Podcast Beta
Questions and Answers
What assumption is crucial when applying Gauss's law to determine the electric field around the central portions of a long wire?
In Gauss's law, why are end effects ignored when considering the electric field around the central portions of a long wire?
What type of charge distribution is assumed for an infinite plane sheet when applying Gauss's law?
In the context of Gauss's law, why do the electric field lines on certain faces of a Gaussian surface not contribute to the total flux?
Signup and view all the answers
What is the direction of the unit vector normal to surface 1 of a Gaussian surface surrounding an infinite plane sheet?
Signup and view all the answers
When considering a Gaussian surface around an infinite plane sheet, which sides contribute to the net flux as per Gauss's law?
Signup and view all the answers
What is the significance of considering a rectangular parallelepiped or cylindrical surface as a Gaussian surface around an infinite plane sheet?
Signup and view all the answers
Why does the electric field not depend on y and z coordinates around an infinite plane sheet when applying Gauss's law?
Signup and view all the answers
Study Notes
Assumptions and Limitations
- The assumption of an infinitely long wire is crucial for applying Gauss's law
- Without this assumption, the electric field (E) cannot be taken as normal to the curved part of the cylindrical Gaussian surface
- Eq. (1.32) is approximately true for the electric field around the central portions of a long wire, where end effects can be ignored
Electric Field due to a Uniformly Charged Infinite Plane Sheet
- The electric field (E) due to a uniformly charged infinite plane sheet does not depend on y and z coordinates
- The direction of E at every point is parallel to the x-direction
- The Gaussian surface can be a rectangular parallelepiped or a cylindrical surface
- The flux through the Gaussian surface is 2EA, and the electric field lines are parallel to the x-axis
- The electric field (E) is given by E = σ / (4πε0r²)
Electric Field due to a Uniformly Charged Thin Shell
- The electric field outside the shell is as if the entire charge of the shell is concentrated at its centre
- The electric field (E) outside the shell is E = q / (4πε0r²)
- The electric field inside the shell is zero (E = 0) due to Gauss's law
- This result confirms the 1/r² dependence in Coulomb's law
Gauss's Law
- Gauss's law states that the total electric flux through a closed surface is proportional to the charge enclosed within the surface
- The term q on the right side of Gauss's law includes the sum of all charges enclosed by the surface
- The electric field on the left side of Gauss's law is due to all charges, both inside and outside the surface
- The Gaussian surface can be chosen to facilitate easier calculation of the electrostatic field when the system has some symmetry
Studying That Suits You
Use AI to generate personalized quizzes and flashcards to suit your learning preferences.
Description
Test your knowledge on Gauss's law and electric fields, including calculating flux through surfaces, finding enclosed charge, and determining electric field vectors. Explore scenarios with different charge distributions and Gaussian surfaces.
