Gauss Law and Divergence Questions and Answers

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.

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.