Electrostatics with Conductors Practice Test

According to the text: In the scenario Capacitors and Charge Storage, a parallel-plate capacitor has two conducting plates separated by an insulating dielectric. When a battery previously established a potential difference $V$ and was then removed, equal and opposite charges $\pm Q$ remained on the plates; the conductors allow charges to move freely until each plate becomes an equipotential surface. The electric field is concentrated in the dielectric region, and energy is stored in the capacitor’s electric field, consistent with $U=\tfrac{1}{2}CV^2$ and $C=\varepsilon A/d$ for plate area $A$ and separation $d$ (SI units: $V$ in volts, $Q$ in coulombs, $C$ in farads). Historically, early capacitors such as Leyden jars demonstrated charge separation using conductors and an insulator. In modern circuits, capacitors smooth voltage, filter signals, and provide short-term energy storage. How does a capacitor store energy according to the passage?

Based on the passage: In the scenario Gauss’s Law in Conductors, a hollow conducting shell carries net charge $+Q$ and contains an empty cavity. At electrostatic equilibrium, free charges move until the electric field inside the conductor’s bulk is zero and the conductor is an equipotential. Using Gauss’s Law, $\oint \vec{E}\cdot d\vec{A}=Q_{\text{enc}}/\varepsilon_0$, a Gaussian surface taken within the conducting material must enclose $Q_{\text{enc}}=0$, so any excess charge cannot reside in the bulk. Therefore, with no charge inside the cavity, the shell’s entire $+Q$ appears on the outer surface; if the shell is spherical, symmetry makes the surface charge density uniform. This principle supports electrostatic shielding of instrument housings and reduces unwanted discharge in high-voltage terminals. Based on the passage, how does Gauss’s Law apply to the distribution of charge on conductors?