Physics 2 Quiz: Choosing Gaussian Surfaces

What this quiz covers

This quiz focuses on Choosing Gaussian Surfaces, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The approach fails to directly yield $\vec{E}$ because the charge distribution is not cylindrically symmetric (it depends on $\phi$), so $\vec{E}$ varies in direction and magnitude on the Gaussian surface, making $\oint \vec{E} \cdot d\vec{A} \neq E(2\pi r L)$. (correct answer)
2. The approach succeeds because the total enclosed charge is $Q_{enc} = \rho_0 \pi R^2 L / 2$, and the cylindrical surface still satisfies the requirement that flux can be computed as $E(2\pi r L)$ for any charge distribution that is uniform in the $z$-direction.
3. The approach fails because the $\cos^2(\phi)$ dependence means the net enclosed charge is zero (the positive and negative lobes cancel), so Gauss's law gives $E = 0$ outside, which contradicts the actual field distribution.
4. The approach succeeds because, even though $\vec{E}$ is not uniform in magnitude on the cylindrical surface, Gauss's law still uniquely determines $E$ at each point on the surface by requiring that the total flux equal $Q_{enc}/\varepsilon_0$, and $Q_{enc}$ can be computed by integrating $\rho$ over the cylinder's cross-section.

Question 2

1. The spherical Gaussian surface is appropriate for both cases, but gives different results: $E = kQ/r^2$ for the solid sphere and $E = kQ/(r^2 - R^2)$ for the shell, because the shell creates a field-free interior that modifies the exterior solution via boundary conditions.
2. The spherical Gaussian surface is appropriate for both cases and gives the same result, $E = kQ/r^2$, because both charge distributions have spherical symmetry and the same $Q_{enc} = Q$ for $r > R$; the internal structure of the distribution does not affect the exterior field. (correct answer)
3. The spherical Gaussian surface is appropriate only for the solid sphere, not the shell, because the shell has no interior charge and Gauss's law requires a nonzero volume charge density within the enclosed region to determine a nonzero exterior field.
4. The spherical Gaussian surface is appropriate for both cases and gives the same enclosed charge $Q_{enc} = Q$, but the resulting field differs because the surface charge distribution of the shell creates a discontinuity in $\vec{E}$ at $r = R$, making the Gaussian surface result valid only asymptotically as $r \to \infty$.

Question 3

1. The claim is correct in principle but uses the wrong surface. A spherical Gaussian surface of large radius $r \gg R$ should be used instead, where the cylinder appears as a point charge, giving a more accurate far-field approximation than the cylindrical surface.
2. The claim is correct. At distances $r \gg R$, the field approaches that of a line charge with linear charge density $\lambda = \frac{3}{4}\rho\pi R^2$, and the cylindrical Gaussian surface yields $E \approx \lambda/(2\pi\varepsilon_0 r)$ because the charge distribution becomes approximately cylindrically symmetric as seen from far away.
3. The claim is incorrect. The offset cavity breaks the exact cylindrical symmetry of the charge distribution, so $\vec{E}$ varies in both magnitude and direction with azimuthal angle on any coaxial cylindrical surface. Gauss's law still gives the correct total flux equal to $Q_{enc}/\varepsilon_0$, but without symmetry this single scalar equation cannot determine $\vec{E}$ at a specific point on the surface. (correct answer)
4. The claim is incorrect. The offset cavity creates a net linear dipole moment in the charge distribution, which means the field at large $r$ falls off faster than $1/r$ (the cylindrical monopole rate), and the Gaussian surface approach overestimates the field by neglecting this correction.

Question 4

1. The enclosed charge is zero
2. E is constant on the end caps
3. E is parallel to the end caps (correct answer)
4. The line charge is infinite

Question 5

1. No surface has uniform E (correct answer)
2. Sphere centered at midpoint
3. Cylinder along dipole axis
4. Pillbox bisecting the charges

Question 6

1. Cylinder through shell center
2. Concentric sphere around shell (correct answer)
3. Cube centered at shell center
4. Sphere tangent to the shell

Question 7

1. Invalid: E not normal to faces
2. Better than a centered sphere
3. Valid and yields E from flux
4. Valid but E varies on faces (correct answer)

Question 8

1. To make E perpendicular
2. To make the side flux zero
3. To enclose a nonzero charge (correct answer)
4. To ensure a closed surface

Question 9

1. The student should have used the displacement field form $\oint \vec{D} \cdot d\vec{A} = Q_{enc,free}$, which directly gives $D = Q/(4\pi r^2)$ and then $E = D/(\kappa\varepsilon_0) = Q/(4\pi\kappa\varepsilon_0 r^2)$. Using $\vec{E}$ with only the free charge on the right-hand side is never valid inside a dielectric medium.
2. There is no error. The result $E = kQ/r^2$ is correct inside the dielectric because the bound surface charges on the inner and outer surfaces of the shell cancel each other exactly, leaving the total enclosed charge equal to $+Q$ and the vacuum form of Gauss's law fully applicable.
3. The spherical Gaussian surface is invalid inside the dielectric because polarization creates a non-uniform bound charge distribution that breaks the spherical symmetry of $\vec{E}$, making the flux integral $\oint \vec{E} \cdot d\vec{A} \neq E(4\pi r^2)$.
4. The student used the vacuum form $\oint \vec{E} \cdot d\vec{A} = Q_{enc,total}/\varepsilon_0$ but inserted only $Q_{free} = +Q$, omitting the bound surface charge $Q_{bound}$ on the inner surface of the dielectric shell at $r = R_2$, which is also enclosed. Including $Q_{bound} = -Q(1-1/\kappa)$ gives $Q_{enc,total} = Q/\kappa$ and $E = Q/(4\pi\kappa\varepsilon_0 r^2)$. (correct answer)

Question 10

1. The surface must be a sphere of radius $r$ centered at the origin, because $\rho$ depends only on radial distance (spherical symmetry), which guarantees that $\vec{E}$ is radial and constant in magnitude on this surface, making $\oint \vec{E} \cdot d\vec{A} = E(4\pi r^2) = Q_{enc}/\varepsilon_0$. (correct answer)
2. The surface must be a sphere of radius $r$ centered at the origin, but only if the student first verifies that $\rho(r)$ is uniform; since this density is non-uniform in $r$, the Gaussian surface approach cannot directly yield $E$, and the flux integral cannot be factored, requiring a full numerical volume integration instead.
3. The surface can be any closed surface enclosing the charge inside radius $r$, because Gauss's law states that the total flux through any closed surface equals $Q_{enc}/\varepsilon_0$, and knowing the total flux is sufficient to determine $E$ at any point on the surface regardless of its shape.
4. The surface must be a sphere of radius $r$ centered at the origin, and additionally $\rho(r)$ must be uniform for the symmetry argument to hold; because $\rho$ is non-uniform here, a cylindrical Gaussian surface should be used instead, since it better captures the radial gradient of the charge density.

Question 11

1. Valid step: The field just outside any conductor is perpendicular to the surface by the boundary condition, ensuring $\vec{E} \cdot d\vec{A} = E_n A$ on the external face with no tangential contribution. Common error: Students use the full conductor result $E_n = \sigma/\varepsilon_0$ when the conductor is isolated, rather than $E_n = \sigma/(2\varepsilon_0)$, which applies when the conductor is in the presence of external fields.
2. Valid step: The face inside the conductor contributes zero flux because the conductor is grounded, and a grounded conductor has no surface charge, so $\sigma = 0$ and $E_n = 0$ outside as well. Common error: Students forget the grounding condition and incorrectly assign a nonzero $\sigma$ to a grounded plane, leading to a nonzero external field.
3. Valid step: The face inside the conductor contributes zero flux because $\vec{E} = 0$ inside a conductor in electrostatic equilibrium, so all flux exits through the external face, giving $E_n = \sigma/\varepsilon_0$. Common error: Students often set $E_n = \sigma/(2\varepsilon_0)$, incorrectly applying the result for an isolated infinite sheet of charge (which has flux through both faces) to this conductor boundary (where one face contributes nothing). (correct answer)
4. Valid step: By Gauss's law, the flux through the pillbox equals $\sigma A/\varepsilon_0$, and since the side walls of the pillbox contribute zero flux ($\vec{E}$ is parallel to them), only the two flat faces matter. Common error: Students include the side-wall flux when the pillbox thickness is nonzero, effectively adding a spurious volume-charge contribution that inflates $Q_{enc}$.

Question 12

1. Concentric spherical Gaussian surfaces work only for $r > b$, because inside the shell (both regions $r < a$ and $a < r < b$) the induced charges on the conductor surfaces create asymmetric fringing fields that break the spherical symmetry needed for the Gaussian surface to be useful.
2. Concentric spherical Gaussian surfaces work only for $r < a$ and $r > b$, not inside the conductor ($a < r < b$), because Gauss's law cannot be applied inside conducting materials where free charges are present and the field is not purely radial.
3. Concentric spherical Gaussian surfaces work in all three regions, but the symmetry argument differs: for $r < a$ and $r > b$ the argument relies on the point charge, while for $a < r < b$ the field is zero by a separate conductor boundary condition rather than by the Gaussian surface construction.
4. Concentric spherical Gaussian surfaces work in all three regions. In each region, the total charge distribution (point charge plus induced surface charges) has spherical symmetry, so $\vec{E}$ is radial and constant in magnitude on any concentric sphere, allowing $E(4\pi r^2) = Q_{enc}/\varepsilon_0$. (correct answer)

Question 13

1. The surface is valid because the net charge distribution has cylindrical symmetry with no angular or axial dependence, ensuring $\vec{E}$ is radial and uniform in magnitude on the curved surface, and zero on the flat end-caps. The enclosed charge per unit length is $+2\pi a \sigma$, since only the inner conductor lies inside the Gaussian surface. (correct answer)
2. The surface is valid because cylindrical symmetry guarantees $\vec{E}$ is radial and uniform in magnitude on the curved surface. The enclosed charge per unit length is zero, however, because Gauss's law depends on the total charge of the entire system — inner plus outer conductor — which sums to zero by design.
3. The surface is valid because the field in the region $a < r < b$ is zero by the cylindrical shell theorem, exactly analogous to the interior of a spherical shell. The enclosed charge per unit length is $+2\pi a \sigma$, but this is irrelevant because the outer shell completely shields the inner conductor from the exterior region.
4. The surface is valid because cylindrical symmetry ensures $\vec{E}$ is radial and uniform on the curved surface. The enclosed charge per unit length is $+\sigma(2\pi a) - \sigma(a/b)(2\pi b) = 0$, since Gauss's law requires including all conductors within the physical system, not just those geometrically inside the chosen surface.

Question 14

1. A rectangular pillbox with faces of area $A$ parallel to the $xy$-plane, with one face at $z = +z_0$ and the other at $z = -z_0$. Because $\rho \propto |z|$ is symmetric about $z = 0$, the field is zero at $z = 0$; by symmetry the bottom face at $z = -z_0$ also contributes zero flux, so only the top face contributes, giving $EA = Q_{enc}/\varepsilon_0$.
2. A rectangular pillbox with faces of area $A$ parallel to the $xy$-plane, with one face at $z = +z_0$ and the other at $z = -z_0$. The symmetry $\rho(-z) = \rho(z)$ requires $\vec{E}(-z_0) = -\vec{E}(z_0)$, so both faces contribute outward flux of magnitude $E(z_0)A$, giving $2E(z_0)A = Q_{enc}/\varepsilon_0$ and correctly yielding $E(z_0)$. (correct answer)
3. A cylindrical Gaussian surface with its axis along the $z$-direction, radius $r$, and length $2z_0$, centered at the origin. The translational symmetry in $x$ and $y$ means $\vec{E}$ is directed radially outward from the $z$-axis, so the curved surface provides all the flux and $E$ is determined from $E(2\pi r)(2z_0) = Q_{enc}/\varepsilon_0$.
4. A spherical Gaussian surface of radius $z_0$ centered at the origin. Because $\rho$ depends only on $|z|$ and the sphere encloses charge that is symmetric about the origin, $\vec{E}$ is radially outward and constant in magnitude over the sphere, giving $E(4\pi z_0^2) = Q_{enc}/\varepsilon_0$.