Gauss's Law - AP Physics C Electricity and Magnetism

A very large, thin, nonconducting sheet has a uniform positive surface charge density $\sigma$ . What is the magnitude of the electric field at a point near the center of the sheet?
Answer: $E = \frac{\sigma}{2\epsilon_0}$ Use a cylindrical Gaussian surface (a 'pillbox') that pierces the sheet, with its flat caps of area $A$ parallel to the sheet. The charge enclosed is $q_{enc} = \sigma A$ . The electric field is perpendicular to the sheet, so flux passes only through the two caps, not the curved side. The total flux is $\Phi_E = EA + EA = 2EA$ . Gauss's law gives $2EA = \sigma A / \epsilon_0$ , which simplifies to $E = \frac{\sigma}{2\epsilon_0}$ . The field is uniform and does not depend on the distance from the sheet.
Gauss's law provides a fundamental relationship between which two physical quantities?
Answer: The net electric flux through a closed surface and the net charge enclosed by that surface. Gauss's law, in integral form, is $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$ . The left side of the equation represents the net electric flux through a closed surface, and the right side involves the net charge enclosed by that surface. Therefore, Gauss's law relates net electric flux and enclosed charge. Distractor B is incorrect because the law relates the integral of the field (flux) to only the enclosed charge. Distractor C relates to work and potential, which is not what Gauss's law directly describes. Distractor D incorrectly refers to force on the surface itself.
The electric field in a certain region of space is given by $\vec{E} = C r^2 \hat{r}$ , where $C$ is a positive constant and $\hat{r}$ is the radial unit vector. What is the total charge enclosed within a spherical surface of radius $R$ centered at the origin?
Answer: $4\pi\epsilon_0 C R^4$ According to Gauss's law, $q_{enc} = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$ . For a spherical surface of radius $R$ , the electric field is constant in magnitude $E = CR^2$ and everywhere parallel to $d\vec{A}$ . The surface integral becomes $E \cdot A = (CR^2)(4\pi R^2) = 4\pi C R^4$ . Therefore, the enclosed charge is $q_{enc} = \epsilon_0 (4\pi C R^4) = 4\pi\epsilon_0 C R^4$ .
A point charge $+Q$ is located at the center of a spherical Gaussian surface of radius $R$ . The net electric flux through the surface is $\Phi_E$ . If the radius of the sphere is doubled to $2R$ while the enclosed charge remains at the center, what is the new net electric flux?
Answer: $\Phi_E$ According to Gauss's Law ( $\Phi_E = q_{enc}/\epsilon_0$ ), the net electric flux through a closed surface depends only on the net charge enclosed by the surface. It does not depend on the size or shape of the Gaussian surface. Since the enclosed charge $+Q$ remains the same, the net electric flux remains unchanged.
A point charge $+q$ is placed at the exact center of a cube of side length $L$ . What is the electric flux through one of the six faces of the cube?
Answer: $\frac{q}{6\epsilon_0}$ According to Gauss's Law, the total electric flux through the entire closed surface of the cube is $\Phi_{total} = \frac{q_{enc}}{\epsilon_0} = \frac{q}{\epsilon_0}$ . Because the charge is at the center of the cube, the situation is symmetric. The flux is distributed equally among the six identical faces of the cube. Therefore, the flux through a single face is $\Phi_{face} = \frac{\Phi_{total}}{6} = \frac{q}{6\epsilon_0}$ .
A conducting sphere of radius $R=0.10\ \text{m}$ is isolated and placed in electrostatic equilibrium. It carries a total charge $Q=+6.0\times10^{-9}\ \text{C}$ uniformly distributed on its surface. Outside the conductor, the field is spherically symmetric, and a spherical Gaussian surface of radius $r>R$ encloses all the charge. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ . Applying Gauss's Law, $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2)=\frac{Q}{\varepsilon_0}.$ Using the situation described, determine the electric field at a point $r=0.30\ \text{m}$ from the sphere's center.
Answer: $5.99\times10^2\ \text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, allowing for the calculation of electric fields in symmetric situations. In this scenario, the charged conducting sphere provides spherical symmetry that simplifies calculations, using a spherical Gaussian surface outside the conductor. Choice A is correct because it correctly applies Gauss's Law for r > R, where E = Q/(4πε₀r²) = (6.0×10⁻⁹)/[4π(8.85×10⁻¹²)(0.30)²] = 5.99×10² N/C. Choice C is incorrect because it appears to have made an order of magnitude error, possibly in the calculation or unit conversion. To help students: Emphasize that for conducting spheres, all charge resides on the surface and the field outside behaves like a point charge. Practice using the correct formula E = kQ/r² or E = Q/(4πε₀r²) and being careful with units and powers of 10.
An infinite, nonconducting plane sheet lies in the $xy$ -plane ( $z=0$ ) and carries a uniform surface charge density $\sigma=+5.0\times10^{-9}\,\text{C/m}^2$ . Take $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$ . Because the plane is infinite and uniformly charged, the electric field must be perpendicular to the plane and have the same magnitude at all points a given distance above or below it. A student chooses a cylindrical "pillbox" Gaussian surface of cross-sectional area $A=0.020\,\text{m}^2$ that straddles the plane, with its flat faces parallel to the plane and located symmetrically at $z=+h$ and $z=-h$ . The flux through the curved side is zero, so $\Phi_E=EA+EA=2EA$ . The enclosed charge is $Q_{\text{enc}}=\sigma A$ . Applying Gauss's Law, $2EA=Q_{\text{enc}}/\varepsilon_0$ , gives the field magnitude on either side. Based on the scenario, what is the magnitude of the electric field at a point just above the plane?
Answer: $2.8\times10^{2}\,\text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields near an infinite charged plane (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, allowing for the calculation of electric fields in symmetric situations. In this scenario, the infinite charged plane provides planar symmetry that simplifies calculations, using a cylindrical 'pillbox' Gaussian surface that straddles the plane. Choice A is correct because it correctly applies Gauss's Law: E = σ/(2ε₀) = (5.0×10⁻⁹)/(2×8.85×10⁻¹²) = 5.0×10⁻⁹/(1.77×10⁻¹¹) = 2.82×10² N/C ≈ 2.8×10² N/C. Choice B would be incorrect if students forgot the factor of 2 in the denominator, a common error. To help students: Emphasize that for infinite planes, E is constant and perpendicular to the plane. Practice setting up pillbox Gaussian surfaces and recognizing that flux only passes through the flat faces.
An insulating sphere of radius $R$ has a total positive charge $Q$ distributed uniformly throughout its volume. What is the magnitude of the electric field $E$ at a distance $r < R$ from the center of the sphere?
Answer: $E = \frac{kQr}{R^3}$ For a Gaussian surface of radius $r < R$ , the enclosed charge is $q_{enc} = Q \frac{V_{enc}}{V_{total}} = Q \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = Q\frac{r^3}{R^3}$ . Applying Gauss's law, $\oint \vec{E} \cdot d\vec{A} = E(4\pi r^2) = \frac{q_{enc}}{\epsilon_0} = \frac{Qr^3}{\epsilon_0 R^3}$ . Solving for $E$ gives $E = \frac{Qr^3}{4\pi\epsilon_0 r^2 R^3} = \frac{1}{4\pi\epsilon_0} \frac{Qr}{R^3} = \frac{kQr}{R^3}$ .
An infinite line of charge with uniform linear density $\lambda$ produces an electric field whose magnitude is proportional to $1/r$ . If the line of charge is finite, with length $L$ , why can Gauss's Law not be easily used to find the electric field at a distance $r$ from its midpoint?
Answer: A finite wire lacks the necessary symmetry to simplify the flux integral. For a finite wire, the electric field lines are not purely radial but curve outwards near the ends. This means that on a cylindrical Gaussian surface, the magnitude of $E$ is not constant, and the field is not everywhere perpendicular to the curved surface. This lack of symmetry prevents us from simplifying the flux integral $\oint \vec{E} \cdot d\vec{A}$ to $E \cdot A$ , making it difficult to solve for $E$ . Gauss's Law is always exact (D is false), but its practical application for finding E relies on symmetry.
A solid, nonconducting sphere of radius $R=0.20\ \text{m}$ has a uniform volume charge density $\rho=+2.0\times10^{-6}\ \text{C/m}^3$ . The charge is fixed in place, and the distribution is perfectly spherically symmetric. A student selects a spherical Gaussian surface of radius $r=0.10\ \text{m}$ (inside the sphere). Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ . For $r<R$ , the enclosed charge is $Q_{\text{enc}}=\rho\left(\frac{4}{3}\pi r^3\right)$ , and the flux is $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2)=Q_{\text{enc}}/\varepsilon_0$ . Using the situation described, what is the magnitude of the electric field at $r=0.10\ \text{m}$ from the center?
Answer: $3.77\times10^3\ \text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, allowing for the calculation of electric fields in symmetric situations. In this scenario, the uniformly charged solid sphere provides spherical symmetry, and we're finding the field inside the sphere using a Gaussian surface at r < R. Choice A is correct because it correctly applies Gauss's Law for r < R, where E = ρr/(3ε₀) = (2.0×10⁻⁶)(0.10)/[3(8.85×10⁻¹²)] = 3.77×10³ N/C. Choice C is incorrect because it's off by a factor of 10, likely due to a calculation error or incorrect handling of the units. To help students: Emphasize the difference between fields inside and outside uniformly charged spheres. Practice recognizing that inside a uniformly charged sphere, E increases linearly with r, while outside it decreases as 1/r².
A conducting sphere of radius $R=0.050\ \text{m}$ is in electrostatic equilibrium and has total charge $Q=+1.2\times10^{-8}\ \text{C}$ on its surface. Inside a conductor in electrostatic equilibrium, the electric field is zero everywhere. A student considers a spherical Gaussian surface of radius $r=0.030\ \text{m}$ located entirely within the conducting material. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ . Although Gauss's Law always holds, the enclosed charge for this interior Gaussian surface is $Q_{\text{enc}}=0$ , so $\oint \vec{E}\cdot d\vec{A}=0$ . Using the situation described, what is the magnitude of the electric field at $r=0.030\ \text{m}$ from the center?
Answer: $0\ \text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, and for conductors in electrostatic equilibrium, the electric field inside is always zero. In this scenario, we're considering a Gaussian surface inside the conducting material where no charge can exist in equilibrium. Choice A is correct because inside any conductor in electrostatic equilibrium, E = 0 N/C everywhere, regardless of the total charge on the surface or the position within the conductor. Choices B, C, and D are incorrect because they suggest non-zero fields inside the conductor, which violates the fundamental property of conductors in electrostatic equilibrium. To help students: Emphasize that this is a fundamental property of conductors - charges redistribute on the surface to ensure E = 0 inside. Practice recognizing conductor problems and remembering that Gauss's Law still applies but Qenc = 0 for any surface inside the conductor.
A solid conducting sphere of radius $R_1$ has a net charge of $+2Q$ . It is surrounded by a concentric conducting spherical shell of inner radius $R_2$ and outer radius $R_3$ , which has a net charge of $-3Q$ . What is the magnitude of the electric field $E$ in the region $R_1 < r < R_2$ ?
Answer: $E = \frac{k(2Q)}{r^2}$ To find the electric field in the region between the sphere and the shell ( $R_1 < r < R_2$ ), we draw a spherical Gaussian surface with radius $r$ . The charge enclosed by this surface is only the charge on the inner sphere, which is $+2Q$ . By spherical symmetry and Gauss's law, the field is equivalent to that of a point charge $+2Q$ at the origin. Therefore, the magnitude of the electric field is $E = \frac{1}{4\pi\epsilon_0} \frac{|+2Q|}{r^2} = \frac{k(2Q)}{r^2}$ .
A conducting sphere of radius $R=0.040\,\text{m}$ carries total charge $Q=+6.0\times10^{-9}\,\text{C}$ uniformly on its outer surface. Take $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$ . In electrostatic equilibrium, the electric field inside the conducting material is zero, and outside the sphere the field is spherically symmetric. A student considers a spherical Gaussian surface centered on the sphere with radius $r=0.020\,\text{m}$ , which lies entirely inside the conductor ( $r<R$ ). The enclosed charge for this Gaussian surface is $Q_{\text{enc}}=0$ because all excess charge resides on the outer surface. Applying Gauss's Law, $E(4\pi r^2)=Q_{\text{enc}}/\varepsilon_0$ , yields the field magnitude at that radius. Based on the scenario, what is the magnitude of the electric field at $r=0.020\,\text{m}$ from the center?
Answer: $0\,\text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields inside a conducting sphere (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, and for conductors in electrostatic equilibrium, the field inside is always zero. In this scenario, the Gaussian surface at r=0.020 m is entirely inside the conductor where no charge exists in the bulk material. Choice A is correct because it correctly recognizes that inside any conductor in electrostatic equilibrium, E = 0 N/C, regardless of the charge on the surface. Choices B, C, and D would be incorrect as they suggest non-zero fields inside a conductor, violating the fundamental property of conductors. To help students: Emphasize that E = 0 everywhere inside a conductor is a fundamental result. Practice recognizing that Gaussian surfaces inside conductors always enclose zero charge because all excess charge migrates to the surface.
An insulating sphere of radius $R$ has a charge density that varies with the distance $r$ from the center as $\rho(r) = \rho_0 (r/R)$ , where $\rho_0$ is a positive constant. What is the magnitude of the electric field at a distance $r < R$ from the center?
Answer: $E = \frac{\rho_0 r^2}{4\epsilon_0 R}$ First, find the enclosed charge $q_{enc}$ within a radius $r$ by integrating the density over the volume: $q_{enc} = \int_0^r \rho(r') 4\pi (r')^2 dr' = \int_0^r \rho_0 \frac{r'}{R} 4\pi (r')^2 dr' = \frac{4\pi\rho_0}{R} \int_0^r (r')^3 dr' = \frac{\pi\rho_0 r^4}{R}$ . By Gauss's Law, $E(4\pi r^2) = \frac{q_{enc}}{\epsilon_0} = \frac{\pi\rho_0 r^4}{\epsilon_0 R}$ . Solving for $E$ gives $E = \frac{\pi\rho_0 r^4}{4\pi r^2 \epsilon_0 R} = \frac{\rho_0 r^2}{4\epsilon_0 R}$ .
A spherical Gaussian surface of radius $R$ is centered at the origin. A point charge $+Q$ is located on the x-axis at $x=2R$ . What is the net electric flux through the surface of the sphere?
Answer: $0$ Gauss's Law states that the net electric flux through a closed surface is proportional to the net charge enclosed by that surface. In this case, the charge $+Q$ is located at $x=2R$ , which is outside the spherical Gaussian surface of radius $R$ . Since the enclosed charge is zero, the net electric flux through the surface is zero.
A student wishes to use Gauss's law to determine the electric field produced by a charge distribution. Which of the following is a necessary property of the Gaussian surface they choose?
Answer: The surface must be a closed three-dimensional surface that encloses a volume. For Gauss's law to be applied, the Gaussian surface must be a closed surface, meaning it must fully enclose a volume with no holes. This is a fundamental requirement for the surface integral to be well-defined. Choice A describes surfaces that are useful for symmetric problems but not a general requirement. Choice B is also a feature of a useful application but not a necessary property of the surface itself. Choice C is incorrect; a Gaussian surface is a mathematical construct, not a physical object.
A point charge $+q$ is placed at the center of a thick, spherical, neutral conducting shell with inner radius $R_1$ and outer radius $R_2$ . What is the surface charge density $\sigma_2$ on the outer surface of the shell?
Answer: $\sigma_2 = +\frac{q}{4\pi R_2^2}$ The electric field inside the conducting material ( $R_1 < r < R_2$ ) must be zero. A Gaussian surface inside the conductor must enclose zero net charge. Therefore, a charge of $-q$ is induced on the inner surface at $R_1$ . Since the shell was initially neutral, by conservation of charge, a charge of $+q$ must appear on the outer surface at $R_2$ . The surface charge density is this charge divided by the outer surface area, so $\sigma_2 = \frac{+q}{4\pi R_2^2}$ .
A cube of side length $L$ is placed with one corner at the origin and its sides aligned with the positive x, y, and z axes. The electric field in the region is given by $\vec{E} = ax\hat{i} + b\hat{k}$ , where $a$ and $b$ are positive constants. What is the net electric flux through the surface of the cube?
Answer: $aL^3$ The net flux is the sum of fluxes through all six faces. The $\hat{k}$ component of the field ( $b\hat{k}$ ) is uniform. It enters the bottom face ( $z=0$ ) and exits the top face ( $z=L$ ), so its net flux is zero. The $\hat{i}$ component ( $ax\hat{i}$ ) contributes flux only through the faces at $x=0$ and $x=L$ . At $x=0$ , $E_x=0$ , so flux is zero. At $x=L$ , $E_x=aL$ and the area vector is $L^2\hat{i}$ , so the flux is $(aL)(L^2) = aL^3$ . The net flux is the sum, which is $aL^3$ .
Gauss's law is theoretically valid for any closed surface and any charge distribution. However, it is only practically useful for calculating the electric field when...
Answer: the charge distribution possesses a high degree of symmetry, simplifying the flux integral. To solve for $E$ using Gauss's law, one must be able to evaluate the flux integral $\oint \vec{E} \cdot d\vec{A}$ . In situations with high spherical, cylindrical, or planar symmetry, we can choose a Gaussian surface where the magnitude of $E$ is constant and the angle between $E$ and $dA$ is constant (often 0 or 90 degrees). This allows $E$ to be factored out of the integral, making it possible to solve for $E$ . Without this symmetry, the integral is generally too complex to solve analytically.
A very long, solid, nonconducting cylinder of radius $R=4.0\ \text{cm}$ has uniform volume charge density $\rho=+8.0\times10^{-7}\ \text{C/m}^3$ . The cylinder is effectively infinite. A student chooses a coaxial cylindrical Gaussian surface of radius $r=2.0\ \text{cm}$ (inside the material) and length $L=0.60\ \text{m}$ . Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$ . For $r<R$ , $Q_{\text{enc}}=\rho(\pi r^2L)$ and Gauss's Law gives $E(2\pi rL)=Q_{\text{enc}}/\varepsilon_0$ . Based on the scenario, determine the electric field magnitude at $r=2.0\ \text{cm}$ .
Answer: $9.04\times10^2\ \text{N/C}$ This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to calculate electric fields (College Board AP Physics C standard). Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface, allowing for the calculation of electric fields in symmetric situations. In this scenario, we're finding the field inside a uniformly charged cylinder at r < R, where the field increases linearly with distance from the axis. Choice A is correct because it correctly applies E = ρr/(2ε₀) = (8.0×10⁻⁷)(0.02)/[2(8.85×10⁻¹²)] = 9.04×10² N/C for the field inside the cylinder. Choice B is incorrect because it appears to have doubled the correct result, possibly confusing the formula for inside versus outside the cylinder. To help students: Emphasize the difference between E ∝ r inside and E ∝ 1/r outside for cylindrical charge distributions. Practice recognizing when r < R versus r > R and applying the correct formula for each region.

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