# Linear Uniform Charge rod

1. Sep 8, 2008

### StephenDoty

A positively charged rod of length 6m ranging from (0,0) to (6,0) has a charge density of 1 X 10^(-5) C/m. What is the magnitude and angle of the electric field at point (4,4)?

Wouldn't the electric field point toward the 2nd quadrant, making Enet= -Ex i + Ey j??

For Ex I got 10019.2 N/C (-i)
and for Ey I got 18721.1 N/C j
So wouldn't the magnitude be 2.1 X 10^4 N/C and the angle 118 degrees from the x-axis????

I am having trouble deciding the angle.
Since to the right is a shorter distance to the point than to the left, would the angle be in the first quadrant at 28 degrees?

Any help would be appreciated.
Thank you.
Stephen

2. Sep 8, 2008

### StephenDoty

I need to know how to find the angle. I have to turn this in soon.

thanks
Stephen

3. Sep 8, 2008

### mezarashi

How are you arriving at these numbers? Also please see the forum guidelines on homework problems.

Properly solved, this problem requires some integration.

4. Sep 8, 2008

### StephenDoty

I know my numbers are right. I am just concerned about the angle. And the integration and derivation that it takes to get the numbers does not help me figure out the angle, which is where my question is at.

If you draw a y axis through the point (4,4) wouldn't the electric field be in the second quad???
Thus making the angle 118 degrees??

5. Sep 8, 2008

### LowlyPion

Yes, it would be directed away from the positive charge toward the second quadrant when you are to the left of the midpoint of the charge density.

6. Sep 8, 2008

### StephenDoty

wouldnt (4,4) from a 6m rod be to the right of the midpoint and not to the left of the midpoint?

So am I right to say the angle is 118 degrees from the x axis or 28 degrees from the y axis?

7. Sep 8, 2008

### LowlyPion

Sorry, my mistake. I mixed up the mid-axis at 3 with your E-field point at 4. So in that case, I'd say the E field is biased toward the right, and not the left. In which case then, no I think it should be first quadrant directed.

8. Sep 8, 2008

### StephenDoty

so the angle would be 28 degrees from the x axis?

9. Sep 8, 2008

### LowlyPion

From the positive y axis into Quadrant I, if that is what comes from your calculation, then that seems about right. So yes.

Disclaimer: I haven't checked your math in calculating the magnitude.

10. Sep 8, 2008

### StephenDoty

since the rod is positive and the test charge is positive and since the point (4,4) is to the right of the midpoint that means that the distance between the point and a dx on the right side is smaller than the distance between the point and a dx on the left side. Thus the E fiel pointing towards the II quad would be stronger than the E field pointing to the right of the I quad. Thus wouldn't the net E field be point in the II quad???????

11. Sep 8, 2008

### LowlyPion

Examine a rod of length 4 whose domain is 2<x<6.

4,4 is along the mid point of the bisector.

That field vector will be normal to the rod and pointing straight up along that line will it not?

Now add the charge whose domain is 0<x<2 the effect of that charge will be away from that. Hence the imbalance in the overall field should be away from the 0<x<2 positive charges. This points the vector at 4,4 into Q1 more, away from Q2 doesn't it?

12. Sep 8, 2008

### StephenDoty

(3,3) would be the bisecting point of a 6m rod. so since it is to the right of the bisecting line wouldn't the E field of a dx from the right would be stronger than a dx from the left of the bisecting line. Thus wouldnt the net E field be pointing toward the left or in the II quad?????

13. Sep 8, 2008

### Redbelly98

Staff Emeritus
(4,4) is to the right of the midpoint, so there is more charge to the left of this point. Therefore the E-field points slightly to the right, i.e. into the 1st quadrant.

Another way to look at it: the charge lying between (2,0) and (6,0) produces no net x-component of E. That leaves the charge lying between (0,0) and (2,0) to produce Ex.

14. Sep 8, 2008

### StephenDoty

No No
My professor made a big deal of saying that if the point is offset from the middle bisect line of the rod then there would be both x and y components.

And as I have been saying, since the point is to the right of the midpoint, the E field of the right dx will force the positive test charge at that point to go away or to the left and up. The E field of the left dx will have a longer r value and thus a weaker E, since dx is the same, going to the right and up. Thus the net E will be going to the left and up.

Last edited: Sep 8, 2008
15. Sep 8, 2008

### StephenDoty

and since it is uniformly charged, the more charge on the left will not matter since the r would be bigger than the r on the right. From the picture that I drew, and with the formula of integrating dE wouldn't the net field be pointing to the left since the E of the left dx would have a much bigger r than that of the dx on the right. Right????

16. Sep 8, 2008

### Redbelly98

Staff Emeritus
You didn't read what I said very carefully. I agree there is a non-zero x component to E. We disagree about the direction.