Gauss's Law

(b) No. All three charges make a field on D. Zero net flux only means what goes in comes out; it says nothing about the field being zero.

(c) Unchanged, \(4q/\varepsilon_0\). Flux depends only on the enclosed charge, not on the size or shape of the surface.

\(A = 0.20\times0.30 = 0.060\ \text{m}^2\), so \(EA = 210\ \text{N}\cdot\text{m}^2/\text{C}\).

The angle in \(\Phi = EA\cos\theta\) is always measured from the normal. Part (b) exists to catch the angle-to-the-plane mistake.

\(q/\varepsilon_0 = 12\times10^{-9}/8.85\times10^{-12} = 1356\ \text{N}\cdot\text{m}^2/\text{C}\).

(a) Whole cube: \(1356\). Six equivalent faces: \(1356/6 = \boxed{226\ \text{N}\cdot\text{m}^2/\text{C}}\) each.

(b) Imagine eight identical cubes meeting at the corner where \(q\) sits. Together they surround \(q\) and receive all \(q/\varepsilon_0\); by symmetry each gets one-eighth:

The three faces that touch the charge's corner have \(\mathbf E\) lying in their planes (the field is radial from that corner), so their flux is 0. The other three faces share the 169 equally: \(\boxed{56.5\ \text{N}\cdot\text{m}^2/\text{C}}\) each, which is \(q/(24\varepsilon_0)\).

The density varies, so \(Q_{\text{enc}}\) must be integrated over thin spherical shells of volume \(4\pi r'^2\,dr'\):

(d) Both give \(\rho_0R/(4\varepsilon_0)\) at \(r = R\) ✓. Inside, \(E\propto r^2\) rather than \(r\): more of the charge is near the surface than in the uniform case. The common error is \(Q_{\text{enc}} = \rho\cdot\tfrac43\pi r^3\), which is only valid for constant \(\rho\).

Radially outward. Because \(E\propto1/r\), the field is strongest right at the wire, so breakdown (corona) starts at the conductor surface. High-voltage lines use thick conductors or bundles of several wires to keep the surface field below this limit.

Each sheet alone makes \(\sigma/(2\varepsilon_0)\), away from a positive sheet and toward a negative one:

Check: outside, the pair looks like a single sheet of net \(+2.0\ \mu\text{C/m}^2\), whose field is \(1.13\times10^{5}\) N/C pointing away on both sides ✓.

(a) A Gaussian sphere inside the metal has \(E = 0\), so it encloses zero net charge: the inner surface has \(+2.0\) nC. The shell is neutral, so the outer surface has \(-2.0\) nC.

(c) The distribution on the inner surface changes (more \(+\) charge gathers on the side nearest the point charge), but its total is still \(+2.0\) nC. The outer surface charge stays uniform and the field outside is unchanged: the metal hides the cavity's details from the outside world.

(d) Grounding lets the outer surface's \(-2.0\) nC drain away. Total enclosed charge inside any sphere outside the shell is then \(-2.0 + 2.0 = 0\), so \(E = 0\) outside. A grounded shield completely hides the charge inside.

(a) Cylindrical symmetry. Gaussian cylinder radius \(r\), length \(\ell\), between the conductors: flux only through the curved side.

All radially outward. The field is \(b/a = 3.75\) times stronger at the inner conductor than at the outer one; the dielectric there is the most stressed.

(c) A cylinder outside the cable encloses \(+\lambda\ell - \lambda\ell = 0\), so \(E = 0\). No field escapes — the cable neither interferes with neighbours nor picks up their (static) fields.

(d) A Gaussian cylinder inside the outer conductor's metal has \(E = 0\), so it encloses zero: the outer conductor's inner surface carries \(-\lambda\), and its outer surface carries nothing. The charges face each other across the dielectric, exactly as in a capacitor (Unit 4).

Check (b) another way: \(kQ/R^2 = 3.0\times10^{6}\) gives \(Q = (3.0\times10^{6})(0.010)/(8.99\times10^{9}) = 3.3\times10^{-6}\) C ✓.

(d) A small radius of curvature reaches the breakdown field with very little charge. On an irregular conductor the sharp points have the highest \(\sigma\) and so the highest \(\sigma/\varepsilon_0\); the air breaks down there first. That is why high-voltage equipment has smooth, large-radius terminals, and why lightning rods are pointed.

Cylindrical symmetry: \(\mathbf E\) radial from the axis, depending only on \(r\). Coaxial Gaussian cylinder of radius \(r\), length \(\ell\); caps contribute no flux.

Both equal \(\rho R/(2\varepsilon_0)\) at \(r = R\) ✓ (no surface charge, so no jump).

(b) With \(\lambda = \rho\pi R^2\): \(E = \lambda/(2\pi\varepsilon_0r)\) outside — identical to a thin line of charge. From outside, a cylindrically symmetric charge looks like a line on its axis (the cylindrical cousin of the shell theorem).

(c) For a short cylinder the field near the ends is not radial and its magnitude varies along the length. Gauss's law is still true, but \(E\) is no longer constant on the curved side (and the caps now have flux), so it cannot be pulled out of the integral. Symmetry, not the law, is what fails.