero, so the flux is generally nonzero as well. Option A is correct; neither the larger charge alone nor the surface’s appearance determines flux.

Gauss’s law gives the total flux through any closed surface as Φ = Q_enclosed/ε₀. It depends on the algebraic value of the charge enclosed, not on the shape, size, or detailed distribution of the surface. If Q_enclosed changes to 3Q_enclosed, then Φ changes to 3Q_enclosed/ε₀ = 3Φ. Thus the net flux becomes three times, assuming the original enclosed charge was not zero.

Gauss’s law states that the net electric flux through any closed surface is Φ = Q_enclosed/ε₀. A dipole contains equal positive and negative charges, so its total enclosed charge is zero. Consequently, the algebraic net flux through the closed surface is zero. This does not mean that the electric field is zero at every point; field lines may enter and leave the surface, with their flux contributions cancelling overall. Gauss’s law still applies.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Since the net enclosed charge is zero, the algebraic total flux is zero. This does not require the electric field to vanish at each point: fields from charges inside and outside may enter and leave different parts of the surface and cancel in the total. Therefore A is correct.

An electric dipole consists of equal positive and negative charges. If the closed surface contains both charges, its net enclosed charge is (+q) + (−q) = 0. By Gauss’s law, Φ = Q_enclosed/ε₀ = 0. The dipole field is not zero at every point, the dipole moment is generally non-zero, and the surface has a finite area. Thus option A gives the correct reason.

Gauss’s law determines the total outward flux through a closed surface from the net charge inside it: Φ = Q_enclosed/ε₀. In this case the enclosed charge is only −q, so Φ = −q/ε₀, which is negative. The external positive charge can affect the field distribution on the surface, but its complete flux contribution is zero. Therefore A is correct.