Gauss Law and Divergence Questions and Answers

Answer: A

The symmetry has to be recognised first because it dictates the surface; the enclosed charge can only be computed once the surface is fixed; and the field follows last. Attempting the steps in any other order leads to an integral that cannot be done.

Answer: B

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: the law holds for any closed surface whatever, and external charges contribute no net flux. Statement 3 confuses the law with its usefulness - it is always true, but it yields the field only where symmetry lets \( \mathbf{D} \) be taken outside the integral.

Answer: C

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: any interior field would drive charge until it vanished, so the charge ends up on the surface. Statement 3 has the direction wrong - the surface field is normal to the surface, since any tangential component would move charge along it and the arrangement would not be in equilibrium.

Answer: D

The flux depends only on the algebraic total of the enclosed charge, so moving a charge about inside changes the field at every point of the surface but not the integral. All three of the other operations alter the enclosed total.

Answer: A

Correct answer: the way the charge is distributed within that radius

For a spherically symmetric distribution only the total charge within the radius matters, however it is arranged - concentrated at the centre, spread uniformly, or in shells. This is Newton’s shell result in electrostatic form, and it is what makes a solid sphere and a point charge indistinguishable from outside.

Answer: B

Correct answer: the enclosed charge is zero

Zero field everywhere on the surface gives zero flux, and zero flux means zero net enclosed charge. It does not mean there is no charge inside - equal positive and negative charges would satisfy it - and it says nothing about charge outside the surface.

Answer: C

Near the ends of a finite line the field has an axial component and its magnitude varies along the cylinder, so it cannot be taken outside the integral. The infinite-line formula is a good approximation only near the middle of a long line.

Answer: D

The enclosed charge fixes the flux, and spreading it over an area growing as the square of the radius reduces the density accordingly. Keeping the two quantities distinct removes most of the confusion in Gauss's law problems.

Answer: A

Differentiating a known flux density gives the charge density directly, whereas the integral form would need a surface integral for every region of interest. Both forms are exact and neither is restricted to symmetrical cases.

Answer: B

Correct answer: zero, all the charge being at the origin

\[ \nabla\cdot\mathbf{D} = \frac{1}{r^2}\frac{d}{dr}\left(r^2 \cdot \frac{k}{r^2}\right) = \frac{1}{r^2}\frac{d}{dr}(k) = 0. \] This is the point charge field, and the divergence vanishes everywhere except at the origin, where the expression is undefined and all the charge in fact resides.