Gauss's Law Practice Test

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 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.