What is the wire's linear charge density from a proton?

In summary: It's important to always check your units and make sure they cancel out to the units you're looking for. That's a good way to catch mistakes. But good job sticking with it and figuring it out!In summary, a proton orbiting a long charged wire at a speed of 1.30*10^6 revolutions per second with a radius of 1.20cm can be used to calculate the wire's linear charge density. Using the equations qE = mw^2r and 9*10^9[2*λ/r]q = mw^2r, the unknown variable λ can be solved for by setting these equations equal to each other and plugging in known values for q, m, r, and w
  • #1
Bigworldjust
54
0

Homework Statement



A proton orbits a long charged wire, making 1.30*10^6 revolutions per second. The radius of the orbit is 1.20cm.

What is the wire's linear charge density?

Homework Equations



- q E = m w^2 r
- 9*10^9 [2 λ /r] q = m w^2 r

The Attempt at a Solution



λ = linear charge dens
w = 2pi/period = 2pi/T
w = d theta/dt = 1.4* 2pi /1

I'm confused on how to solve for λ here though. Can anyone help me out?
 
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  • #2
You said you know:
F=m v^2/r=m r ω^2=mr(2π/t)^2
F=q E= q 2 k λ/r

You correctly said that you have to set them equal to each other
2qkλ/r=4π^2mr/t^2
You were given r and t. q, k, and m are known physical constants. The only unknown variable is λ. Just solve for it and plug in numbers.
 
  • #3
frogjg2003 said:
You said you know:
F=m v^2/r=m r ω^2=mr(2π/t)^2
F=q E= q 2 k λ/r

You correctly said that you have to set them equal to each other
2qkλ/r=4π^2mr/t^2
You were given r and t. q, k, and m are known physical constants. The only unknown variable is λ. Just solve for it and plug in numbers.

Ah, okay, so I have it set up like this:

(2*1.6*10^-19*9*10^9*λ)/(.012)=(4π^2*1.67*10^-27*.012)/(1.30*10^6)^2

I got 1.95056*10^-33 nC/m but that seems to be wrong, anywhere I went wrong?
 
Last edited:
  • #4
Don't plug in numbers until the final answer. Solve for λ, then plug in numbers. Also, just because it's small doesn't mean it's wrong.
 
  • #5
frogjg2003 said:
Don't plug in numbers until the final answer. Solve for λ, then plug in numbers. Also, just because it's small doesn't mean it's wrong.
Alright so I got λ = 2π^2mr^2/kqt^2. I plugged in the numbers and got the same answer as before though, lol.
 
  • #6
Ok, you got the same answer. Why did you think it was wrong?
 
  • #7
frogjg2003 said:
Ok, you got the same answer. Why did you think it was wrong?

I submitted it through Mastering Physics and it said it was wrong, lol. I'm not sure if its the units because the answer is suppose to be in nC/m, but I am pretty sure it cancels out accordingly. Hmm..
 
  • #8
I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"
 
  • #9
frogjg2003 said:
I see the problem. You weren't given the period, you were given the rotation speed. "1.30*10^6 revolutions per second"

Oh, so I have to convert the rotation speed to its period? So it would be 1/1300000?
 
  • #10
Or you could use the frequency 1.6*10^6 instead of 1/period. Either way, you're multiplying by (1.3*10^6)^2 instead of dividing.
 
  • #11
frogjg2003 said:
Or you could use the frequency 1.6*10^6 instead of 1/period. Either way, you're multiplying by (1.3*10^6)^2 instead of dividing.

Hmm I got 5.57*10^-33 and that still seems to be wrong, lol.
 
  • #12
Let's go back to the equation that you found:
[tex]\lambda = \frac{2\pi^2mr^2}{kqt^2}[/tex]
Replace t by 1/f to get
[tex]\lambda = \frac{2\pi^2mr^2f^2}{kq}.[/tex]
Now, what you have been plugging in as t (1.6*10^6s) isn't t, it's the frequency f (1.6*10^6 Hz). Also, make sure that when you're plugging in units, you're using only base units. Only m should be in kg, r should be in m, f should be in Hz, k should be in N*m^2/C^2. Your answer should be in C/m.
 
  • #13
frogjg2003 said:
Let's go back to the equation that you found:
[tex]\lambda = \frac{2\pi^2mr^2}{kqt^2}[/tex]
Replace t by 1/f to get
[tex]\lambda = \frac{2\pi^2mr^2f^2}{kq}.[/tex]
Now, what you have been plugging in as t (1.6*10^6s) isn't t, it's the frequency f (1.6*10^6 Hz). Also, make sure that when you're plugging in units, you're using only base units. Only m should be in kg, r should be in m, f should be in Hz, k should be in N*m^2/C^2. Your answer should be in C/m.
Alright I got 8.439 nC/m and that is still wrong, lol. Damn, I don't know where I'm going wrong. I'm plugging in everything:

I got:
m = 1.67*10^-27 kg
r = .012m
f = 1.6*10^6 Hz
k = 9*10^9 N*m^2/C^2
q = 1.6*10^-19 C
 
  • #14
That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.
 
  • #15
frogjg2003 said:
That's because I had a typo and I wrote 1.6 MHz instead of 1.3 MHz.

Ah, I finally got it. Thanks a lot for all of your help, I really appreciate it.
 
  • #16
You're welcome.
 

Related to What is the wire's linear charge density from a proton?

1. What is a linear charge density?

A linear charge density refers to the amount of electric charge per unit length along a line or wire. It is typically represented by the symbol λ and has units of coulombs per meter (C/m).

2. How is the linear charge density of a wire calculated?

The linear charge density of a wire can be calculated by dividing the total charge of the wire by its length. This can be represented by the formula λ = Q/L, where Q is the total charge and L is the length of the wire.

3. What is the unit of measurement for linear charge density?

The unit of measurement for linear charge density is coulombs per meter (C/m). This unit is a combination of the unit for electric charge (coulomb) and the unit for length (meter).

4. How does a proton contribute to the wire's linear charge density?

A proton, being a fundamental particle with a positive charge, contributes to the overall linear charge density of a wire. Its contribution depends on the number of protons present in the wire and the wire's length.

5. Can the linear charge density of a wire be negative?

Yes, the linear charge density of a wire can be negative if the wire has an overall negative charge. This can occur if there are more negatively charged particles (such as electrons) present in the wire than positively charged particles (such as protons).

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