# Homework Help: Electric Field outside a cylinder

1. Aug 9, 2007

### t_n_p

The problem statement, all variables and given/known data
Immediately outside a very long cylindrical wire of radius Ri = 0.5 mm, the electric field (in air) is E = 40 kV/m,
away from the wire’s surface. Use clearly labelled diagrams, explain any principle/law used, for all parts.
a) What is the charge per unit length along the wire?

The attempt at a solution

I'm not sure which formula I should apply, can somebody direct me?

2. Aug 9, 2007

### learningphysics

There's an important law that relates the charge to electric flux (which is field strength*area)... can you think of the law? That's what you need to use here.

3. Aug 9, 2007

### t_n_p

Gausses?..............

4. Aug 9, 2007

### learningphysics

Yes. That's what you need to use here. Take an arbitrary length x of the cylinder... Then use gauss' law over that cylinder over that length... You should be able to solve for charge per unit length...

5. Aug 9, 2007

### t_n_p

I'm still confused. Which formula exactly should I be using?

6. Aug 9, 2007

### learningphysics

It's Gauss' law... what does Gauss' law say?

7. Aug 9, 2007

### t_n_p

Ok, so I should be using

Φ = EA = Q/εo
= 40*2*pi*0.0005*x = Q/εo

Therefore Q/x=1.112*10^-12 C/m

Correct?

8. Aug 9, 2007

### learningphysics

Yes, looks correct to me.

9. Aug 9, 2007

### t_n_p

Cool!
How about if I wanted to find the field at a radius of 2mm from the centre of the wire?

Using Gauss again, I know all values but x (length of wire), Q, and E (which is what I want to find).

If I use the Q/x value I just found out from the previous question, I get 1.24*10^-25 V/m.

Sounds like quite a far out number, could it be right?

10. Aug 9, 2007

### learningphysics

Oops... before for E, you needed to use 40,000... not 40 (you have to convert to volts). So that would give Q/x=1.112*10^-9 C/m

For this part, I get 10,000V/m or 10kv/m.

11. Aug 9, 2007

### t_n_p

Good spot with the kV/m!

I got 10kv/m for part 2 as well, just a silly arithmatic error!

Thanks ALOT!

12. Aug 9, 2007

### learningphysics

You're welcome.

13. Aug 9, 2007

### t_n_p

Managed to do one part by myself (good acheivement by my standards) before being getting stuck again.

How is it possible to calculate the potential diff betweem r=2 and r=0.5?
As the length of the wire is supposedly the same and the E for r=0.5 is 40000 and the E for r=2 is 10000, is it simply 40000-10000 = 30000V?

14. Aug 9, 2007

### learningphysics

No. The potential diff is found using an integral. Your text should have that integral.

15. Aug 9, 2007

### t_n_p

I think this is the one?

http://img245.imageshack.us/img245/1862/untitledkr7.jpg [Broken]

where f = 2
and i = 0.5

I'm not sure how to do the actual mathematics part of it and there is no example in the book.

edit: my understanding is that the test charge is moving parallel to E and hence angle will be 0degrees.

Last edited by a moderator: May 3, 2017
16. Aug 9, 2007

### learningphysics

Yes, you take a path radially outward from the wire... and E is parallel to that path. cos 0=1. So it is just -integral Edr

Last edited by a moderator: May 3, 2017
17. Aug 9, 2007

### t_n_p

So basically...
next steps

=-∫Edr (with terminals 0.5 and 2)

=-∫Er (with terminals 0.5 and 2)

= -([E*0.5]-[E*2])

Do I use the corresponding E values for each radial distance?

Am I even on the right track?

18. Aug 9, 2007

### learningphysics

No, you need a formula for E in terms of r... ie you need E(r)... then you'd calculate:

=-∫E(r)dr (with terminals 0.5 and 2)

19. Aug 9, 2007

### t_n_p

Well in my textbook I have found

E=(λ)/(2pi*εo*r)

where lambda is Q/x.

Is this equation still suitable even though lambda has been brought in?

20. Aug 9, 2007

### learningphysics

Not sure... do exactly what you did to calculate the field at 2mm... except instead of 2mm just leave the radius as r. So solve for E, and you'll have an expression in terms of r.

21. Aug 9, 2007

### t_n_p

Yeah the formula: E=(λ)/(2pi*εo*r)

Is what I did to find E @ 2mm. Just wondering if I use the lambda value (Q/x) I found in the first question (i.e. 1.112*10^-9)

If yes, then my equation in terms of E and r will look like this..

E = (1.112*10^-9)*(1/5.56r)

hence
-integral Edr becomes
-∫[(1.112*10^-9)*(1/5.56r)]dr

Vf-Vi = -3.75*10^10.
Is a negative figure feasible? (if not, will swapping the terminals, i.e. make f=0.5 and i=2, fix this?)

Last edited: Aug 9, 2007
22. Aug 9, 2007

### learningphysics

The denominator seems off by a power of 10...

I get E = 20/r

What is -∫(20/r)dr

Get the formula first before you plug in the numbers...

Last edited: Aug 9, 2007
23. Aug 9, 2007

### t_n_p

silly me.

E = (1.112*10^-9)*(1/5.56*10^-11*r)
E = 20/r

Vf-Vi = -20∫r dr with terminals (0.5 and 2)
Vf-Vi = -20 [r²/2] with terminals (0.5 and 2)
Vf-Vi = -20 [2-0.125]
Vf-Vi = -37.5V

Hope that's right!

24. Aug 9, 2007

### learningphysics

Nope. You should be integrating 1/r not r. also remember to convert mm to m.

25. Aug 9, 2007

### t_n_p

So many silly errors!
when u say convert mm to m, you mean the terminals correct?

Changing all that gives final answer -27.73V
Still unsure if negative value is valid!

Last edited: Aug 9, 2007