# Calculate the electric field strength and the direction on the z-axis resultant

1. Aug 25, 2008

### SilverBullet

1. The problem statement, all variables and given/known data
Calculate the electric field strength and the direction on the z-axis resultant from 3 line charges of blah blah and blah located at (x,y), (x1,y1), (x2,y2) respectively.

2. Relevant equations
I assume I just use E = k.Q/r^2

3. The attempt at a solution
So I work out the electric fields and add them together. I don't understand the z-axis part

2. Aug 25, 2008

### Defennder

You might want to type out what you wrote in your post as "blah blah and blah". In particular, how are the lines of charge oriented? If they are all parallel to the z-axis, just exploit symmetry for the answer.

3. Aug 25, 2008

### Redbelly98

Staff Emeritus
They're only interested in the field at the z-axis.

I suggest looking up the field of a line charge. You're equation is for point charges.

4. Aug 25, 2008

### SilverBullet

Oops, I was trying to keep the question less restricted. I don't seem to be able to edit my post but it should read: " resultant from 3 line charges of 2nC/m, 1nC/m and -1.5nC/m located at (x=y=1), (x=-1,y=2), (0,2) respectively."

Sorry but could you elaborate on your first sentence. And oops, F = [1/(2)(pi)(E)][QQ/r]. Is that right?

Thanks for the replies btw!

EDIT: Ok I used that formula above to work out the forces and I got -2.7x10^-8, -3.8x10^-8, 1.6x10^-8. Adding them together comes out with -4.9x10-8. I used the distance formula to work out the distance between the points. I don't know if this is even right in any way.

Last edited: Aug 25, 2008
5. Aug 26, 2008

### Redbelly98

Staff Emeritus
Sorry that I've had very little free time in the past 1-1/2 or 2 days, but now I'm back.

Space is 3 dimensional. To specify where any point is located, we must use a coordinate system with 3 coordinates. It is customary to label those coordinates x, y, and z. Moreover, each coordinate has an axis associated with it. We call these the x-axis, y-axis, and z-axis.

It would be possible to calculate the electric field any point in space, given the 3 line charges described in the problem statement. But they don't want you to calculate the E-field at just any point; they would like that point (where the E-field is to be calculated) to be located on the z-axis. To keep things simple, you may just use the origin (x=y=z=0) if you wish.

Getting warmer. We don't want a force, we want an electric field. Also, the equation you want should contain the charge density in it, so that you can plug in those nC/m values you gave earlier.

No. The electric field is a vector, requiring you to do vector addition which accounts for the directions of the 3 different vectors you are adding. Don't just add up the numbers.

6. Aug 28, 2008

### SilverBullet

Ok, I understand this fine.
Hmm. The only other equation I know for electric field strength is E = F/Q which won't work cause we don't have a force.

7. Aug 28, 2008

### Defennder

You need to calculate the E-field due to a line of charge here. Use Gauss law to get that. You don't have to use E=F/q. I'm going to assume that the charged lines are parallel to the z-axis. Just superpostion all the E-field contributions from the charged lines to get the resultant.

8. Aug 30, 2008

### Redbelly98

Staff Emeritus
It seems odd that they would expect you derive the E-field of a line charge as part of this homework problem. Typically, deriving the field would be a problem unto itself. For something like this, I really expect that the field has already been present to the students, either in the main text of the book or perhaps as an earlier HW problem (eg., "Show that the electric field of a uniform line of charge is E = _____")

Or, perhaps the book has given the force between a point charge and a line charge. This could be where SilverBullet got the formula in Post #4:

"F = [1/(2)(pi)(E)][QQ/r]"

Then one could use E = F/Q to get the field, except that this formula is wrong as written. There should be just Q (not QQ or Q^2), and the line charge should appear in there somewhere. I think this formula was incorrectly copied from wherever it came, and checking more carefully could reveal the right formula.

If I'm wrong about what the text book says, then Defennder's suggestion of using Gauss' Law to get the field is the best way to go.

9. Aug 30, 2008

### Staff: Mentor

This is still insufficient to determine the the E(z)-vector since it does not provide the orientation of the lines of charge, unless one is to assume that they are all perpendicular to the displacement vector running from the origin to the coordinate given, which would make sense since that would be the closest point then. One has to determine the three E-field vectors for each line as a function of z on the z-axis.

Also, presumably, these lines are coplanar and that plane is taken at z=0.

10. Aug 30, 2008

### Redbelly98

Staff Emeritus
I assumed the field is constant all along the z-axis, and one must find its magnitude and direction. A single value, and the angle. That would mean the lines of charge are all parallel to the z-axis.

This sounds like an undergrad electrostatics problem, I don't think they expect the students to calculate something that varies along the z-axis.

edit:

That defines a line at x=y=1, parallel to the z-axis. And similarly for the other lines.

11. Aug 30, 2008

### Staff: Mentor

Ah - yes. Pardon my post then. I interpreted the phrase "the direction on the z-axis resultant" as being "along" the Z-axis as opposed to "at" the z-axis.

12. Sep 1, 2008

### SilverBullet

I don't have a book or notes unfortunately. It's from an exam paper. Thanks, I will try later