Electric Field outside a cylinder

[t_n_p]

Homework Statement
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?

 

[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.

 

[t_n_p]

Gausses?.....

 

[learningphysics]

Gausses?.....

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...

 

[t_n_p]

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

 

[learningphysics]

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

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

 

[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?

 

[learningphysics]

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?

Yes, looks correct to me.

 

[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?

 

[learningphysics]

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?

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.

 

[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!

 

[learningphysics]

Good spot with the kV/m!

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

Thanks ALOT!

You're welcome.

 

[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?

 

[learningphysics]

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?

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

 

[t_n_p]

I think this is the one?

http://img245.imageshack.us/img245/1862/untitledkr7.jpg

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.

 

[learningphysics]

I think this is the one?

http://img245.imageshack.us/img245/1862/untitledkr7.jpg

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.

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

 

[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?

 

[learningphysics]

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?

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)

 

[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?

 

[learningphysics]

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?

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.

 

[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

solving leads to...
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?)

 

[learningphysics]

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)

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...

 

[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!

 

[learningphysics]

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!

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

 

[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!

 

[learningphysics]

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

Yeah.

Don't feel bad about the errors. With experience, you'll make less and less of them.

 

[t_n_p]

Sorry edited my previous post, just as you were writing your reply. Hope I'm correct!

 

[learningphysics]

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!

Yup, that's what I got also. I'm pretty sure the negative is valid.

One thing I was wrong about... you didn't have to convert the terminals to m... you'll still get the right answer... because it is the ratio of the lengths that matters... but converting to m is better just in case...

One way to check that the negative makes sense is... think about what is happening with energy... since the field E is positive in the outward direction (direction of increasing r)... that means that as a positive test charge moves outward, the field E does positive work on it... when positive work is done on the test charge, that means it is gaining kinetic energy... but another way of looking at this is... potential energy is being converted to kinetic energy... so the positive test charge loses potential energy going outward, so it makes sense that we get a negative for potential difference...

 

[t_n_p]

Yup, that's what I got also. I'm pretty sure the negative is valid.

One thing I was wrong about... you didn't have to convert the terminals to m... you'll still get the right answer... because it is the ratio of the lengths that matters... but converting to m is better just in case...

Thanks! It seems as though I am once again stumped.

A hollow cylinderical metal sheath with inner radius 2mm is now placed around the wire to form a coaxial cable. what will be the charge on the inner surface of this sheath?

Really not sure where to start

 

[learningphysics]

Thanks! It seems as though I am once again stumped.

A hollow cylinderical metal sheath with inner radius 2mm is now placed around the wire to form a coaxial cable. what will be the charge on the inner surface of this sheath?

Really not sure where to start

The cylindrical sheath is a conductor... how do the charges arrange themselves on a conductor... your calclations from before are still valid here... the field at 2mm is still the same (just right before the sheath). What's the field on the inside of the sheath (between the inner and outer radius)?

 

[t_n_p]

Well, you say the sheath is a conductor.
So I would say that the field on the inside of the sheath between the inner and outer radius is negative.

Where do I go from there?

 

[lightgrav]

you say the field within the material is negative ...
do you mean pointed inward?
what would happen to an electron in that material,
if there was an inward-pointing E-field?

 

[t_n_p]

well e fields go from + to -, so if there was an inward pointing e field and electrons are in the material, they would be repelled?

 

[lightgrav]

?? repelled ? ... from what ?
(they can't see what kind of charge is on the central wire)
do they move inward or outward?
(the Force on a qharge = qE ...)

 

[lightgrav]

Sorry ...
what direction *did* E (used to) point ,
before the sheath was put there?

Where would an electron go (they move freely, in a conductor)
if it was immersed in such a 10 kV/m field ?