Gauss Law and Divergence Questions and Answers

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.

Answer: D

Differentiating each component with respect to its own coordinate and adding gives \( 3 + 2 = 5 \), a uniform density throughout the region. A z component would add a third term, and its absence simply means nothing depends on z.

Answer: A

Correct answer: \( 4r \)

In spherical coordinates the radial divergence carries a metric factor: \[ \nabla\cdot\mathbf{D} = \frac{1}{r^2}\frac{d}{dr}(r^2 D_r) = \frac{1}{r^2}\frac{d}{dr}(r^4) = \frac{4r^3}{r^2} = 4r. \] Differentiating \( D_r \) alone would give \( 2r \), which is the offered trap - the \( r^2 \) must go inside the derivative.

Answer: B

The inner surface takes minus five nanocoulombs, and earthing supplies whatever charge is needed to hold the sphere at zero potential, which here means the outer surface carries nothing. The external field is therefore zero, which is complete screening.