Net Electric Field between two line charges (different densities)

In summary, we are given two infinite line charges, one with a charge density of 4.8 * 10^(-6) C/m and the other with a charge density of -2.4 * 10^(-6) C/m, both placed parallel to each other at different points on the x-axis. Using the relevant equation lambda / (2 * PI * epsilon - nought * radius) and considering the direction of the electric fields, we can find the net electric field at two given points on the y-axis. The net electric field at y = 0.2m is 6.5 * 10^5 N/C and at y = 0.6m is 1.62 *
  • #1
LostTime77
7
0

Homework Statement


We are given that one infinite line of charge is placed horizontally on the x axis. The other line charge is placed parallel to the first at y = 0.4m. The first line has charge density 4.8 * 10^(-6) C/m, and the second -2.4 * 10^(-6) C/m. Find the net electric field at the following points. A) At y = 0.2m and B) At y = 0.6m.

Homework Equations


The relevant equation would be lambda / (2 * PI * epsilon - nought * radius).

Where epsilon - nought = 8.85 * 10^(-12) C^2 / (N*m^2)

The Attempt at a Solution



Well.. since we are said to find the net electric field. I assume that we have to sum the two fields at the given point. So we find the field at the given point with respect to the first line, and then the second, then add them. Also do the same for the second point. However, the answer they give in the back of the book is 8.05 * 10^5 N/C, which is a single answer to a two part question. (Young and Freedman University Physics Ed 12) I am at a total loss as how they can get a single answer to a two part question...

Anyhow, taking the zero point as the origin (0.2m) I did (4.8 * 10^(-6)) / (2 * 3.14... * 8.85 * 10^(-12) * 0.2m) + (-2.4 * 10^(-6)) / (2 * 3.14 * 8.85 * 10^(-12) * -0.2m)). I tried all variations of this, including negating the 0.2m in both equations and adding them. I did the same type of calculation for the y = 0.6m charge as well, except for line one 0.2 is changed to 0.6 and for line two it is kept at 0.2m.

Any help would be appreciated. Thank you!
 
Physics news on Phys.org
  • #2
Those two parts will certainly have different answers. (I might have that book. What's the problem number?)
 
  • #3
The chapter is 22 and the problem is 21, for edition 12.


Thanks
 
  • #4
LostTime77 said:
The chapter is 22 and the problem is 21, for edition 12.
Must be a typo. Note that it says "toward negatively charged sphere", which has nothing to do with this problem.
 
  • #5
They seem to have a ton of typos in this book. The book is great.. it's just the answer key that pretty much sucks 10% of the time. However, is the method I am using in the problem, correct?
 
  • #6
I'm not quite sure I follow what you're doing (especially regarding signs). Describe the magnitude and direction of each field contribution.
 
  • #7
Ok well.. I am finding the sum of the two contributions of both electric fields due to the charged lines (wires). I am taking the point y=0.0m to be the origin so that the origin lies on the origin of the Cartesian plane.

The contribution from the first line charge (plugging into the equation) would be :

4.316 * 10^5 N/C. The first line charge is positive. Since the point is above the first line charge, the radius is 0.2m - 0.0m. The radius from the first line charge is 0.2m. This makes sense.

For the second line charge :

-2.158 * 10^5 N/C. The second line charge at (y = 0.4m) is negative. The radius would be 0.4m - 0.2m = 0.2m. The electric field contribution is negative.

If I sum the two contributions I should get the net electric field:

= 2.158 * 10^5 N/C. This makes sense to me. The electric field would be a positive value at y = 0.2m which makes sense. The bottom line charge is stronger than the top line charge.
 
  • #8
LostTime77 said:
The contribution from the first line charge (plugging into the equation) would be :

4.316 * 10^5 N/C. The first line charge is positive. Since the point is above the first line charge, the radius is 0.2m - 0.0m. The radius from the first line charge is 0.2m. This makes sense.
What direction does it point?

For the second line charge :

-2.158 * 10^5 N/C. The second line charge at (y = 0.4m) is negative. The radius would be 0.4m - 0.2m = 0.2m. The electric field contribution is negative.
What does "negative" mean? What direction does it point?

Hint: With respect to the line charge, which way does the field point for a positive line charge? For a negative line charge? Figure out the actual direction of the fields first, then assign the proper sign wth respect to your y-axis.
 
  • #9
I think I get what you are saying. Basically the positive line charge pushes the probe at that point away because its a positive charge. The negative line charge pulls the probe toward it because it is negative. Therefore the contribution of the positive line charge is positive (the probe is being pushed directly upward) and with respect to the same probe, the contribution from the negative line charge should be positive as well because the line charge is negative (pulling toward). So to get the positive out of the negative contribution should I change the radius to be 0.2m - 0.4m = -0.2m? Then would I sum the fields again to get something on the order of 6.5 * 10 ^5N/C?
 
  • #10
LostTime77 said:
Therefore the contribution of the positive line charge is positive (the probe is being pushed directly upward) and with respect to the same probe, the contribution from the negative line charge should be positive as well because the line charge is negative (pulling toward).
Exactly.
So to get the positive out of the negative contribution should I change the radius to be 0.2m - 0.4m = -0.2m?
No. Radius is always positive. Don't try to plug into equations blindly. (The negative sign in the general equation just means toward the line charge; you have to give the field the proper sign according to its actual direction and your coordinate system.)

The negative line chage exerts a positive field contribution (in part a). Period.
Then would I sum the fields again to get something on the order of 6.5 * 10 ^5N/C?
Yes. (In what direction?)
 
  • #11
It would be in the direction of the negative line charge (positively upward), with respect to where it is , directly between the two line charges.
 
  • #12
Right. But express it as acting in the +y direction.
 
  • #13
Right. The final answer would be 6.5 * 10 ^ 5 N/C (estimate, Ill do the actual calculation out) in the +y direction. I think I can figure out part b of the problem, given the help.

Thank you!
 

FAQ: Net Electric Field between two line charges (different densities)

What is the equation for calculating the net electric field between two line charges with different densities?

The equation for calculating the net electric field between two line charges with different densities is given by: E = (k * λ1 * λ2) / d, where k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), λ1 and λ2 are the linear charge densities of the two line charges, and d is the distance between the two line charges.

How does the direction of the net electric field change with varying densities of the line charges?

The direction of the net electric field depends on the relative densities of the two line charges. If the two line charges have the same density, the net electric field will be zero. If the densities are different, the direction of the net electric field will be towards the line charge with higher density.

Can the net electric field between two line charges ever be zero?

Yes, the net electric field between two line charges can be zero if the two line charges have the same density and are placed at equal distances from each other. In this case, the electric fields created by the two line charges cancel each other out, resulting in a net electric field of zero.

How does the distance between the two line charges affect the net electric field?

The distance between the two line charges is directly proportional to the net electric field. As the distance increases, the net electric field decreases and vice versa. This means that the net electric field becomes weaker as the two line charges move farther away from each other.

Can the net electric field between two line charges be negative?

Yes, the net electric field between two line charges can be negative. This happens when the two line charges have opposite densities and the direction of the net electric field is towards the line charge with lower density. In this case, the net electric field is considered negative because it is in the opposite direction of the electric field created by the line charge with higher density.

Back
Top