Gauss Law and Divergence Questions and Answers

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.

Answer: C

A Gaussian sphere outside both encloses a net charge of zero, so the flux and hence the field vanish. This is the spherical counterpart of the coaxial cable's external screening, and it is why such an arrangement is a capacitor rather than a radiator.

Answer: D

The law is exact for any closed surface whatever. Symmetry is needed only to solve for the field from it: without symmetry the integral still equals the enclosed charge, but the field cannot be extracted from under the integral sign.

Answer: A

A uniform field has zero divergence, so the charge density is zero throughout. The charge producing it lies elsewhere, on the plates that bound the region, which is exactly the situation inside a parallel plate capacitor.

Answer: B

Correct answer: False, the flux depending only on the charge enclosed by the surface

The area grows as \( r^2 \) and the flux density falls as \( 1/r^2 \), so the two cancel exactly and the total flux is unchanged. That cancellation is the whole content of Gauss’s law, and it is why the surface may be of any size or shape at all.

Answer: C

Correct answer: False, the net flux being fixed by the enclosed charge alone

A charge outside the surface sends flux in through one part and out through another, and the two are exactly equal, so its net contribution is zero. The field on the surface may therefore be large and varying while the integral of the flux over the whole surface vanishes.

Answer: D

Zero enclosed charge gives zero net flux, and the symmetry of the situation then forces the field itself to vanish rather than merely to cancel on average. The argument fails if a charge is placed inside the cavity, and then the field there is not zero.

Answer: A

Because the symmetry lets the field be taken outside the integral, the result depends only on the total charge and not on how it is distributed radially. That is exactly why the point charge formula serves for the earth's gravitational field as well.