Electric Field: Gauss' Law

In summary: E_1 = q/(\epsilonAcos(0)) because electric field and area vector are parallelE_2 = q/(\epsilonAcos(180)) because electric field and area are in perpendicular directions.adding them:E_1 = (r1+r2)*(D/2)E_2 = -(r1+r2)*(D/2)E_1+E_2 = DandE_1 = \lambda/(\epsilon*2\pir)+\lambda/(\epsilon*2\pir)= \lambda/(\epsilon\pir
  • #1
Electra
2
0
1. As a member of a team of storm physicists, you are attempting to replicate lightning by chargeing two long cables stretched over a canyon. one cable will attain a highly positive and uniform density of -[itex]\lambda[/itex] and the other will attain the same amount of charge density, but opposite in sign(i.e.-[itex]\lambda[/itex]. Since the appareance of lightning directly depends on the electric field strength created by charge separation, it is important to derive an expression for electric field strength at all points between the two cables (albeit near the midpoint of the wires).

Given Data:
[itex]\lambda[/itex]= +/- 40 [itex]\mu[/itex]C
D=30.0m
r=15.0 m




2. (a) The cables are sufficiently long as to be considered infinitely long. Calculate the magnitude of the electric field strength between the two cables as a function of [itex]\lambda[/itex] and r(the distance from the positively charged cable. Use epsilon as the permittivity of free space and assume the wires are separated y a Distance D
(b) if the charge density separation between teh two cables is 40[itex]\mu[/itex]C and the distance between the two cables is 30.0m, calculate the electric field midway between the two cables.




3. The Attempt at a Solution :
well its asking for electric field between two line charges so fine the net vector sum of the electric fields
(a) Using Gauss' Law
E_1 = q/([itex]\epsilon[/itex]Acos(0)) because electric field and area vector are parallel
E_2 = q/([itex]\epsilon[/itex]Acos(180)) because electric field and area are in diff directions

adding them:
E_1 = [itex]\lambda[/itex]l/([itex]\epsilon[/itex]2[itex]\pi[/itex]rl*1) A=2pi(r)(l)
E_2 = -[itex]\lambda[/itex]l/([itex]\epsilon[/itex]2[itex]\pi[/itex]rl*-1)

E_1+E_2 = [itex]\lambda[/itex]/([itex]\epsilon[/itex]*2[itex]\pi[/itex]r)+[itex]\lambda[/itex]/([itex]\epsilon[/itex]*2[itex]\pi[/itex]r)
= [itex]\lambda[/itex]/([itex]\epsilon[/itex][itex]\pi[/itex]r)

(b) inputting the values for lambda, r and epsilon
= 40*10^-6/ (8.85*10^-12*[itex]\pi[/itex]*(30/2))= 9.59*10^4 N/C

Yet no matter how many times i plug that in, it keeps saying I'm wrong.

 
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  • #2
Pleasee, i urgently need help :(
 
  • #3
The equation is a bit more complicated than the equation to answer the second part of the problem. The second part of the problem is simple Gauss' Law. Just remember that r=D/2
 
  • #4
1. I see nothing wrong with the way you did this, and I get the same answer. Who says it's the wrong answer?

2. Same deal except you use two cylindrical surfaces of different radii. r1 + r2 = D
 
  • #5




I would recommend checking your calculations and units to ensure accuracy. It is important to double check all calculations, especially when dealing with equations involving multiple variables. Additionally, make sure that your units are consistent throughout the calculation. It may also be helpful to check your answer using an online calculator or consulting with a colleague to verify your results. If the issue persists, it may be necessary to review the concept of Gauss' Law and its application to line charges.
 

1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle and exerts a force on other charged particles within its range. It is a vector field, meaning it has both magnitude and direction.

2. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the flux of an electric field through a closed surface to the charge enclosed by that surface. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

3. How is Gauss' Law used to calculate electric fields?

Gauss' Law can be used to calculate electric fields in situations with high symmetry, such as a point charge or a charged sphere. By choosing a Gaussian surface that encloses the charge distribution, the electric flux can be calculated and then used to determine the electric field.

4. What is a Gaussian surface?

A Gaussian surface is an imaginary closed surface used in Gauss' Law calculations. It is chosen to have a symmetrical shape that makes it easier to calculate the electric flux. The choice of Gaussian surface is arbitrary, as long as it encloses the charge distribution being studied.

5. Can Gauss' Law be applied to any charge distribution?

Yes, Gauss' Law is a general law that can be applied to any charge distribution. However, in order for it to be useful in calculating electric fields, the charge distribution must have a high degree of symmetry so that a Gaussian surface can be chosen that simplifies the calculations.

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