Gauss Law and Divergence Questions and Answers

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.

Answer: D

Correct answer: 10 nC

Gauss’s law gives the total flux as the enclosed charge, so it is 10 nC regardless of the radius, the shape of the surface or the medium. The radius of 1 m is given only to make the next calculation possible and plays no part in this one.

Answer: A

Correct answer: 0.796 nC/m2

\[ D = \frac{Q}{4\pi r^2} = \frac{10}{4\pi \times 1} = 0.796\ \mathrm{nC/m^2}. \] The flux is spread over the sphere’s area, and it is this division by \( 4\pi r^2 \) that produces the inverse square law - the flux itself does not fall off at all.

Answer: B

Correct answer: 1 nC

By symmetry the charge at the centre sends equal flux through each of the six faces: \[ \frac{6}{6} = 1\ \mathrm{nC}. \] The symmetry argument is what makes this trivial; without the charge at the centre the faces would carry different amounts and the calculation would be a genuine integration.

Answer: C

Only the algebraic sum matters, and \( 8 - 3 + 2 = 7 \) nanocoulombs. Where the individual charges sit within the surface makes no difference to the total flux, though it makes a great deal of difference to the field at any particular point.

Answer: D

Correct answer: 50.3 nC

\[ Q = \sigma \times 4\pi r^2 = 1 \times 4\pi \times 4 = 50.3\ \mathrm{nC}. \] The surface area of a sphere of radius 2 m is \( 16\pi \), or 50.3 square metres, so the numbers coincide here because the density is one nanocoulomb per square metre.

Answer: A

Correct answer: 113 V/m

First the charge: \[ Q = \rho \times \tfrac{4}{3}\pi r^3 = 3 \times 4.19 = 12.57\ \mathrm{nC}. \] Then the field at the surface, where the sphere acts as a point charge at its centre: \[ E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{12.57 \times 10^{-9}}{1.113 \times 10^{-10}} = 113\ \mathrm{V/m}. \]

Answer: B

Inside a uniformly charged sphere the field rises linearly with radius, so at half the radius it is half the surface value. The interior is field free only for a conductor or a hollow shell, not for a solid insulating sphere of charge.

Answer: C

Using \( E = 2 \times 9 \times 10^{9}\rho_L/\rho \) gives \( 2 \times 9 \times 10^{9} \times 2 \times 10^{-9}/(2 \times 10^{-3}) = 1.8 \times 10^{4} \) V/m. Fields of tens of kilovolts per metre are entirely ordinary inside a cable.