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

1. 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)

2. 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

3. 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?
 
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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