#3 Slide Notes

Recap of Gauss’s Law

• Gauss’s Law Overview

  • States that the electric flux through a closed surface is proportional to the charge enclosed within that surface.

  • The mathematical expression for Gauss's Law is given by: $\oint<em>S \vec{E} \cdot d\vec{A} = \frac{Q</em>{enc}}{\epsilon_0}$

  • When the charge is outside a Gaussian surface, the total flux through the surface is zero.

Applications of Gauss’s Law

1. Field of a Sphere of Charge

• Gaussian Surface: Consider a spherical Gaussian surface of radius 'r' within a uniformly charged sphere of radius 'R'.

• Flux Integral: The electric field strength inside the sphere is calculated using Gauss's Law.

• Charge Enclosed: The charge enclosed within the Gaussian surface impacts the calculated electric field.

• Electric Field Strength:

  • The strength of the electric field inside a uniformly charged sphere (r < R) increases linearly with the distance 'r' from the center of the sphere, given by $E = \frac{1}{4\pi\epsilon_0} \frac{Qr}{R^3}$.

  • Outside the uniformly charged sphere ($r \ge R$), the electric field is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.

2. Electric Field of a Long Charged Wire

• Modeling: A long charged wire is modeled as an infinitely long line of charge.

• Gaussian Surface: Use a cylindrical Gaussian surface of length 'l' around the charged wire. The wire has a linear charge density denoted by $\lambda$.

• Calculation Using Gauss’s Law:

  • The electric field at a distance 'r' from the wire can be derived from the symmetry of the problem.

  • Substituting these values into Gauss’s Law yields a formula relating electric field strength to distance from the wire: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.

3. Electric Field of a Plane of Charge

• Uniformly Charged Plane: The plane is assumed to have a uniform surface charge density, denoted by $\eta$, and extends infinitely in all directions.

• Planar Symmetry: The electric field generated points either straight toward or away from the plane.

• Gaussian Surface: A cylindrical Gaussian surface of length 'L' and faces of area 'A' is used, centered on the plane of charge.

  • No electric flux passes through the walls of the cylinder because the electric field is perpendicular to them.

  • The electric field is perpendicular to both end faces of the cylinder.

• Flux Calculation: The electric flux through the cylinder is evaluated to determine the electric field generated by an infinite charged plane.

  • The electric field strength due to an infinite plane of charge is given by $E = \frac{\eta}{2\epsilon_0}$.

Electric Flux and Charge Relationships

• Example Problem: Electric flux is demonstrated through two Gaussian surfaces, prompting the comparison of enclosed charges.

• Possible Charge Relations:

  • A. q1 = 2q; q2 = q

  • B. q1 = q; q2 = 2q

  • C. q1 = 2q; q2 = −q

  • D. q1 = 2q; q2 = −2q

  • E. q1 = q/2; q2 = q/2

• Quick Check: Evaluate the relationships to reinforce understanding of Gauss's Law and electric flux.