Gauss Law and Divergence Questions and Answers

Answer: B

A coaxial Gaussian cylinder encloses the fixed charge on the inner conductor and has an area growing linearly with radius, so the field falls as the first power. Outside the outer conductor the enclosed charge is zero and so is the field.

Answer: C

A Gaussian cylinder drawn beyond the outer conductor encloses both charges, whose sum is zero, so the flux and hence the field vanish. This complete external screening is why coaxial cable neither radiates nor picks up interference.

Answer: D

Correct answer: \( \nabla \cdot \mathbf{D} = \rho_v \)

This is Gauss’s law at a point: the divergence of the flux density equals the volume charge density there. The integral form follows by the divergence theorem. Note that it is \( \mathbf{D} \) rather than \( \mathbf{E} \) that appears, which is what makes the equation hold in any medium without a permittivity in it.

Answer: A

The divergence theorem equates the volume integral of the divergence to the closed surface integral of the flux, so equating both to the enclosed charge gives the two forms of the same law. Stokes's theorem performs the same office for Ampere's law.

Answer: B

Correct answer: no net charge density

Zero divergence means no source or sink at that point, so the charge density vanishes there. It says nothing at all about the field, which may be strong and varying - the field of charges lying outside the region has zero divergence throughout it.

Answer: C

Correct answer: the volume charge density at that point

Divergence is a local operation, so it reports the source strength at the point itself and nothing about what lies elsewhere. The total charge enclosed by a surface is what the integral form gives, and the two are connected by the divergence theorem rather than being the same statement.

Answer: D

Any field within the metal would drive a current, and in equilibrium none flows, so the field must vanish. Gauss's law then requires that no net charge exist within the metal either, so all of it resides on the surface.

Answer: A

Charge of opposite sign is induced on the inner surface and an equal charge of the original sign on the outer, so an earthed shell removes the latter and leaves no external field. Unearthed, the outer charge remains and the field outside is that of the original charge.

Answer: B

A small Gaussian box with one face in the metal and one outside receives flux through the outer face only, since the interior field is zero. That gives the whole of the density over the permittivity, twice what an isolated sheet would give.

Answer: C

A Gaussian surface drawn wholly within the metal must enclose zero net charge because the field there is zero, and that forces the induced wall charge to cancel what is inside. An equal charge of the original sign then appears on the outer surface.