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

In summary, the problem asks for the calculation of the electric field strength and direction on the z-axis resulting from 3 line charges with different charge densities located at specific points. The equation E = k.Q/r^2 is not applicable here as it is for point charges. Instead, the field of a line charge needs to be used, which can be derived using Gauss's law. Superposition can be used to add the contributions of each line charge to get the resultant field. It is important to note that the formula F = [1/(2)(pi)(E)][QQ/r] is incorrect and may have been copied incorrectly from another source.
  • #1
SilverBullet
4
0

Homework Statement


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.


Homework Equations


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


The Attempt at a Solution


So I work out the electric fields and add them together. I don't understand the z-axis part
 
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  • #2
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
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
Defennder said:
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.

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."

Redbelly98 said:
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.

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:
  • #5
Sorry that I've had very little free time in the past 1-1/2 or 2 days, but now I'm back.

SilverBullet said:
Sorry but could you elaborate on your first sentence.

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.

And oops, F = [1/(2)(pi)(E)][QQ/r]. Is that right?

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.

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.

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
Redbelly98 said:
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.

Ok, I understand this fine.
Redbelly98 said:
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.

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
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
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 textbook says, then Defennder's suggestion of using Gauss' Law to get the field is the best way to go.
 
  • #9
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."
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
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:

... located at (x=y=1), ...

That defines a line at x=y=1, parallel to the z-axis. And similarly for the other lines.
 
  • #11
Redbelly98 said:
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.

That defines a line at x=y=1, parallel to the z-axis. And similarly for the other lines.
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
I don't have a book or notes unfortunately. It's from an exam paper. Thanks, I will try later
 

1. How do you calculate the electric field strength on the z-axis resultant?

The electric field strength on the z-axis resultant can be calculated by dividing the force exerted on a charge by the magnitude of the charge. This results in a vector quantity with units of newtons per coulomb (N/C).

2. What factors affect the electric field strength on the z-axis resultant?

The electric field strength on the z-axis resultant is affected by the magnitude and distance of the charges creating the field. It is also influenced by the medium in which the charges are located, as different materials have different permittivity values.

3. How is the direction of the electric field on the z-axis resultant determined?

The direction of the electric field on the z-axis resultant is determined by the direction of the force that would be exerted on a positive test charge placed at that point. The field lines point away from positive charges and towards negative charges.

4. Can the electric field strength on the z-axis resultant be negative?

Yes, the electric field strength on the z-axis resultant can be negative. This indicates that the direction of the field is opposite to the direction of a positive test charge. Negative electric field strengths can occur when there are more negative charges than positive charges present.

5. How is the electric field strength on the z-axis resultant represented graphically?

The electric field strength on the z-axis resultant can be represented graphically by using electric field lines. These lines represent the direction and strength of the field at different points in space. The denser the lines, the stronger the field at that point.

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