Gauss's Law and its Applications

The governing concept is electrostatic induction. A charged object placed near a neutral conductor causes its mobile electrons to redistribute: the nearer side acquires an induced charge of opposite sign, while the farther side has the like-sign effect. Because the opposite charge is closer, its attractive force is stronger than the farther repulsive force, giving a net attraction. The conductor remains neutral overall, so A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Qinside/ε₀. More field lines leaving than entering means the outward flux is positive. Therefore Qinside is positive, so the net enclosed charge has a positive sign. Individual positive and negative charges may both be present, but the positive contribution is larger. Thus A is correct; B reverses the sign and C ignores the nonzero net flux.

Gauss’s law relates outward electric flux through a closed surface to the enclosed charge: Φ = Q_enclosed/ε₀. If fewer field lines leave than enter, the net outward flux is negative. Therefore Q_enclosed is negative. Physically, the surface contains a net sink of field lines, characteristic of an overall negative charge. A positive charge would produce more outward lines than inward lines.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. If the charges are +q and −q, their algebraic sum is Q_enclosed = +q − q = 0. Therefore Φ = 0, regardless of the surface’s shape or the individual field lines crossing it. Option A is correct; options B, C, and D incorrectly use one charge or ignore cancellation of the equal opposite charges.

By Gauss’s law, the net flux through a closed surface is determined only by the algebraic charge enclosed within it. Field lines from an external charge can enter the surface in some regions and leave it in others. The corresponding inward and outward flux contributions cancel in the total, so the external charge contributes zero net closed flux. Option A is correct; the other options confuse local field with net flux.

By Gauss’s law, the net flux through a closed surface is Φ = Q_net,inside/ε₀. Only charge enclosed by the surface contributes to this total. The external negative charge can change the electric field at individual points on the surface, but its net contribution to closed-surface flux is zero. Hence option A is correct; options B and C incorrectly include an outside charge.

By Gauss’s law, the net flux through a closed surface is Φ = Q_enclosed/ε₀. Since only the positive charge +q lies inside the surface, Q_enclosed = +q and therefore Φ = +q/ε₀, which is positive. The external negative charge can affect the electric field at points on the surface, but its field contributes zero net flux through the complete closed surface. Hence option A is correct.

The total flux through a closed surface is governed by Gauss’s law, Φ = Q_enclosed/ε₀. When the surface contains only the dipole’s negative charge −q, the enclosed charge is negative, so Φ = −q/ε₀. The positive charge outside may change the local field distribution on the surface, but it cannot change the net flux produced by the enclosed charge. Therefore the total flux is negative.

For a closed surface, Gauss’s law gives Φ = Q_enclosed/ε₀. An external charge can create an electric field at the surface, and some field lines may enter it, but those lines must leave again because the charge is not enclosed. The inward and outward contributions cancel, so the net flux is zero. Option A is correct; the other options wrongly claim that no field exists, the area vanishes or the charge becomes neutral.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Since ε₀ is positive, a positive net flux implies Q_enclosed is positive. This means the algebraic sum of charges inside the surface is positive, although individual positive and negative charges may both be present. Therefore option A is correct; zero or negative enclosed charge would give zero or negative net flux.

According to Gauss’s law, Φ = Q_enclosed/ε₀ for a closed surface. Because ε₀ is positive, negative net flux requires the enclosed net charge Q_enclosed to be negative. Field lines therefore enter the surface more strongly than they leave it in the net sense. Option A is correct. The conclusion concerns the algebraic net charge, not necessarily every individual charge; positive or zero net charge would produce positive or zero flux.

The correct answer is A, No. Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Thus, zero net flux implies zero algebraic enclosed charge, not zero electric field at every point. External charges can produce a field on the surface, and internal positive and negative charges can also give cancelling flux while the local field remains nonzero. Therefore, flux and field are different quantities.

Gauss’s law states that the net electric flux through any closed surface is Φ = Q_enclosed/ε₀. The result depends on the algebraic net charge enclosed, not directly on the surface shape or the detailed distribution of the field. If Q_enclosed changes to 2Q_enclosed, then Φ′ = 2Q_enclosed/ε₀ = 2Φ. Thus option A is correct, provided the statement refers to doubling the net enclosed charge rather than merely adding equal opposite charges.

According to Gauss’s law, the net flux through a closed surface is Φ = Q_enclosed/ε₀. Consequently, if the enclosed net charge is unchanged, the total flux is unchanged, even when the surface is stretched, compressed, or given an irregular shape. The electric field and local flux density may vary from point to point, but their surface integral remains fixed. Therefore option A is correct; the other statements are not generally valid.

Gauss’s law determines the net flux through a closed surface from the net charge enclosed: Φ = Q_enclosed/ε₀. External charges can produce a nonzero electric field at many points on the surface, but their field lines enter and leave the closed surface in balanced contributions because those charges are outside. Since Q_enclosed = 0, the algebraic net flux is zero. Hence option A is correct, although the local field itself need not be zero.

Gauss’s law says the net flux through a closed surface depends only on the net charge enclosed by that surface. An external charge can certainly produce an electric field on the surface, so its field is not absent. However, its field lines cross the closed surface inward at some places and outward at others, giving equal and opposite contributions whose algebraic sum is zero. Therefore A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. The external charge can produce electric field at different points of the surface, but it is not included in Q_enclosed. Its incoming and outgoing contributions cancel in the net total. Since the enclosed charge is zero, option A, zero flux, is correct; the sign cannot be fixed as positive or negative.

The governing principle is Gauss’s law: Φ = Q_net/ε₀ for any closed surface. If the enclosed charges are +q and −q, their algebraic sum is Q_net = +q − q = 0. Hence the total flux is Φ = 0, even though the electric field may be nonzero at many points on the surface. Option A is correct; the other choices ignore the cancellation of equal opposite charges.

Gauss’s law relates net flux to net enclosed charge, Φ = Q_enclosed/ε₀; it does not state that the field must vanish at every point. Equal positive and negative enclosed charges, or suitable external fields, can produce zero net flux while the field remains nonzero on portions of the surface. Flux is an integrated quantity involving E·dA. Therefore option A is correct; the other choices confuse net flux with local field.

A dipole consists of charges +q and −q, so its net charge is Q_enclosed = +q − q = 0. Gauss’s law then gives Φ = Q_enclosed/ε₀ = 0 for any closed surface containing both charges. However, the dipole field is generally nonzero at points on and outside the surface, and its dipole moment p = qd need not vanish. Thus option A is the only valid conclusion.

Gauss’s law states that the total electric flux through a closed surface is Φ = Q_enclosed/ε₀. If each positive charge has magnitude q and each negative charge has magnitude −q, the enclosed charge is q + q − q − q = 0. Consequently, Φ = 0/ε₀ = 0. The individual fields need not vanish everywhere, but their net closed-surface flux is zero because the net enclosed charge is zero. Thus option A is correct.

Gauss’s law gives the total flux through any closed surface as Φ = Q_enclosed/ε₀, regardless of the surface’s shape or size. An electric dipole contains equal positive and negative charges, so when both remain enclosed, its net charge is +q − q = 0. Therefore the total flux is zero before and after changing the surface shape. Local field distribution can change, but the total closed-surface flux cannot. Option A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. The surface shape and the external charges do not alter the net flux; only the algebraic charge enclosed matters. Therefore, if the enclosed charge changes from Q to 2Q, the flux changes from Q/ε₀ to 2Q/ε₀. Hence option A is correct. Option B would correspond to halving the charge, while C and D are incorrect unless the net enclosed charge is zero or unchanged.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. If the enclosed charge is negative, the net flux is negative. With the outward area-vector convention, outward flux is positive and inward flux is negative; therefore, the overall tendency is for more field lines to enter the surface than leave it. This makes option A correct; option B describes positive enclosed charge, while C gives zero net charge and D is unnecessarily absolute.

Gauss’s law determines the total flux through a closed surface from the algebraic sum of charges enclosed by that surface: Φ = Q_enclosed/ε₀. Since the positive charge lies inside, it contributes to Q_enclosed. The negative charge just outside produces an electric field on the surface, but it is not enclosed and therefore does not affect the net flux. Thus option A is correct. Option B would apply only if both charges were enclosed.