Gauss Law and Divergence Questions and Answers

Answer: A

Both statements are true and the reason explains the assertion (option A). The divergence theorem is a mathematical identity for any vector field, so it holds for static and time-varying fields alike.

The reason gives its use: with Gauss's law, the flux of \(\overline{D}\) through a closed surface (or of \(\overline{E}\) times the permittivity) equals the enclosed charge, so the theorem is used to find that charge. Options B, C and D deny one of the statements.

Answer: A

Written \( \oint_s \mathbf{D} \cdot d\mathbf{s} = Q_{enc} \), it involves the flux density and so needs no permittivity on the right. Expressed in terms of the field intensity instead, the permittivity appears as a divisor on the charge.

Answer: B

Every line of flux from an external charge that enters the surface must also leave it, so the net contribution cancels. This is why only the enclosed charge appears on the right hand side, however much charge lies just outside.

Answer: C

The law holds for any closed surface whatever, which is what makes it so general. A surface chosen to match the symmetry of the problem lets the field be taken outside the integral, and that is the only reason for preferring one shape to another.

Answer: D

On such a surface the dot product becomes a constant magnitude times an area, and the field can be solved for. Where the field is tangential the contribution is zero, which disposes of the end faces of a cylinder round a line charge.

Answer: A

The field is radial and of constant magnitude at a given radius, so a coaxial cylinder makes the integral trivial and its flat ends contribute nothing because the field is tangential there. A sphere would suit only a point-symmetric source.

Answer: B

Correct answer: a rectangular box straddling the sheet

The field of an infinite sheet is uniform, perpendicular to the sheet and outward on both sides. A box with two faces parallel to the sheet therefore has flux through those two faces only - the sides being parallel to the field - and the enclosed charge is the sheet area they span. That gives \( E = \sigma/2\varepsilon_0 \) immediately.

Answer: C

The enclosed charge is fixed while the area of the sphere grows as the square of the radius, so their ratio falls as the inverse square. The same reasoning gives the inverse first power for a cylinder and a constant for a box straddling a sheet.

Answer: D

A concentric sphere drawn inside the shell encloses no charge, so the flux through it is zero, and the spherical symmetry then forces the field itself to vanish. Outside the shell the field is the same as that of a point charge at the centre.

Answer: A

The volume within radius r grows as the cube, and with the density constant so does the charge. Dividing by an area growing as the square leaves a field rising linearly, which is the standard result for the interior of such a sphere.