The electric field of infinite line charges and infinite sheet of charges

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Homework Help Overview

The discussion revolves around calculating the electric field generated by infinite line charges and infinite sheets of charge in free space. The original poster presents two questions involving specific charge distributions and asks for assistance in determining the electric field at given points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of formulas for electric fields due to line charges and sheets of charge, questioning the definitions of variables such as \(\rho\) and \(\sigma\). Some participants suggest showing calculations step-by-step to clarify misunderstandings.

Discussion Status

There is an ongoing exploration of the correct application of formulas and the superposition principle. Participants are actively questioning assumptions and seeking clarification on the calculations provided by the original poster.

Contextual Notes

Some participants express confusion regarding the correct interpretation of the variables in the equations and the application of the superposition principle. There is a mention of differing results from calculations, indicating potential misunderstandings in the approach to the problems.

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Homework Statement


Q1,Infinite uniform line charges of 5 nC/m lie along the x and y axes in free space. Find E at (a), P(0, 0 , 4); and (b), P (0 ,3 , 4)


Q2, Three infinite uniform sheets of charge are located in free space as follows: 3 nC/m2 at z= -4, 6 nC/m2 at z=1, -8 nC/m2 at z=4. Find E at point P (2,5,-5)

Homework Equations


E=ρL/2∏ε0ρ
Q1,I used this got: a), E=22.5z V/m
b), E=10.8y + 14.4z V/m
Q2, I used E= ρ/2ε0

The Attempt at a Solution


Q1, The right answer is a), E=45z V/m b), E = 10.8y+36.9z V/m


I really confused in this topic, So wish your guys help me. Thanks a lot.
 
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Is the [itex]\rho[/itex] in the denominator of your Eline function supposed to be the density? Or is it the displacement from the line of charge? Also, it would be helpful if you could show your calculations step-by-step because I get the correct answer for (a) using

[itex]E = \frac{\lambda}{2\pi\epsilon_{0}r}[/itex]

as you had there, combined with the superposition principle. And I got the correct answer for (b) using the superposition principle and the same function you listed of

[itex]E = \frac{\sigma}{2\epsilon_{0}}[/itex]

So I don't know if you just aren't superposing the charge distributions' respective fields properly or if you made some incorrect substitutions. So post your calculations step-by-step so we can actually see where you might be going wrong.
 
In Q1, the ρ is the density i think.
Here is the step for Q1:
a), E=ρL/2∏ε0ρ
= (5*10-9/2*∏*ε0*42)*4z
=22.5z V/m

b), E={5*10-9/2*∏*ε0*(32+42)}*(3y+4z)
=10.8y + 14.4z V/m


For Q2, i still can't got the answer that using the E=σ/2ϵ0,
in this question i just calculate each of three with the formula above and then add them to find the E. could you show me the step?
 
wjswanson said:
is the [itex]\rho[/itex] in the denominator of your eline function supposed to be the density? Or is it the displacement from the line of charge? Also, it would be helpful if you could show your calculations step-by-step because i get the correct answer for (a) using

[itex]e = \frac{\lambda}{2\pi\epsilon_{0}r}[/itex]

as you had there, combined with the superposition principle. And i got the correct answer for (b) using the superposition principle and the same function you listed of

[itex]e = \frac{\sigma}{2\epsilon_{0}}[/itex]

so i don't know if you just aren't superposing the charge distributions' respective fields properly or if you made some incorrect substitutions. So post your calculations step-by-step so we can actually see where you might be going wrong.

In Q1, the ρ is the density i think.
Here is the step for Q1:
a), E=ρL/2∏ε0ρ
= (5*10-9/2*∏*ε0*42)*4z
=22.5z V/m

b), E={5*10-9/2*∏*ε0*(32+42)}*(3y+4z)
=10.8y + 14.4z V/m


For Q2, i still can't got the answer that using the E=σ/2ϵ0,
in this question i just calculate each of three with the formula above and then add them to find the E. could you show me the step?
 

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