Electric Flux and Electric Dipole

The area vector of a surface is defined as a vector normal, or perpendicular, to that surface. If the electric field is parallel to the surface, it is therefore perpendicular to the area vector. Hence the angle θ between E and the area vector is 90°. This also gives Φ = EA cos 90° = 0. Zero degrees would describe a field along the normal, not along the surface.

The governing distinction is that electric field is a vector field, so it has magnitude and direction at every point. Electric flux measures the field passing through a specified surface and is obtained from Φ = E A cos θ for a uniform field; it is a scalar. Therefore A is correct. B wrongly calls flux a vector, while D confuses a surface quantity with a point quantity.

Electric flux through a plane surface is Φ = E · A = EA cos θ. The area vector points east while the electric field points west, so the vectors are antiparallel and θ = 180°. Consequently, cos 180° = −1 and Φ = −EA, which is negative. Thus A is correct. The flux is not zero because the vectors are not perpendicular, and it is not positive because their directions oppose each other.

For a uniform field through a plane surface, electric flux is given by Φ = EA cos θ, where θ is the angle between the electric field and the area vector. If the field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos θ = 0. Thus flux is zero although E is non-zero. Hence A is correct.

Electric flux through a plane surface is Φ = EA cos θ, where θ is the angle between the electric field and the area vector. Maximum positive flux occurs at θ = 0°, because cos 0° = 1. Zero flux occurs at θ = 90°, because cos 90° = 0. Therefore the rotation changed the angle from zero to ninety degrees, making A correct; 180° gives maximum negative flux.

For a plane surface in a uniform electric field, Φ = EA cos θ. If the angle θ is unchanged, cos θ is constant. Doubling the area gives a factor of 2, and doubling the field gives another factor of 2; therefore the new flux is (2E)(2A)cos θ = 4EA cos θ = 4Φ. Hence option A is correct, while B includes only one doubling effect.

For a closed surface, the differential area vector dA is conventionally directed outward at every point. Flux is Φ = ∮ E · dA. When the electric field has an outward component, E · dA is positive; when it points into the enclosed region, the dot product is negative. Tangential components contribute zero locally. Therefore outward-going field is counted positive and inward-going field negative, making option A the correct sign convention.

Electric flux through a plane surface is given by Φ = EA cos θ, where θ is the angle between the electric field and the area vector. For maximum positive flux, cos θ must equal 1, which occurs at θ = 0°. Thus the area vector must point in the same direction as the electric field. The opposite direction would give maximum negative flux, while a perpendicular orientation gives zero flux.

Flux is Φ = EA cos θ, with θ measured between the electric field and the area vector, which is normal to the surface. If the field is parallel to the surface, it is perpendicular to the area vector, so θ = 90° and cos 90° = 0. Therefore no field component crosses the surface and the flux is zero, even when E is very large. Flux is not restricted to closed surfaces.

For a uniform field through a plane surface, flux is Φ = EA cos θ. The most negative possible value occurs when cos θ = −1, which requires θ = 180°. Therefore the electric field must be exactly opposite to the area vector, giving Φ = −EA. If the field is along the area vector, flux is maximum positive; if it is parallel to the surface, θ = 90° and flux is zero.

Electric flux through a plane surface is Φ = E A cos θ, where θ is the angle between the electric field and the chosen area vector. North and south are opposite directions, so θ = 180° and cos 180° = −1. Hence Φ = −EA, which is negative. It is not zero because the field is perpendicular to the surface, and it is not positive because the vectors are oppositely directed.

For every small element of a closed surface, the area vector is conventionally chosen along the outward normal. With this convention, an electric field directed outward has E · dA > 0 and contributes positive flux, whereas an inward-directed field contributes negative flux. This makes the interpretation of field lines leaving or entering the closed surface consistent with Gauss’s law. The convention does not make the field vanish or change the surface shape.

Net charge and dipole moment describe different properties. For a dipole, the charges are +q and −q, so their algebraic sum is zero. However, the dipole moment has magnitude p = qd and direction from the negative charge to the positive charge. If the charges are separated by a non-zero distance d, p is non-zero even though the net charge vanishes. Same-sign charges or zero separation would not describe this dipole.

The area vector of a surface is defined perpendicular to that surface. If the surface itself is parallel to the electric field, the field direction must be perpendicular to the area vector. Thus the angle θ between E and the area vector is 90°. This also explains why the flux Φ = EA cos 90° becomes zero. Zero degrees would describe a field normal to the surface, not parallel to it.

An electric dipole consists of equal and opposite charges, +q and −q, separated by a non-zero distance d. Their algebraic sum is zero, so the net charge is zero. However, the dipole moment is a vector with magnitude p = qd and direction from the negative charge to the positive charge. Since d is non-zero, p is non-zero. Therefore A is correct; same-sign charges do not form the standard dipole, zero separation gives zero moment, and moment is not another name for net charge.

For a closed surface, the conventional area vector dA is directed outward at every point. Electric flux is Φ = ∮ E·dA. When the electric field crosses the surface outward, E and dA have the same direction, so their dot product is positive and the total outward contribution is positive. Thus option A is correct; inward flux is negative, while zero applies only when the net flux cancels or no field crosses.

The area vector of a closed surface is defined outward. For field entering the surface, the electric field vector points opposite to this outward normal, so E·dA is negative. Consequently, the entering contribution to electric flux is negative, making option A correct. Outward flux is positive; it is not automatically zero or infinite. The sign describes direction relative to the chosen outward normal, not the field’s size alone.