Solving Gauss's Law for Electric Field at y=0.2m & y=0.6m

[carloqthegrea]

Homework Statement

A very long uniform line of charge has charge per unit length 4.8*10^-6 Coulomb/meter and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.4*106-6 coulomb/meter and is parallel to the x-axis at y=.400 meters. What is the net electric field (magnitude and direction) at the following points on the y-axis: a). y=0.200 metes b). y=0.600 meters?

Homework Equations

not sure

The Attempt at a Solution

no idea? NEED HELP PLEase!

 

[chrisk]

Relevant Equation:

\vec{E}=K\int_{-\infty}^{\infty}\frac{dq}{r^2}\hat{r}

Express dq and r in terms of a single varible and use symmetry.

 

[carloqthegrea]

I still don't know what to do can you give me more details.

 

[chrisk]

Draw a picture of one line charge along the x axis. Choose a differential element of charge, dq, somewhere on the x-axis away from the origin and choose a point on the y axis. Draw a line connecting the point on the y-axis to the charge element thus forming a right triange. This is the distance r. Use the properties of a right triangle to express r in terms of the x coordinate and the y coordinate (the y coordinate will be a constant so call it h for height above the x axis). Also, express dq in terms of x and the charge density. Note, if you choose the charge element on the other side of the y-axis the same distance from this axis the horizontal components of the E fields from the two charges cancel so you only have to determine the vertical component. Use intergration to sum up all the vertical components contributed from all the charge elements. Apply this method to the other line of charge then sum the two resultant E fields at the point h.

 

[gabbagabbahey]

Draw a picture of one line charge along the x axis. Choose a differential element of charge, dq, somewhere on the x-axis away from the origin and choose a point on the y axis. Draw a line connecting the point on the y-axis to the charge element thus forming a right triange. This is the distance r. Use the properties of a right triangle to express r in terms of the x coordinate and the y coordinate (the y coordinate will be a constant so call it h for height above the x axis). Also, express dq in terms of x and the charge density. Note, if you choose the charge element on the other side of the y-axis the same distance from this axis the horizontal components of the E fields from the two charges cancel so you only have to determine the vertical component. Use intergration to sum up all the vertical components contributed from all the charge elements. Apply this method to the other line of charge then sum the two resultant E fields at the point h.

I took "long line of charge" to essentially mean an infinitely long line of charge. In that case, Gauss' Law is much easier than using your method.

@carloqthegrea---Just use a Gaussian cylinder of length 'L' and radius 'r', with the x-axis as its axis of symmetry. What is the charge enclosed by the cylinder? Can you reason that \vec{E} is axial and uniform over your Gaussian surface? What does that make the flux through the surface?

 

[chrisk]

The problem states to find the net E field at a point on the y axis. This is the contribution of each charge element considered as a point charge. Guass's Law does not apply to this problem e.g. how would one find the net electric field created by three point charges at a particular point in space?

 

[gabbagabbahey]

The problem states to find the net E field at a point on the y axis. This is the contribution of each charge element considered as a point charge. Guass's Law does not apply to this problem e.g. how would one find the net electric field created by three point charges at a particular point in space?

You don't have three point charges--you have an infinitely long continuous charge distribution. Gauss' Law works just fine and this is a common introductory level application of Gauss' Law. See for example, 'Griffiths Introduction to Electrodynamics' 3rd ed. Prob. 2.13.

 

[chrisk]

It can be done both ways. Sorry, gabbagabbahey. See the below link.

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html

 

[carloqthegrea]

sweet that was plenty of help, I solved the problem thanks guys.