Electric Fields and Gauss's Law Notes

The Electric Field

  • Definition: E = \frac{F}{q_0} (SI unit: N/C)
  • Electric field lines help visualize the electric field vector E.
    • E is tangent to the electric field lines at any point.
    • The magnitude of E is proportional to the density of the field lines.
    • Lines begin on positive charges and end on negative charges.

Electric Dipole

  • A system of two equal charges of opposite sign (\pm q) separated by a distance d.
  • Electric dipole moment ($\vec{p}$):
    • Magnitude: p = qd
    • Direction: From -q to +q

Electric Field due to Dipole

  • Using superposition, the electric field E generated by the electric dipole is:
    • E = \frac{1}{4\pi\epsilon0} \frac{q}{(z - d/2)^2} - \frac{1}{4\pi\epsilon0} \frac{q}{(z + d/2)^2}
    • If d \ll z, then E \approx \frac{1}{2\pi\epsilon_0} \frac{p}{z^3}

Torque on Dipoles in Electric Fields

  • In a uniform electric field E, the net force on a dipole is zero.
  • The torque is \tau = p \times E .

Continuous Charge Distribution

  • Divide the charge distribution into elements of volume dV with charge dq = \rho dV (
  • The electric field dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} where r is the distance from dq to point P.
  • Sum all contributions: E = \int dE

Electric Field of a Uniformly Charged Ring

  • At a point P on the axis of a ring of radius R with total charge q:
    • E = \frac{z q}{4 \pi \epsilon_0 (z^2 + R^2)^{3/2}}, where z is the distance from the ring center to point P.

Electric Field Lines

  • Always begin on a positive charge and end on a negative charge.
  • The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.

Parallel Plate Capacitor

  • Charge density \sigma = q/A
  • Electric field E = \frac{q}{\epsilon0 A} = \frac{\sigma}{\epsilon0}

Electric Flux

  • \Phi_E = E A \cos \phi
  • A measure of the “flow” of the electric field through a surface.

Gauss's Law

  • \PhiE = \oint E \cdot dA = \frac{q}{\epsilon0}
  • The electric flux through a closed surface equals the net charge q enclosed by the surface divided by \epsilon_0.

Applying Gauss' Law

  1. Sketch the charge distribution.
  2. Identify the symmetry.
  3. Choose a Gaussian surface that simplifies flux calculation.
  4. Use Gauss's law to find E.

Electric Field of a Long, Uniformly Charged Rod

  • Using a cylindrical Gaussian surface:
    • E = \frac{\lambda}{2 \pi \epsilon_0 r}, where \lambda is the linear charge density.

Electric Field of a Thin, Infinite, Nonconducting Sheet

  • Using a cylindrical Gaussian surface:

    • E = \frac{\sigma}{2 \epsilon_0}, where \sigma is the surface charge density.

    Electric Field of Two Parallel Conducting Infinite Planes

  • E = \frac{\sigma}{\epsilon_0}

Electric Field Inside a Conductor

  • Any excess charge resides on the surface.
  • The electric field E is 0 at any point within the conducting material.
  • Electric field lines are perpendicular to the surface.

Electric Field of a Spherical Shell of Charge q and Radius R

  • Inside the shell (r < R): E = 0
  • Outside the shell (r > R): E = \frac{q}{4 \pi \epsilon_0 r^2}

Electric Field of a Uniformly Charged Sphere of Radius R and Charge q

  • Outside the sphere (r > R): E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}
    • Inside the sphere (r < R): E = \frac{1}{4 \pi \epsilon_0} \frac{q}{R^3} r