AP Physics C Electricity and Magnetism Quiz: Gausss Law

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?

A charge is uniformly distributed over the surface of a very long, thin, hollow cylinder of radius $R$. Which of the following Gaussian surfaces would be most appropriate to calculate the electric field at a distance $r > R$ from the axis of the cylinder?

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 $R_1 < r < R_2$?

A solid, uncharged conducting sphere of radius $R$ is placed in a uniform external electric field. After electrostatic equilibrium is reached, what is the net electric flux through a spherical Gaussian surface of radius $r < R$ located concentric with the conducting sphere?

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 $r=2.0\ \text{cm}$.

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?

A long, thin insulating rod can be modeled as an infinite line of charge with uniform linear charge density $\lambda=+2.5\times10^{-8}\,\text{C/m}$. Assume cylindrical symmetry about the line, and take $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$. A student selects a coaxial cylindrical Gaussian surface of radius $r=0.040\,\text{m}$ and length $L=0.60\,\text{m}$. The electric field is radial and constant on the curved surface, and the flux through the end caps is zero because $\vec{E}$ is parallel to those surfaces. Thus $\Phi_E=E(2\pi rL)$. The enclosed charge is $Q_{\text{enc}}=\lambda L$. Using Gauss's Law, $E(2\pi rL)=Q_{\text{enc}}/\varepsilon_0$, the student solves for $E$ at the chosen radius. Using the situation described, determine the electric field at a distance $0.040\,\text{m}$ from the line of charge.

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?

An infinite, nonconducting sheet lies in the $xy$-plane and carries a uniform surface charge density $\sigma=+4.0\times10^{-6}\ \text{C/m}^2$. Because the sheet is infinite, the electric field is perpendicular to the sheet and has the same magnitude at all points a fixed distance above or below it. A student uses a thin cylindrical "pillbox" Gaussian surface of cross-sectional area $A=0.020\ \text{m}^2$ that straddles the sheet, with its flat faces parallel to the sheet. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$. The flux through the curved side is zero because $\vec{E}$ is parallel to that surface, so Gauss's Law becomes $\oint \vec{E}\cdot d\vec{A}=EA+EA=2EA=Q_{\text{enc}}/\varepsilon_0$, where $Q_{\text{enc}}=\sigma A$. Using the situation described, what is the magnitude of the electric field at a point $0.10\ \text{m}$ above the sheet?

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.

An infinite insulating sheet carries uniform surface charge density $\sigma=-2.5\times10^{-6}\ \text{C/m}^2$ and lies in the $xy$-plane. Because of planar symmetry, $|\vec{E}|$ is constant and points perpendicular to the sheet on both sides. A student uses a pillbox Gaussian surface of area $A=0.010\ \text{m}^2$ that straddles the sheet. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$. Gauss's Law gives $2EA=|\sigma|A/\varepsilon_0$, so $E=|\sigma|/(2\varepsilon_0)$. Using the situation described, what is the magnitude of the electric field $0.20\ \text{m}$ above the sheet?

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?

A long, straight, solid insulating cylinder of radius $R=0.020\,\text{m}$ carries a uniform volume charge density $\rho=+2.0\times10^{-5}\,\text{C/m}^3$. The cylinder is much longer than any distance of interest, so end effects are negligible and the electric field depends only on the radial distance $r$ from the axis. Take $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$. A student selects a coaxial cylindrical Gaussian surface of radius $r=0.010\,\text{m}$ (so $r<R$) and length $L=0.50\,\text{m}$. Using Gauss's Law, $\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0}$, and the symmetry result that flux passes only through the curved surface so $\Phi_E=E(2\pi rL)$. The enclosed charge is $Q_{\text{enc}}=\rho(\pi r^2L)$. Using the situation described, determine the electric field at a distance $r=0.010\,\text{m}$ from the cylinder's axis.

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?

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?

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?

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?

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?

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?

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?