Physics 2 Quiz: Gausss Law For Electric Fields

What this quiz covers

This quiz focuses on Gausss Law For Electric Fields, 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. Between the planes: $E = \frac{3\sigma}{2\epsilon_0}$ directed rightward; to the right of Plane 2: $E = \frac{\sigma}{2\epsilon_0}$ directed leftward. (correct answer)
2. Between the planes: $E = \frac{\sigma}{2\epsilon_0}$ directed rightward; to the right of Plane 2: $E = \frac{\sigma}{2\epsilon_0}$ directed leftward.
3. Between the planes: $E = \frac{3\sigma}{2\epsilon_0}$ directed leftward; to the right of Plane 2: $E = \frac{\sigma}{2\epsilon_0}$ directed rightward.
4. Between the planes: $E = \frac{\sigma}{\epsilon_0}$ directed rightward; to the right of Plane 2: $E = \frac{\sigma}{\epsilon_0}$ directed leftward.

Question 2

1. $Q_{enc} = \epsilon_0 E_0 L^2$, found by multiplying the field gradient $E_0$ by the cross-sectional area $L^2$ and $\epsilon_0$, omitting the length $L$ that appears when the enclosed volume is properly accounted for.
2. $Q_{enc} = 2\epsilon_0 E_0 L^3$, found by noting that each face perpendicular to $\hat{x}$ has flux $E_0(L/2)L^2$, and summing both faces gives $2 \times E_0(L/2)L^2 = E_0 L^3$... but a factor of 2 is included because both faces contribute positively.
3. $Q_{enc} = 0$, because the electric field $\mathbf{E} = E_0 x\hat{x}$ is antisymmetric about the origin, making the flux through the left face equal and opposite to the flux through the right face, yielding zero net flux.
4. $Q_{enc} = \epsilon_0 E_0 L^3$, found by computing the net flux through the two faces perpendicular to $\hat{x}$ (the flux through the other four faces is zero since $\mathbf{E} \cdot \hat{n} = 0$ on them) and applying Gauss's law. (correct answer)

Question 3

1. The student's conclusion is wrong because Gauss's law only applies to charge distributions with planar, cylindrical, or spherical symmetry. A dipole has neither, so Gauss's law cannot be applied at all in this configuration.
2. The student correctly applies Gauss's law to conclude the net flux is zero, but incorrectly equates zero net flux with zero field. Gauss's law relates the total flux through a closed surface to enclosed charge; it does not imply $E = 0$ pointwise unless $\mathbf{E}$ is uniform in magnitude and normal to the surface, allowing it to factor out of the integral. (correct answer)
3. The student's error is in the choice of Gaussian surface: a sphere centered between the charges encloses both $+q$ and $-q$, so the enclosed charge is not zero but $+q - q = 0$... wait, that is zero. The error is that the surface must be centered on one charge only for Gauss's law to be valid.
4. The student's argument is flawed because Gauss's law requires the Gaussian surface to pass through the region where the field is to be determined. Since the student wants $E$ on the surface, not inside it, the law is being misapplied and a different surface geometry must be used.

Question 4

1. $r$
2. $1/r^2$
3. $r^2$
4. $r^3$ (correct answer)

Question 5

1. inward $kq/r^2$
2. zero field magnitude
3. outward $2kq/r^2$
4. outward $kq/r^2$ (correct answer)

Question 6

1. 2 times (correct answer)
2. 4 times
3. 1/2 times
4. 16 times

Question 7

1. $1/r$
2. $r^2$
3. $r$ (correct answer)
4. $1/r^2$

Question 8

1. zero field magnitude
2. $\sigma/\varepsilon_0$ (correct answer)
3. $\sigma/(2\varepsilon_0)$
4. $2\sigma/\varepsilon_0$

Question 9

1. $\frac{E(R)}{E(2R)} = 1$, because both expressions for $E$ in their respective regions yield the same value when evaluated at $r = R$ and $r = 2R$ respectively, since the field is continuous at the surface.
2. $\frac{E(R)}{E(2R)} = 2$, because $E(R) = \rho_0 R/(3\epsilon_0)$ inside and $E(2R) = \rho_0 R/(6\epsilon_0)$ outside, giving a ratio of 2. (correct answer)
3. $\frac{E(R)}{E(2R)} = \frac{3}{4}$, because $E(R) = \rho_0 R/(3\epsilon_0)$ and $E(2R) = 4\rho_0 R/(9\epsilon_0)$, using a uniform-density approximation for the exterior field.
4. $\frac{E(R)}{E(2R)} = \frac{1}{2}$, because outside the cylinder the field falls as $1/r$ and doubling the radius halves the field, while the field at the surface uses the interior formula evaluated at $r = R$, which gives the same result as the exterior formula at $r = 2R$.

Question 10

1. $E = 0$, because the outer conducting shell completely screens the interior region from the inner conductor's charge, just as a grounded shell would, making the field zero between the conductors.
2. $E = \frac{2\lambda}{2\pi\epsilon_0\left(\frac{a+b}{2}\right)}$, because the Gaussian surface at $r = (a+b)/2$ encloses the inner conductor's $+\lambda$ plus an additional $+\lambda$ induced on the inner surface of the outer shell, doubling the enclosed charge.
3. $E = \frac{\lambda}{2\pi\epsilon_0\left(\frac{a+b}{2}\right)}$, because only the inner conductor's charge contributes within the Gaussian surface; the outer shell's charges reside on its inner and outer surfaces but the net charge enclosed by a surface between the conductors is only $+\lambda$. (correct answer)
4. $E = \frac{\lambda}{\pi\epsilon_0\left(a+b\right)}$, because the field between coaxial conductors is computed using the average radius $(a+b)/2$ in the denominator and a factor of $\pi\epsilon_0$ rather than $2\pi\epsilon_0$.

Question 11

1. $E = 0$ inside the conductor; outer surface charge density $\sigma_{outer} = \frac{2Q}{4\pi b^2}$, because the $-Q$ point charge induces $+Q$ on the inner surface, leaving $+2Q$ on the outer surface from the shell's net charge. (correct answer)
2. $E = 0$ inside the conductor; outer surface charge density $\sigma_{outer} = \frac{4Q}{4\pi b^2}$, because the $-Q$ point charge induces $+Q$ on the inner surface and the full $+3Q$ of the shell migrates to the outer surface, giving a total outer charge of $+4Q$.
3. $E = 0$ inside the conductor; outer surface charge density $\sigma_{outer} = \frac{3Q}{4\pi b^2}$, because the net charge of the shell is $+3Q$ and this charge resides entirely on the outer surface regardless of the interior point charge.
4. $E = \frac{kQ}{r^2}$ directed inward inside the conductor material; outer surface charge density $\sigma_{outer} = \frac{2Q}{4\pi b^2}$, because the point charge $-Q$ creates a non-zero field even within the conducting material when a free charge is present at the center.

Question 12

1. The claim is incorrect: Gauss's law requires knowing $\rho(r)$ explicitly to evaluate $Q_{enc}$ at an arbitrary $r$ inside the shell, and without this the flux integral cannot be solved — knowing only the total charge $Q$ is insufficient for intermediate radii.
2. The claim is incorrect: although Gauss's law applies, the spherical symmetry is broken by the non-uniform distribution, so the electric field is not purely radial and the surface integral $\oint \mathbf{E}\cdot d\mathbf{A}$ cannot be equated to $E\cdot 4\pi r^2$.
3. The claim is partially correct: Gauss's law can determine $E(r)$ for $r > R_2$ from the total charge $Q$, but inside the shell the enclosed charge at radius $r$ depends on the distribution, so $E(r)$ for $R_1 < r < R_2$ cannot be found without $\rho(r)$. (correct answer)
4. The claim is correct: because the charge density depends only on $r$, the system retains full spherical symmetry, and Gauss's law with a spherical surface of radius $r$ gives $E(r) = Q_{enc}/(4\pi\epsilon_0 r^2)$ where $Q_{enc}$ is uniquely determined by $Q$ and the geometry alone.

Question 13

1. $E = 0$ everywhere in this region, because the total charge per unit length of the system is zero and the electric field must therefore vanish between the two cylindrical surfaces.
2. $E = \frac{\rho R^2}{8\epsilon_0 r}$ directed radially inward, because only the inner solid cylinder contributes to the enclosed charge and $\rho < 0$, making the field point inward with magnitude that decreases as $1/r$. (correct answer)
3. $E = \frac{|\rho| R^2}{8\epsilon_0 r}$ directed radially outward, because the magnitude of the enclosed negative charge produces an inward-pointing field whose magnitude equals that of an equivalent positive charge pointing outward.
4. $E = \frac{\rho R^2}{4\epsilon_0 r}$ directed radially inward, because the enclosed charge from the inner cylinder is $\rho \pi (R/2)^2 L$ per length $L$, and the factor of 2 in the denominator is omitted when applying the cylindrical Gauss's law.

Question 14

1. $\rho(r) = \frac{2\epsilon_0 E_0 r}{R^2}$ for $r < R$, with a positive surface charge density at $r = R$, found by differentiating $E(r)$ directly with respect to $r$ and multiplying by $\epsilon_0$, without accounting for the full spherical divergence.
2. $\rho(r) = \frac{4\epsilon_0 E_0 r}{R^2}$ for $r < R$, with a negative surface charge density at $r = R$ because the field just outside ($E_0 R/r$) is less than just inside ($E_0$) at $r = R$, indicating a surface charge compensates the discontinuity.
3. $\rho(r) = \frac{3\epsilon_0 E_0}{R^2}$ for $r < R$, with no surface charge at $r = R$, found by applying $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$ using only the radial derivative of $r^2 E$ without the full divergence in spherical coordinates.
4. $\rho(r) = \frac{4\epsilon_0 E_0 r}{R^2}$ for $r < R$, with no surface charge at $r = R$ because the field is continuous across the boundary. (correct answer)