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

In summary: Is the \rho 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 the superposition principle. And i got the correct answer for (b) using the superposition principle and the same function you listed ofe = \frac{\lambda}{2\pi\epsilon_{0}r}as you had there, combined with the superposition principle. And i got the correct answer for (c) using the superposition principle and the functione = \frac{\sigma}{
  • #1
dada130500015
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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)

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5


Hello,

Thank you for sharing your work and questions. It seems like you are on the right track with your calculations, but there may be some errors in your equations and units conversions.

For Q1, the correct equation for the electric field due to an infinite line charge is E = λ/(2πε0ρ), where λ is the linear charge density and ρ is the distance from the line charge. So for part a), using the given linear charge density of 5 nC/m, the equation should be E = (5 nC/m)/(2πε0*4) = 39.8 z V/m. For part b), the distance from the line charges is √(3^2 + 4^2) = 5, so the equation should be E = (5 nC/m)/(2πε0*5) = 31.8 z V/m.

For Q2, the correct equation for the electric field due to an infinite sheet of charge is E = σ/2ε0, where σ is the surface charge density. So for point P, the equation should be E = (6 nC/m^2)/(2ε0) = 3 nC/m^2. This is a constant electric field in the z direction with magnitude 3 N/C.

I hope this helps clarify things for you. It's important to double check your equations and units to ensure accuracy in your calculations. Keep up the good work!
 

1. What is an infinite line charge?

An infinite line charge is a hypothetical charge that extends infinitely in one direction. It is often used in theoretical physics and calculations involving the electric field.

2. How does an infinite line charge affect the electric field?

An infinite line charge creates a radial electric field that decreases with distance from the charge. The direction of the electric field is determined by the sign of the charge, with positive charges creating an outward field and negative charges creating an inward field.

3. What is an infinite sheet of charge?

An infinite sheet of charge is a hypothetical charge that extends infinitely in two dimensions. It is often used in theoretical physics and calculations involving the electric field.

4. How does an infinite sheet of charge affect the electric field?

An infinite sheet of charge creates a uniform electric field in the region surrounding it. The magnitude of the electric field is constant and the direction is perpendicular to the sheet of charge.

5. What is the difference between the electric field of an infinite line charge and an infinite sheet of charge?

The main difference is the shape of the electric field. The electric field of an infinite line charge is radial, while the electric field of an infinite sheet of charge is uniform. Additionally, the electric field of an infinite sheet of charge is constant in magnitude, while the electric field of an infinite line charge decreases with distance from the charge.

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