Finding the electric field of a charged rod

  • #1
328
0

Homework Statement



photo.jpg


Homework Equations



See above

The Attempt at a Solution



I messed with this for hours. I also searched all over the web and can't find anything that is similar. None of the problems I could find gave a value, they just gave lambda. Anyway, I have no idea what I'm doing wrong here. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
130
13
Your method is correct but the units are not written. Note that the units of charge density given are pC not C.

1pC=10-12C
 
  • Like
Likes 1 person
  • #3
ehild
Homework Helper
15,526
1,901

Homework Statement



photo.jpg


Homework Equations



See above

The Attempt at a Solution



I messed with this for hours. I also searched all over the web and can't find anything that is similar. None of the problems I could find gave a value, they just gave lambda. Anyway, I have no idea what I'm doing wrong here. Any help would be greatly appreciated.
Your method is correct, but:

The linear charge density is λ=-3.5 pC/m. What do you think pC means?

Plug in the value of the constant k, to get the final result. What is k?

You need to the include the units in the final result.


ehild
 
  • Like
Likes 1 person
  • #4
328
0
Your method is correct but the units are not written. Note that the units of charge density given are pC not C.

1pC=10-12C
Wow, thank you. I had no idea, nor did our professor ever bother to bring that up when he assigned this homework. I actually though it may have been a ρ symbolizing the density. It threw me off because I couldn't find any similar problem that defined λ.

I really appreciate the feedback. I made the adjustment and my answer checks out. I know I had my sign wrong (went from the point to dx instead of dx to the point for r), but I wasn't so much worried about that as I was figuring out what I was doing wrong.
 
  • #5
ehild
Homework Helper
15,526
1,901
I really appreciate the feedback. I made the adjustment and my answer checks out. I know I had my sign wrong (went from the point to dx instead of dx to the point for r), but I wasn't so much worried about that as I was figuring out what I was doing wrong.

According to the vectorial form of the Coulomb force , the electric field at a point x on the x axis, produced by a charge Q at x=X0 is

[tex]F=k\frac{Q}{|x-X_0|^2} \frac{x-X_0}{|x-X_0|}[/tex]
So you have to measure the distance from the charge to the point, you did it well, but you have too multiply kdq/(x-0.1)2 with the sign of x-0.1. It is negative: you missed a - sign.

YOu need to be familiar with the prefixes of SI units : http://www.physics.nist.gov/cuu/Units/prefixes.html

ehild
 
  • Like
Likes 1 person
  • #6
328
0
According to the vectorial form of the Coulomb force , the electric field at a point x on the x axis, produced by a charge Q at x=X0 is

[tex]F=k\frac{Q}{|x-X_0|^2} \frac{x-X_0}{|x-X_0|}[/tex]
So you have to measure the distance from the charge to the point, you did it well, but you have too multiply kdq/(x-0.1)2 with the sign of x-0.1. It is negative: you missed a - sign.

YOu need to be familiar with the prefixes of SI units : http://www.physics.nist.gov/cuu/Units/prefixes.html

ehild
Got it and thanks for the link to the prefixes too.
 

Related Threads on Finding the electric field of a charged rod

Replies
10
Views
4K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
3
Views
7K
Replies
7
Views
646
Replies
7
Views
37K
Replies
1
Views
3K
Replies
13
Views
16K
Replies
11
Views
2K
  • Last Post
Replies
1
Views
7K
Top