Electromagnetism Help: Calc Electric Field w/ Gauss's Law

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SUMMARY

This discussion focuses on applying Gauss's Law to derive the electric field produced by an infinite sheet of charge with a surface density of 1 C/m², confined to the xy plane. Participants emphasize the importance of selecting an appropriate Gaussian surface, suggesting either a cylindrical surface or a rectangular prism. The key takeaway is that the divergence of the electric field is zero for all points not in the xy plane, confirming the uniform nature of the electric field produced by the infinite sheet.

PREREQUISITES
  • Understanding of Gauss's Law, specifically the equation \(\int {E.dS} = {Q}/{\epsilon}\)
  • Familiarity with electric field concepts and vector calculus
  • Knowledge of charge density and its implications in electric field calculations
  • Ability to visualize and manipulate Gaussian surfaces in electrostatics
NEXT STEPS
  • Study the derivation of electric fields from different charge distributions using Gauss's Law
  • Learn about the properties of divergence in vector fields, particularly in electrostatics
  • Explore examples of Gaussian surfaces for various symmetries, including planar, cylindrical, and spherical
  • Practice problems involving electric fields and charge distributions to reinforce understanding
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mastering electrostatics, particularly in applying Gauss's Law to calculate electric fields from charge distributions.

andyfreesty1e
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Homework Statement


Use Gauss’s law to obtain a vector expression for the electric field produced by an
infinte sheet of charge with a surface density of 1 C m−2 , confined to the xy plane. Hence
show that the divergence of this electric field is zero for all points not in the xy plane.


Homework Equations


\int {E.dS} = {Q}/{epsilon}
del.E = rho/epsilon


The Attempt at a Solution

 
Physics news on Phys.org
try picking a useful guassian surface
 
i still don't really know where to go with it, any other hints you can give?
 
Hi andyfreesty1le, welcome to PF!:smile:

First figure out what kind of symmetry the charge distribution (and hence the electric field it produces) possesses (e.g. spherical, cylindrical, planar, etc.). Then read your textbook to find examples of what kind of Gaussian surface you would use for problems with that type of symmetry.
 
Ok, so according to my textbook, i would use cylindrical gaussian surface. It tells me how to find the electric field, but i can't see how to work out the electric field vector.
 
No, you use cylindrical Gaussian surfaces when the charge distribution is cylindrically symmetric. A plane does not possesses cylindrical symmetry.
 
you could use a cylindrical surface, or a thin rectangular prism surface equivalently

place it so the flat ends of either the cylinder or prism ar parallel to the plane

consider the case when it contains the plane and when it doesn't, then relate the charge contained to the flux through the surface in each case. It may also help to consider wthen the thisness of the prism shrink to infinitesimal
 
andyfreesty1e said:

Homework Statement


Use Gauss’s law to obtain a vector expression for the electric field produced by an
infinte sheet of charge with a surface density of 1 C m−2 , confined to the xy plane. Hence
show that the divergence of this electric field is zero for all points not in the xy plane.


Homework Equations


\int {E.dS} = {Q}/{epsilon}
del.E = rho/epsilon


The Attempt at a Solution


You can use a cylinder. Or you can use a "box".

Gauss's Law states that the charge inside a volume is equal to the electric field integrated over the surface of that volume.

So you're looking for the electric field.

First, imagine a box being intersected by the charged plane.

Imagine the "height" of the box goes to zero, so we don't have to concern ourselves with the field coming out of the sides.

If the box has dimensions x^2, what is the charge inside it?

Hint: charge density is charge per area. Multiply that by the area enclosed.

Now, integrate the electric field by the surface area of the volume.

Then solve for E.
 

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