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:
  • $E = \frac{1}{4\pi\epsilon<em>0} \frac{q}{(z - d/2)^2} - \frac{1}{4\pi\epsilon</em>0} \frac{q}{(z + d/2)^2}$
  • 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 $E = \frac{q}{\epsilon<em>0 A} = \frac{\sigma}{\epsilon</em>0}$

Electric Flux

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

Gauss's Law

• $\Phi<em>E = \oint E \cdot dA = \frac{q}{\epsilon</em>0}$
• 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): $E = \frac{1}{4 \pi \epsilon_0} \frac{q}{R^3} r$