Gauss's Law and its Applications

Gauss’s law gives the total flux through a closed surface as Φ = Q_enclosed/ε₀. Moving the positive charge changes its position and therefore changes the field strength and direction at different points of the surface. However, the same charge remains enclosed, so Q_enclosed is unchanged. Consequently, the integrated total flux remains q/ε₀. Hence option A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. An external charge can produce a substantial electric field on the surface, so the local flux through different patches changes. However, because the charge lies outside, the total number of field lines entering the surface equals the number leaving it. Their signed contributions cancel, giving zero net flux from that charge. Thus option A is correct; the other choices incorrectly deny the field or the meaning of a closed surface.

Gauss’s law states that the net outward electric flux through a closed surface is Φ = Q_enclosed/ε₀. Field lines leaving the surface represent positive outward flux, while lines entering represent negative flux. If fewer lines leave than enter, the net flux is negative. Therefore Q_enclosed must also be negative. The conclusion concerns the algebraic net charge inside the surface, not the total amount of positive and negative charge separately. Thus A is correct.

By Gauss’s law, the total flux through a closed surface depends only on the algebraic net charge enclosed: Φ = Q_net/ε₀. Let each charge magnitude be q. The four positive charges contribute +4q and the three negative charges contribute −3q, giving Q_net = +4q − 3q = +q. Since ε₀ is positive, the total flux is positive. Its exact value is q/ε₀, independent of the shape of the closed surface, so A is correct.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Taking the magnitude of each charge as q, the enclosed charge is 2q − 5q = −3q. Therefore Φ = −3q/ε₀, so the outward flux is negative. It is not zero because the enclosed charges do not cancel, and it is not infinite for finite charge.

For any closed surface, Gauss’s law gives Φ = Q_enclosed/ε₀. Thus the total flux depends on the net charge enclosed, not on the surface’s shape or area. Changing the shape can change the electric field at individual points and the flux through separate portions, but the algebraic total remains unchanged when the enclosed charge is fixed. Hence option A is correct.

A dipole consists of equal positive and negative charges, so its net enclosed charge is q + (−q) = 0. By Gauss’s law, the net flux through the closed surface is therefore zero. Equal entering and leaving field lines describe zero net flux; they do not imply that the electric field is zero at every point or that the dipole moment vanishes. Thus A is correct.

Gauss’s law relates the net flux through a closed surface only to the net charge enclosed: Φ = Q_enclosed/ε₀. Since the enclosed charge is zero, the total flux is zero. Charges outside the surface can nevertheless produce a nonzero electric field at points on the surface; their entering and leaving contributions cancel in the total. Therefore option A correctly separates local field from net closed-surface flux.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Therefore positive total flux means that the algebraic net charge enclosed by the surface is positive. It does not mean the field points outward at every point; local inward contributions may also exist. External charges can affect the field pattern but contribute zero net flux through a closed surface. Hence option A is correct.

By Gauss’s law, the net flux through a closed surface is Φ = Q_enclosed/ε₀ and depends only on the net charge enclosed. Charges outside the surface can alter the electric field at individual points, but their total flux through the closed surface is zero. Negative flux therefore indicates net negative enclosed charge, not the absence of positive external charge. Positive charges may still exist outside, so option A is correct.

Gauss's law gives the relation Φ = Q_enclosed/ε₀, but its strongest examination use appears when a charge distribution has symmetry. A student must identify spherical, cylindrical, or planar symmetry, select a suitable Gaussian surface, and then simplify the flux calculation. Therefore option A describes the intended skill. The other choices concern unrelated quantities and do not test application of Gauss's law or electric-flux reasoning.

Step 1: In Gauss's law, total flux is proportional to net enclosed charge. Step 2: If enclosed charge is doubled, total flux also doubles. Step 3: In such questions, check net enclosed charge before surface shape.

Step 1: Total closed flux is decided only by net charge enclosed by the surface. Step 2: Lines from an external charge may enter and leave the surface. Step 3: Their algebraic contribution to total flux becomes zero.

Step 1: Gauss's law relates total flux to net enclosed charge. Step 2: Changing the shape does not change the enclosed charge. Step 3: Therefore total flux remains same, even if local field changes.

Step 1: Total flux depends on the charge enclosed by the closed surface. Step 2: Changing the radius does not change the enclosed charge. Step 3: Field may decrease, but total flux remains unchanged.

Step 1: Field lines may cross the surface due to external charges. Step 2: In Gauss's law, total flux is determined by net enclosed charge. Step 3: Since enclosed charge is zero, total flux is zero.

Step 1: Field of a long charged line is radial. Step 2: Area vectors of cylinder end caps are along the line. Step 3: The radial field does not cross the end caps, so their flux is zero.

Step 1: Field of an infinite sheet is perpendicular to the sheet. Step 2: On the curved surface of the pillbox, this field runs parallel to the surface. Step 3: Since it does not cross that surface, flux through it is zero.

Step 1: A point charge has spherical symmetry around it. Step 2: Every point on the sphere is at the same distance from the centre. Step 3: Thus field magnitude is same and calculation becomes simple.

Step 1: For a point charge, a spherical Gaussian surface is used. Step 2: Area of the sphere depends on square of radius. Step 3: Since total flux is fixed by charge, field decreases inversely as square of distance.

Step 1: For a long line charge, a cylindrical Gaussian surface is chosen. Step 2: Curved area of the cylinder is proportional to radius. Step 3: Hence field decreases inversely with distance.

Step 1: For an infinite sheet, a small cylindrical pillbox is chosen. Step 2: Flux mainly comes through the two flat faces. Step 3: Due to symmetry, the field does not depend on distance.

Step 1: Inside a conductor in electrostatic condition, electric field is zero. Step 2: By Gauss's law, zero field gives zero total flux. Step 3: Therefore the Gaussian surface inside encloses zero net charge.

Step 1: Inside a conductor at electrostatic condition, electric field is zero. Step 2: If excess charge were inside, flux through an internal closed surface would not be zero. Step 3: Hence excess charge resides on the conductor surface.

Step 1: In electrostatic condition, free charges in a conductor remain at rest. Step 2: If a tangential component existed, they would move. Step 3: Therefore the field just outside must be perpendicular to the surface.