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Electric Fields and Gauss' Law - Physics with Calculus

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Termini in questo insieme (20)

  • What is the general expression for the electric field from a charge distribution?
    The electric field \(\vec{E} = k \int \frac{dq}{(r')^2} \hat{r}'\) where dq is the charge element and \(\hat{r}'\) is the unit vector from the charge.
  • How is the electric force on a point charge related to the electric field?
    The force \(\vec{F} = q \vec{E}\), where q is the charge experiencing the field \(\vec{E}\).
  • What does Gauss's Law relate in electrostatics?
    Gauss's Law relates the electric flux through a closed surface to the total charge enclosed: \(\oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\epsilon_0}\).
  • Define electric flux through a surface.
    Electric flux is the amount of electric field passing through a surface, calculated as \(\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta\).
  • What is the significance of choosing a Gaussian surface with symmetry?
    Choosing a Gaussian surface where the electric field magnitude is constant and perpendicular simplifies the integral, allowing \(E\) to be factored out.
  • What is the shape of the Gaussian surface used for a point charge?
    A sphere centered on the point charge, where the electric field is radial and constant in magnitude at a fixed radius.
  • What is the formula for the electric field outside a uniformly charged sphere of total charge Q?
    Outside the sphere, \(E = \frac{Q}{4 \pi \epsilon_0 r^2}\), identical to a point charge at the center.
  • How do you calculate the total charge enclosed q_in inside a Gaussian surface?
    Integrate the charge density over the volume inside the Gaussian surface: \(q_{in} = \int \rho dV\).
  • What is the electric field inside a hollow cylindrical shell with charge density ρ between radii R1 and R2 for r < R1?
    Inside the hollow region (r < R1), the enclosed charge is zero, so \(E = 0\).
  • How is the electric field outside a long charged cylindrical shell calculated?
    Using a cylindrical Gaussian surface, \(E = \frac{\rho (R_2^2 - R_1^2)}{2 \epsilon_0 r}\), where r is the radial distance from the axis.
  • What is the electric field from an infinite charged sheet with surface charge density σ?
    The field is constant and perpendicular to the sheet: \(E = \frac{\sigma}{2 \epsilon_0}\).
  • Why does the electric field from a conducting sheet differ from that of an infinite charged sheet?
    In a conductor, the field inside is zero, so the field emerges only on one side, unlike the infinite sheet where it emerges on both sides.
  • What happens to charges inside a cavity within a conductor when a charge +q is placed inside?
    Negative charges move to the cavity surface to cancel the field inside the conductor, and an equal positive charge appears on the conductor's outer surface.
  • How does Gauss's Law apply in a 2D world with point charges?
    The closed surface is a circle, and the electric flux is through the circumference, adapting Gauss's Law to 2D geometry.
  • What is the integral form of Gauss's Law for electric flux?
    \(\Phi = \oint \vec{E} \cdot d\vec{A}\), representing the total electric flux through a closed surface.
  • How do you simplify the flux integral when the electric field is constant over the surface?
    The flux becomes \(\Phi = E \times A\), where A is the total surface area.
  • What is the electric field at radius r outside a uniformly charged sphere of radius R0?
    The field is \(E = \frac{Q_{TOT}}{4 \pi \epsilon_0 r^2}\), where Q_TOT is the total charge.
  • How do you calculate the charge enclosed inside a cylindrical volume with charge density ρ between R1 and R2?
    Integrate over the volume: \(q_{in} = \rho \pi (R_2^2 - R_1^2) L\), where L is the cylinder length.
  • What is the electric field inside the charged part of a cylindrical shell at radius r?
    The field is \(E = \frac{\rho}{2 \epsilon_0} \frac{r^2 - R_1^2}{r}\).
  • How does the electric field behave for a variable charge density ρ(r) in a sphere?
    Gauss's Law still applies if ρ depends only on radius; the Gaussian surface remains a spherical shell.