Finding the electric field of a charged rod

In summary: I will definitely keep that in mind for future problems. In summary, the conversation discusses a homework problem involving the linear charge density and the units of pC and C. The expert provides guidance on how to correctly solve the problem and reminds the student to include units in the final result. The expert also suggests becoming familiar with the prefixes of SI units for future problems.
  • #1
Ascendant78
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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.
 
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  • #2
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
 
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  • #3
Ascendant78 said:

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
 
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  • #4
consciousness said:
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
Ascendant78 said:
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
 
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  • #6
ehild said:
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.
 

1. How do you find the electric field of a charged rod?

To find the electric field of a charged rod, you can use the formula E = k*q*x/r^3, where k is the Coulomb's constant, q is the charge of the rod, x is the distance from the rod, and r is the length of the rod.

2. What is the Coulomb's constant?

The Coulomb's constant is a proportionality constant that relates the force between two point charges to the product of their charges and the distance between them. It is represented by the letter k and has a value of approximately 8.99 x 10^9 Nm^2/C^2.

3. How does the distance from the charged rod affect the electric field?

The electric field is inversely proportional to the distance from the charged rod. This means that the farther away you are from the rod, the weaker the electric field will be. The electric field follows an inverse square law, which means that the electric field strength decreases by a factor of 1/distance^2.

4. Can the electric field of a charged rod be negative?

Yes, the electric field of a charged rod can be negative. The direction of the electric field is determined by the sign of the charge. If the charge on the rod is negative, the electric field will point towards the rod and be negative. If the charge is positive, the electric field will point away from the rod and be positive.

5. How do you calculate the electric field at a specific point due to multiple charged rods?

To calculate the electric field at a specific point due to multiple charged rods, you can use the principle of superposition. This means that you can find the electric field at the point by adding the individual electric fields from each rod using vector addition. The total electric field will be the sum of the individual electric fields at that point.

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