Charge Density of wire with potential difference

In summary, the problem involves a charged particle moving away from a fixed wire at different distances and speeds. To find the charge density of the wire, one must use the relation \DeltaK = -\DeltaU and \DeltaU = q*\DeltaV, where \DeltaV is the potential difference. However, the electric field in this case is not uniform, so the simplified equation V = -Ed cannot be used. Instead, one must integrate the electric field as a function of distance to find the potential difference. The "d" in the equation represents a differential.
  • #1
Nexest
5
0
Problem:

Consider a long charged straight wire that lies fixed and a particle of charge +2e and mass 6.70E-27 kg. When the particle is at a distance 1.91 cm from the wire it has a speed 2.80E+5 m/s, going away from the wire. When it is at a new distance of 4.01 cm, its speed is 3.20E+6 m/s. What is the charge density of the wire?

Alright so this is how I approached it but didn't get it right:

The difference in kinetic energy at two given distances should be equal to the opposite difference in potential energy ([tex]\Delta[/tex]K = -[tex]\Delta[/tex]U), which when divided by the given charge of the particle should give the potential difference. Right? ([tex]\Delta[/tex]U = q*[tex]\Delta[/tex]V)

The potential difference is equal to -E*d where d is the distance between the two given points and E is the electric field of the wire given by: lambda/2*pi*epsilon*r.

With these you should be able to solve for lambda but I keep getting the answer wrong and I think its because I'm using the wrong r in the equation for the electric field. I thought r should be the distance from the wire to the farthest given point but I wasn't sure...

Can anyone help?
 
Last edited:
Physics news on Phys.org
  • #2
Without showing your work, I would guess you may be finding the potential incorrectly. It looks like you are treating E as a constant with the expression V=-E*d. Recall,

[tex]V=-\int \mbox{Edr}[/tex]

where r is the perpendicular distance from the wire. Your approach to the problem appears correct.
 
  • #3
Yes I think I found my problem, I can't use the simplified V = -Ed equation because E is not a uniform electric field in this case.
 
  • #4
Can you please explain it in more detail ? How do you calculate the potential difference then if you can't consider it a uniforn electric field ? and what is d ?
 
  • #5
*unicorn* said:
Can you please explain it in more detail ? How do you calculate the potential difference then if you can't consider it a uniforn electric field ? and what is d ?
I'm guessing you don't know calculus. chrisk explained it. You must integrate the electric field as a function of distance. The "d" means "differential". If you don't know calculus, then I don't expect you will have any idea what this means.
 
  • #6
ok.. nevermind.. i got it.. thanks !
 

Related to Charge Density of wire with potential difference

1. What is charge density?

Charge density is a measure of the amount of electric charge per unit volume of a material. It is typically expressed in units of coulombs per meter cubed (C/m³).

2. How is charge density related to potential difference?

The charge density of a wire is directly proportional to the potential difference applied across it. This means that as the potential difference increases, the charge density also increases, and vice versa.

3. What is the formula for calculating charge density?

The formula for charge density is ρ = Q/V, where ρ is the charge density, Q is the amount of charge, and V is the volume of the material.

4. How does charge density affect the flow of current in a wire?

Charge density plays a significant role in determining the flow of current in a wire. A higher charge density means there is a greater concentration of charges in the wire, which leads to a stronger electric field and a faster flow of current.

5. How does the charge density of a wire affect its resistance?

The charge density of a wire is directly proportional to its resistance. This means that as the charge density increases, the resistance also increases. This is because a higher charge density results in more collisions between the charges and the atoms of the material, making it more difficult for the charges to flow through the wire.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
9K
  • Advanced Physics Homework Help
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
16
Views
3K
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
9
Views
6K
  • Advanced Physics Homework Help
Replies
1
Views
950
  • Advanced Physics Homework Help
Replies
2
Views
9K
  • Classical Physics
Replies
3
Views
302
Back
Top