Corrected exercises: Gauss's law (203-SN2-RE)

Area of a face: $A=(0.200)^{2}=0.0400$ m$^{2}$. The two faces perpendicular to the field give $\Phi=\pm EA=\pm 350\times 0.0400=\pm 14.0$ N·m$^{2}$/C, one counted negative because the field ENTERS there and the other positive because it LEAVES there, the normal of a closed surface being oriented outward by convention. The other four faces are parallel to the field, their normal is perpendicular to it, $\cos 90°=0$ and their flux is zero. Total: $-14.0+14.0=0$. Gauss's law confirms it at once: the cube encloses no charge, so $\Phi=\dfrac{0}{\varepsilon_{0}}=0$.

The TOTAL flux does not change, it stays zero: Gauss's law depends only on the enclosed charge, which is still zero, and not at all on the orientation. The flux through EACH face does change, however: the two faces that used to be perpendicular now receive only $EA\cos 30.0°=12.1$ N·m$^{2}$/C in absolute value, and two of the faces that used to be parallel now receive $EA\cos 60.0°=7.00$ N·m$^{2}$/C. The sum stays zero, with four contributions instead of two. This is the first lesson of the chapter: the law is about the total, never about the detail.

$\Phi_{\text{tot}}=\dfrac{q}{\varepsilon_{0}}=\dfrac{6.00\times 10^{-9}}{8.854\times 10^{-12}}=678$ N·m$^{2}$/C. By symmetry, with the charge at the centre, the six faces are strictly equivalent and share the flux equally: $\dfrac{678}{6}=113$ N·m$^{2}$/C each. No integral is needed, and that is exactly the point of the law: integrating the field of a point charge directly over a square face is a painful exercise whose result is this number.

A corner is shared by EIGHT identical cubes that would fill the space around the charge. By symmetry, each receives one eighth: $\Phi=\dfrac{q}{8\varepsilon_{0}}=\dfrac{678}{8}=84.7$ N·m$^{2}$/C. The split among the faces is subtler, and that is the real question. The THREE faces that meet at the corner contain the charge in their own plane: the field is parallel to the face everywhere on it, so their flux is ZERO. The other three faces therefore share everything, equally by symmetry: $\dfrac{84.7}{3}=28.2$ N·m$^{2}$/C each. This result, one eighth spread over three faces, is a classic exam question and it can only be obtained by the solid-angle argument, never by direct calculation.

The total flux is EXACTLY ZERO, however close the charge is and whatever its value: it is not enclosed. Every field line that enters the cube leaves it, and the positive and negative contributions cancel exactly. The field itself is by no means zero on that face: at one centimetre it equals $\dfrac{k\times 50\times 10^{-9}}{(0.0100)^{2}}=4.49\times 10^{6}$ N/C, enormous. This is the distinction never to lose sight of: a zero flux says nothing about the field point by point, it only says that the inward and outward balance is even.

$\rho=\dfrac{Q}{\frac{4}{3}\pi R^{3}}=\dfrac{45.0\times 10^{-9}}{\frac{4}{3}\pi(0.100)^{3}}=\dfrac{45.0\times 10^{-9}}{4.189\times 10^{-3}}=1.07\times 10^{-5}$ C/m$^{3}$. The word INSULATING in the problem is decisive: in a conductor all the charge would flee to the surface, and there would be no volume density at all.

Spherical Gaussian surface of radius $r<R$. The enclosed charge is that of the inner ball: $Q_{\text{enc}}=Q\dfrac{r^{3}}{R^{3}}$, since the density is uniform and the volumes are in the ratio of the cubes. Then $E(4\pi r^{2})=\dfrac{Q r^{3}}{\varepsilon_{0}R^{3}}$, so $E=\dfrac{kQr}{R^{3}}$. At $r=4.00$ cm: $E=\dfrac{8.988\times 10^{9}\times 45.0\times 10^{-9}\times 0.0400}{10^{-3}}=1.62\times 10^{4}$ V/m. The field grows because two effects compete and one wins: the enclosed charge increases as $r^{3}$, the area of the Gaussian surface only as $r^{2}$, so the quotient grows as $r$. This is not an exception to the usual decrease, it is what happens when you add source faster than you move away from it. Gravity inside the Earth follows exactly the same law.

From the inside expression: $E(R)=\dfrac{kQR}{R^{3}}=\dfrac{kQ}{R^{2}}=\dfrac{404.5}{0.0100}=4.04\times 10^{4}$ V/m. From the outside expression: $\dfrac{kQ}{R^{2}}$, identical. The match is continuous, and that is the mandatory check for any field defined piecewise: if the two branches do not meet at $r=R$, there is an error. At $r=25.0$ cm: $E=\dfrac{404.5}{0.0625}=6.47\times 10^{3}$ V/m. The three points on the curve are indeed $16.2$, $40.4$ and $6.47$ kV/m, with the sharp peak at the surface.

Outside, everything happens as if the charge were a point charge: $V(0.250)=\dfrac{404.5}{0.250}=1.62\times 10^{3}$ V. At the surface: $V(R)=\dfrac{404.5}{0.100}=4.04\times 10^{3}$ V. Inside, $V=\dfrac{kQ}{2R^{3}}(3R^{2}-r^{2})$: at $r=4.00$ cm, $V=\dfrac{404.5}{2\times 10^{-3}}(3\times 0.0100-0.00160)=2.022\times 10^{5}\times 0.0284=5.74\times 10^{3}$ V. At the centre: $V=\dfrac{3}{2}\dfrac{kQ}{R}=1.5\times 4.045\times 10^{3}=6.07\times 10^{3}$ V. The potential is therefore greatest at the centre while the field is zero there, which is consistent: $E=-\dfrac{dV}{dr}$, and a function at its maximum has a zero derivative.

Everything changes inside, nothing outside. In a conductor in equilibrium, all the charge moves to the surface, so for $r<R$ the enclosed charge is zero and $E=0$ everywhere inside, in particular at $r=4.00$ cm. Outside, on the other hand, the enclosed charge is still $Q$: the field at $r=25.0$ cm is UNCHANGED, $6.47\times 10^{3}$ V/m. This is the most important point of the question, and it illustrates the limit of the law: the outside field cannot possibly tell a solid sphere from a hollow one, a conductor or a point charge, as long as the total charge is the same. Finally, since the field is zero throughout the volume, the potential there is CONSTANT and equal to the surface value: $V=4.04\times 10^{3}$ V at the centre, against $6.07\times 10^{3}$ V in the insulating case.

The distribution is invariant under translation along the axis and under rotation about it, provided end effects are neglected, which is what the infinite-cable assumption does. The field can therefore only be RADIAL and its magnitude can only depend on $r$. The suitable surface is a coaxial cylinder of radius $r$ and length $L$, closed by two end caps. On the curved wall, $\vec{E}$ is everywhere perpendicular to the surface and of constant magnitude: the flux is $E(2\pi rL)$. On the two end caps, $\vec{E}$ is radial and therefore PARALLEL to the cap, the normal is perpendicular to it, and $\cos 90°=0$: their contribution is zero. This is exactly why the length $L$ will end up cancelling, and why an arbitrary choice of $L$ causes no problem.

At $r=3.00$ mm, between the two conductors: $Q_{\text{enc}}=\lambda L$, so $E=\dfrac{\lambda}{2\pi\varepsilon_{0}r}=\dfrac{2k\lambda}{r}=\dfrac{2\times 8.988\times 10^{9}\times 3.20\times 10^{-9}}{3.00\times 10^{-3}}=1.92\times 10^{4}$ V/m. At $r=1.00$ mm, INSIDE the metal of the core: $E=0$, a property of a conductor in equilibrium, and consistent with the law since all the charge of the core is on its surface, so $Q_{\text{enc}}=0$. At $r=5.00$ mm, inside the metal of the sheath: $E=0$ for the same reason. At $r=8.00$ mm, outside: the surface encloses $+\lambda L-\lambda L=0$, so $E=0$. The field exists only in the annulus between the two conductors, and nowhere else.

A Gaussian surface drawn INSIDE the metal of the sheath, between $b$ and $c$, sees a zero field there, hence a zero flux, hence a zero enclosed charge. It encloses the $+\lambda$ carried by the core and the charge of the inner surface of the sheath: the latter is therefore $-\lambda=-3.20$ nC/m. Since the sheath carries $-\lambda$ in total, its outer surface carries $-\lambda-(-\lambda)=0$. No charge outside, which is precisely what cancels the field at $r=8.00$ mm.

$U=\displaystyle\int_{a}^{b}\dfrac{2k\lambda}{r}dr=2k\lambda\ln\dfrac{b}{a}=2\times 8.988\times 10^{9}\times 3.20\times 10^{-9}\times\ln 3.00=57.5\times 1.0986=63.2$ V, the core being at the higher potential since it carries the positive charge. Capacitance per unit length: $\dfrac{C}{L}=\dfrac{\lambda}{U}=\dfrac{3.20\times 10^{-9}}{63.2}=5.06\times 10^{-11}$ F/m, that is $50.6$ pF/m. Check with the general formula: $\dfrac{2\pi\varepsilon_{0}}{\ln(b/a)}=\dfrac{5.563\times 10^{-11}}{1.0986}=5.06\times 10^{-11}$ F/m. Fifty picofarads per metre is the real order of magnitude of an instrumentation cable, and it is what limits the usable length with a high-impedance source.

No, the signal is not disturbed, and that is the whole point of a coaxial cable. The sheath is a closed conductor around the core: it is a cylindrical Faraday cage, and any external field induces in it a redistribution of charges that cancels the field inside. The core is therefore electrically blind to the outside world, and conversely it does not radiate. If the sheath is CUT over a few centimetres, the cage is no longer closed: the external field enters through the gap, the shielding is lost locally, and the cable both picks up noise and emits it. This is why a poor shield continuity at a connector is enough to ruin an entire cabling, and why sensitive measurements are always made with connectors that grip the shield all the way around rather than with a simple braid twisted together at one point.

By symmetry, the field is perpendicular to the plane, directed away from it if $\sigma>0$, and its magnitude can only depend on the distance $|x|$. Take as Gaussian surface a cylindrical pillbox through the plane, of cross-section $A$, whose two end caps are at equal distances on either side. The curved wall is parallel to the field, so its flux is zero; each end cap gives $EA$. The total flux is therefore $2EA$, and the enclosed charge $\sigma A$: $2EA=\dfrac{\sigma A}{\varepsilon_{0}}$, hence $E=\dfrac{\sigma}{2\varepsilon_{0}}$. For plane 1: $\dfrac{4.00\times 10^{-9}}{2\times 8.854\times 10^{-12}}=226$ N/C. The independence from distance comes from $A$ cancelling: moving away, you lose intensity per charge but gain as much visible surface, an infinite plane having no length scale of its own to say what “far” means.

Each plane creates $226$ N/C. BETWEEN the planes, the two fields point the same way, from the positive plane toward the negative one: they add, $E=2\times 226=452$ N/C, directed along $+x$. OUTSIDE, on each side, the two fields are opposite and of equal magnitude: $E=0$. This is the ideal parallel-plate capacitor, with a uniform field confined between the plates and nothing outside. We recover $E=\dfrac{\sigma}{\varepsilon_{0}}$, which agrees with the capacitor formula.

Plane 1 creates $226$ N/C, plane 2 creates $\dfrac{6.00\times 10^{-9}}{2\times 8.854\times 10^{-12}}=339$ N/C, both repulsive since both are positive. Region $x<0$: both fields point along $-x$, total $-565$ N/C, that is $565$ N/C to the left. Region $0<x<d$: plane 1 pushes toward $+x$ at $226$ and plane 2 pushes toward $-x$ at $339$; total $-113$ N/C, that is $113$ N/C to the LEFT, toward the more weakly charged plane. Region $x>d$: both push toward $+x$, total $+565$ N/C. The reassuring check is that the jump in the field across each plane is indeed $\dfrac{\sigma}{\varepsilon_{0}}$: from $-565$ to $-113$ across plane 1, that is $452=\dfrac{\sigma_{1}}{\varepsilon_{0}}$, and from $-113$ to $+565$ across plane 2, that is $678=\dfrac{\sigma_{2}}{\varepsilon_{0}}$.

The formula remains an excellent approximation as long as the distance to the plane is SMALL compared with the dimensions of the plate and with the distance to the edges. In other words, an observer close enough sees the plate fill the whole horizon and cannot tell that it is finite. Near the edges, the symmetry is broken: the field lines bend outward, the field stops being perpendicular, it is no longer uniform, and it concentrates there since the surface density rises on an edge. These are the edge effects, and they explain why the real capacitance of a parallel-plate capacitor is always slightly HIGHER than $\dfrac{\varepsilon_{0}A}{d}$, the fringing field storing energy too.

The field just outside a conductor is $E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{4.00\times 10^{-9}}{8.854\times 10^{-12}}=452$ N/C, that is DOUBLE the result of question a) for the same surface density. The paradox is no paradox once the Gauss argument is taken apart. For the isolated sheet, the pillbox extends on both sides and the flux leaves through TWO end caps, hence the factor 2 in the denominator. For the conductor, the pillbox has one end cap outside and the other INSIDE the metal, where the field is zero: the flux leaves through a single end cap, $EA=\dfrac{\sigma A}{\varepsilon_{0}}$, with no factor 2. Another reading, just as valid: a conducting plate carries $\sigma$ on EACH of its two faces, that is $2\sigma$ in all, and two sheets of $\sigma$ superposed do give $2\times\dfrac{\sigma}{2\varepsilon_{0}}=\dfrac{\sigma}{\varepsilon_{0}}$. Both formulas are therefore correct, they simply do not refer to the same $\sigma$, and you must always specify whether the stated density is that of one face or of the whole object.

Cut the sphere into shells of thickness $dr'$, of volume $4\pi r'^{2}dr'$, each of constant density: $Q=\displaystyle\int_{0}^{R}\rho_{0}\dfrac{r'}{R}4\pi r'^{2}\,dr'=\dfrac{4\pi\rho_{0}}{R}\displaystyle\int_{0}^{R}r'^{3}dr'=\dfrac{4\pi\rho_{0}}{R}\cdot\dfrac{R^{4}}{4}=\pi\rho_{0}R^{3}$. Numerically: $\pi\times 2.50\times 10^{-6}\times(0.120)^{3}=1.36\times 10^{-8}$ C, that is $13.6$ nC. Note that the result is $\pi\rho_{0}R^{3}$ and not $\dfrac{4}{3}\pi\rho_{0}R^{3}$: the density being zero at the centre and greatest at the edge, the volume-weighted average is $\dfrac{3}{4}\rho_{0}$.

The same calculation stopped at $r$ gives $Q_{\text{enc}}(r)=\dfrac{\pi\rho_{0}r^{4}}{R}$. Gauss's law on a sphere of radius $r$: $E(4\pi r^{2})=\dfrac{\pi\rho_{0}r^{4}}{\varepsilon_{0}R}$, so $E=\dfrac{\rho_{0}r^{2}}{4\varepsilon_{0}R}=\dfrac{k\pi\rho_{0}r^{2}}{R}$. The field now grows as $r^{2}$ and not as $r$: with the density increasing outward, the charge builds up faster than in the uniform case. At $r=6.00$ cm: $E=\dfrac{8.988\times 10^{9}\times\pi\times 2.50\times 10^{-6}\times(0.0600)^{2}}{0.120}=2.12\times 10^{3}$ V/m.

At $r=R$: $E=\dfrac{k\pi\rho_{0}R^{2}}{R}=k\pi\rho_{0}R=8.988\times 10^{9}\times\pi\times 2.50\times 10^{-6}\times 0.120=8.47\times 10^{3}$ V/m. Check from the outside: $\dfrac{kQ}{R^{2}}=\dfrac{8.988\times 10^{9}\times 1.357\times 10^{-8}}{0.0144}=8.47\times 10^{3}$ V/m. The two agree, the match is good. At $r=30.0$ cm, outside, everything happens as for a point charge whatever the internal distribution: $E=\dfrac{8.988\times 10^{9}\times 1.357\times 10^{-8}}{0.0900}=1.36\times 10^{3}$ V/m.

The uniform formula would give $E=\dfrac{kQr}{R^{3}}=\dfrac{8.988\times 10^{9}\times 1.357\times 10^{-8}\times 0.0600}{1.728\times 10^{-3}}=4.24\times 10^{3}$ V/m, exactly DOUBLE the correct answer. The discrepancy is no accident: the ratio of the two expressions is exactly $\dfrac{R}{r}$, which equals 2 precisely at $r=\dfrac{R}{2}$. The origin of the error is conceptual. The formula $\dfrac{kQr}{R^{3}}$ assumes that the enclosed charge grows as $r^{3}$, which is true only if the density is uniform; here it grows as $r^{4}$, the matter being concentrated toward the outside, so there is less charge than expected in the central region. Nothing outside gives the difference away, both distributions giving the same field there: Gauss's law cannot distinguish two distributions of the same total charge seen from outside.

We want $E$ constant, so $\dfrac{Q_{\text{enc}}(r)}{r^{2}}$ constant, so $Q_{\text{enc}}(r)\propto r^{2}$. Differentiating, $\dfrac{dQ}{dr}=\rho(r)4\pi r^{2}\propto r$, hence $\rho(r)\propto\dfrac{1}{r}$. The density must therefore DIVERGE at the centre like $\dfrac{1}{r}$, which is mathematically acceptable, the integral $\int\dfrac{1}{r}r^{2}dr$ converging without difficulty, but physically unrealistic since the density would become infinite there. The question is nonetheless instructive: it shows that a given field law completely determines the density that produces it, and that is exactly the meaning of the differential form of the law, $\vec{\nabla}\cdot\vec{E}=\dfrac{\rho}{\varepsilon_{0}}$, which reconstructs the source from the field.

Gauss's law is still true but becomes unusable, because the distribution NO LONGER has spherical symmetry: the cavity is off-centre, and no axis or centre leaves the configuration invariant. On any closed surface you draw, the magnitude of $\vec{E}$ is not constant and its direction is not known, so $E$ cannot be pulled out of the integral. The way out is the principle of superposition: the field created by a distribution is the sum of the fields created by its pieces, and nothing requires these pieces to exist physically. So we write the object with the hole as the SUM of two solid and perfectly symmetric objects: a complete sphere of radius $R$ and density $+\rho$, plus a small sphere of radius $5.00$ cm and density $-\rho$ exactly filling the cavity. Where they overlap, the densities cancel, and we indeed recover a hole.

Each solid sphere creates, at an interior point located by the vector $\vec{r}$ from ITS centre, a field $\vec{E}=\dfrac{\rho\vec{r}}{3\varepsilon_{0}}$, which is simply $\dfrac{kQr}{R^{3}}$ rewritten with the density. Let $O$ be the centre of the large sphere, $O'$ that of the cavity, and $P$ any point of the cavity. The large sphere creates $\dfrac{\rho\,\overrightarrow{OP}}{3\varepsilon_{0}}$ there and the small negative sphere creates $-\dfrac{\rho\,\overrightarrow{O'P}}{3\varepsilon_{0}}$. The sum is $\dfrac{\rho}{3\varepsilon_{0}}\left(\overrightarrow{OP}-\overrightarrow{O'P}\right)=\dfrac{\rho\,\overrightarrow{OO'}}{3\varepsilon_{0}}$, by vector addition (the triangle rule). The point $P$ has DISAPPEARED from the result: the field depends only on the vector joining the two centres, so it is the same everywhere in the cavity, in direction and in magnitude. This is a remarkable and thoroughly counterintuitive result, obtained in three lines only because the interior field of a uniform sphere is linear in $\vec{r}$.

$E=\dfrac{\rho d}{3\varepsilon_{0}}=\dfrac{6.00\times 10^{-6}\times 0.0700}{3\times 8.854\times 10^{-12}}=\dfrac{4.20\times 10^{-7}}{2.656\times 10^{-11}}=1.58\times 10^{4}$ V/m, directed from $O$ toward $O'$, that is away from the centre of the large sphere. Note what does NOT appear: neither the radius of the cavity nor that of the large sphere. Only the density and the distance between the centres matter.

Total charge: $Q=\rho\dfrac{4}{3}\pi\left(R^{3}-r_{c}^{3}\right)=6.00\times 10^{-6}\times\dfrac{4}{3}\pi\left(3.375\times 10^{-3}-1.25\times 10^{-4}\right)=8.17\times 10^{-8}$ C, that is $81.7$ nC. At $50.0$ cm, we are far compared with all the dimensions of the object, and it can be treated as a point charge: $E=\dfrac{kQ}{r^{2}}=\dfrac{8.988\times 10^{9}\times 8.17\times 10^{-8}}{0.250}=2.94\times 10^{3}$ V/m. We must be honest about the status of this result, however: unlike the perfectly spherical case, it is only an APPROXIMATION, valid because $50.0$ cm is large compared with $R$. The off-centre cavity gives the object a dipole moment that adds a correction in $\dfrac{1}{r^{3}}$, negligible here but not zero.

If $\vec{d}=\vec{0}$, the expression gives $\vec{E}=\vec{0}$: the field is zero throughout the centred cavity. This is perfectly consistent, and it is even a check of the derivation, since we recover by another route the classic result of Gauss's law: a uniform spherical shell creates no field in its inner cavity, because a Gaussian surface drawn in the central void encloses no charge. The off-centre case is therefore the generalization of the centred case, and it shows that the shell's zero field is not due to some mysterious property of the void but to the exact cancellation of two contributions which, as soon as the symmetry is broken, stop compensating each other.

Insulating sheet: $E=\dfrac{\sigma}{2\varepsilon_{0}}=\dfrac{8.00\times 10^{-9}}{2\times 8.854\times 10^{-12}}=452$ N/C, on both sides. Conductor: $E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{8.00\times 10^{-9}}{8.854\times 10^{-12}}=904$ N/C just outside, and zero inside the metal. The ratio is exactly 2, and you must be able to explain it, not just observe it.

For the sheet, the Gaussian pillbox is SYMMETRIC: its two end caps are on either side of the plane, in vacuum, and the field leaves through both. The total flux is $2EA$, equal to $\dfrac{\sigma A}{\varepsilon_{0}}$, hence the factor 2 in the denominator. For the conductor, the pillbox is ASYMMETRIC, and that is the whole point: one end cap is outside, in vacuum, the other is buried INSIDE THE METAL, where the field is strictly zero. The inner end cap therefore lets no flux through, nor does the curved wall since the field is parallel to it there, and all the flux leaves through the single outer end cap: $EA=\dfrac{\sigma A}{\varepsilon_{0}}$, with no factor 2. The factor 2 is thus not a property of the material, it is the number of faces through which the flux escapes.

An ISOLATED charged conducting plate carries charge on BOTH its faces, each at density $\sigma$. A point just outside therefore sees two sheets, the near face and the far face, each creating $\dfrac{\sigma}{2\varepsilon_{0}}$ in the same direction outside: the total is $2\times\dfrac{\sigma}{2\varepsilon_{0}}=\dfrac{\sigma}{\varepsilon_{0}}$. Inside the metal, on the other hand, the two contributions are opposite and cancel, which gives back $E=0$. The two explanations, that of b) by the Gaussian surface and this one by superposition, are strictly equivalent. The practical moral is to always ask whether the stated $\sigma$ is that of ONE face or that of the whole plate, because the two answers differ by a factor of two.

$P=\dfrac{\sigma^{2}}{2\varepsilon_{0}}=\dfrac{(8.00\times 10^{-9})^{2}}{2\times 8.854\times 10^{-12}}=\dfrac{6.40\times 10^{-17}}{1.771\times 10^{-11}}=3.61\times 10^{-6}$ Pa. This is tiny, three millionths of a pascal against a hundred thousand pascals for the atmosphere, and it is a reminder that electrostatic forces only become mechanically noticeable at much higher fields. The pressure is directed OUTWARD whatever the sign of the charge because $\sigma$ enters it SQUARED: physically, the surface charges, all of the same sign, repel each other and tend to push the surface apart. Changing the sign of all the charges changes nothing about their mutual repulsion.

A bump with a radius of curvature ten times smaller carries, at the common potential, a surface density about TEN times greater, exactly as for the smaller of two spheres joined by a wire, the classic point effect. The local field, proportional to $\sigma$, is therefore ten times more intense. And the pressure, proportional to $\sigma^{2}$, is a HUNDRED times greater. This last factor explains a concrete phenomenon: a tip subjected to an intense field feels a much stronger pull than the rest of the surface, and on a heated metal it can stretch until it tears off, which triggers an arc. This is the mechanism of field-emission breakdown in vacuum tubes and accelerators, and the reason why electrodes meant for very high voltages are electropolished.

False, and this is the most frequent confusion in the chapter. A zero flux means that the BALANCE between what goes in and what comes out is zero, not that nothing passes through. The cube of exercise 1 in a uniform field is the minimal example: the field is $350$ N/C everywhere, it enters through one face and leaves through the opposite one, and the total flux is zero. Correct statement: a zero flux only indicates that the enclosed charge is zero, and it says nothing about the field point by point. The converse is true: if the field is zero everywhere on the surface, so is the flux.

False as written, and a word is missing. Gauss's law is a general law of electrostatics, valid for ANY charge distribution and ANY closed surface, with no symmetry assumption whatsoever. What is restricted to the three symmetries is its ability to give the field: only they allow $E$ to be pulled out of the integral. Correct statement: the law is always true, but it is only COMPUTATIONALLY useful if a symmetry provides a surface on which the magnitude of the field is constant and its direction known. Elsewhere, it remains usable as a constraint, as in exercise 6, where it provides the field of each sphere in the superposition.

The conclusion is right, the justification is wrong. Outside charges do create a field on the surface, and it is not small: a charge of $50$ nC one centimetre from a face produces more than four million volts per metre, as in 1e). What is zero is their contribution to the TOTAL FLUX: their field lines cross the closed surface from side to side, entering on one side and leaving on the other, and the two contributions cancel exactly. Correct statement: outside charges contribute to the field but not to the net flux, and that is why only $Q_{\text{enc}}$ appears on the right-hand side.

False. Outside, a spherical Gaussian surface encloses the same total charge in both cases, and the spherical symmetry is the same: the law therefore gives strictly the same field, $\dfrac{kQ}{r^{2}}$. Correct statement: seen from outside, any distribution with spherical symmetry is indistinguishable from an equal point charge placed at the centre, whether it is hollow, solid, uniform or of varying density as in exercise 5. This is what makes the law so powerful and, at the same time, what marks its limit: it cannot be used to probe the internal structure of an object from outside. It is also for the gravitational version of the same theorem that Newton had to prove that a planet attracts as if all its mass were at its centre.

The conclusion is right, the reason is wrong and even absurd: at the centre of a charged sphere, you are not far from any charge, you are surrounded by all of it. The right explanation is that of the law. As you move toward the centre, the Gaussian surface encloses only a fraction of the charge, that of the inner ball, and this fraction decreases as $r^{3}$ while the area only decreases as $r^{2}$: the quotient therefore decreases as $r$. Correct statement: the field decreases toward the centre because the charge outside the Gaussian surface stops contributing, its effects cancelling exactly, and not because you would be moving away from anything.