Question 2 · Física II URV Juny 2023

a. What is the value of the electric field inside the conductor? Justify your answer.

Solución: $\vec{E}_{\text{in}} = 0$ A conductor contains free charge carriers (electrons). If a non-zero electric field existed inside, an electrostatic force $\vec{F} = -e\vec{E}$ would act on them, causing current. In electrostatic equilibrium, no net macroscopic motion occurs, so charges redistribute onto the surface until the internal field is zero everywhere.

b. What is the value of the electric field just outside the surface of the conductor? Provide the mathematical derivation to support your answer.

Solución: $\vec{E} = \frac{\sigma}{\varepsilon_0}\hat{n}$ (or magnitude $E = \frac{\sigma}{\varepsilon_0}$) • Direction: The tangential component must be zero ($E_\parallel = 0$); otherwise surface charges would accelerate along the surface. Thus $\vec{E} = E_\perp \hat{n}$. • Magnitude (Gauss's Law): Take a small pillbox with cross-sectional area $\Delta A$ straddling the surface. - Inner face: $\Phi_{\text{inner}} = 0$ (since $E = 0$ inside). - Curved side wall: $\Phi_{\text{sides}} = 0$ (since $\vec{E}$ is parallel to walls). - Outer face: $\Phi_{\text{outer}} = E\Delta A$. Enclosed charge is $Q_{\text{enc}} = \sigma\Delta A$. By Gauss's law: $$\oint_S \vec{E}\cdot d\vec{A} = E\Delta A = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{\sigma\Delta A}{\varepsilon_0} \implies E = \frac{\sigma}{\varepsilon_0}$$