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Physics 2 Units

  • Thread starter TypeFun
  • Start date
  • #1
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Homework Statement


1) What are the units for the charge, Q?
2) What are the units for ke*Q?
3) What are the units for - ke*Q / R, R=radius (m)
4) What are the units for 2*ke*[tex]\lambda[/tex]*ln R
5) What are the units for - 2*ke*[tex]\lambda[/tex]

2. The attempt at a solution
1) C, Coulomb's
2) N*m2/C
3) N*m/C
4) ?
5) ?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
2,685
20
What are the units of lambda?
 
  • #3
32
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cubic millimeter
mm3

I think...
 
  • #4
23
1
I'm assuming this all has to do with electricity, correct? Lambda in electricity isn't mm3...double check your notes or Google possibly? Let me know if you can't get it.
 
  • #5
32
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It does.
Lambda = 9.99 V / [2*ke*ln(.0860 m / .0055 m)]

ke = Coulomb's Constant, C
I think that makes its V/C*m
 
  • #6
2,685
20
Lambda is charge per unit density I believe - so it can't be mm3.

(Someone correct me if I'm wrong here)
 
  • #7
32
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This was the formula we were given to calculate Lamda:

Lambda = 9.99 V / [2*ke*ln(.0860 m / .0055 m)]
 
  • #8
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Well I've never seen that before in my life... if it was given in class I'd go with it. Lambda denotes charge per unit length, by the way jared. Charger per unit area is sigma and charge per unit volume is rho. I think O.O
 
  • #9
2,685
20
Well I've never seen that before in my life... if it was given in class I'd go with it. Lambda denotes charge per unit length, by the way jared. Charger per unit area is sigma and charge per unit volume is rho. I think O.O
Ah, electrics not my strong suit.
 
  • #10
23
1
Ah, electrics not my strong suit.
I'm still in the process of learning circuit analysis so I'm nowhere near an expert here.
 
  • #11
2,685
20
I'm still in the process of learning circuit analysis so I'm nowhere near an expert here.
Currently in the electronics modules myself, figured it was best to give some help rather than nothing. Hence the hint towards units of lambda.
 
  • #12
23
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Electricity is so much fun! Well, now that we've successfully redirected this topic thoroughly, I wonder if he got the right answer in the end.
 
  • #13
2,685
20
Well I'm waiting for him to combine his equation units for lambda and employ them in his question.
 
  • #14
32
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Assuming, I should use the formula given by the teacher. the units are: V/C*m, right?
 
  • #15
2,685
20
Assuming, I should use the formula given by the teacher. the units are: V/C*m, right?
First, what are the units for ke?

Then, plug them into your equation for lambda and solve out.

What you are left with is the units for lambda. I'd recommend you show your workings so we can keep tabs on it for you.

Something between your question 2 and your lambda equation results isn't adding up here.
 
  • #16
23
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The units for ke should be kg*m/s2C2
ke times a C2 (incorrect, fixed below) yields a Newton.

AH Correction: kg*m3/s2C2
forgot that there is a 1/r2 term in the equation for electrostatic force.
ke * C2/m2 yields a Newton.
 
Last edited:
  • #17
2,685
20
ke: kg*m3/s2C2
forgot that there is a 1/r2 term in the equation for electrostatic force.
ke * C2/m2 yields a Newton.
Which makes the answer to part 2 wrong. You see where I'm going with this?
 
Last edited:
  • #18
23
1
I completely messed that up, then fixed it.
kg*m3/s2C2 are the terms for ke
You are correct. The answer to part two is wrong. As well as the others. You may wanna recheck your work, TypeFun. I'm gonna look over it myself and see what I can come up with.
 
  • #19
SammyS
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It does.
Lambda = 9.99 V / [2*ke*ln(.0860 m / .0055 m)]

ke = Coulomb's Constant, C
I think that makes its V/C*m
This appears to be the result of solving an equation for Lambda (λ). It doesn't explain what λ is nor how to calculate it in general.

λ linear charge density, i.e. charge per unit length. Usually has units of coulombs/meter.
 
  • #20
32
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So the correct unit is a Newton?
 

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