Electric Field outside a cylinder

In summary, the electric field immediately outside a long cylindrical wire with radius Ri = 0.5 mm and E = 40 kV/m is related to the charge per unit length along the wire by Gauss' law. To find the field at a radius of 2mm from the center of the wire, the formula E=(λ)/(2pi*εo*r) is used and the resulting value is 10 kV/m. To calculate the potential difference between r=0.5 and r=2, the formula Vf-Vi = -integral Edr is used with the corresponding E values for each radial distance. After solving,
  • #36
Before the sheath was there, E was away from wires surface.

Hence electron would be attracted to the wire?
 
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  • #37
yes ... (there are +++Q on the wire, that's why E points away from it)
so the electrons end up on the
inside of the sheath , or the outside of the sheath?

When do they stop accumulating there?

that is, what causes them to go there,
and what has to be =0
before they'll stop moving to the inside?
 
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  • #38
They want to move away from the wire, but at the same time, there are electrons inside the sheath so they will end up on the outside of the sheath. meaning the charge on the inner surface of the sheath will be neutral?
 
  • #39
??

positive Qharges, the ones that are ON the inner wire,
are "trying" to get away from the other + charges on the wire,
(trying to move along E, to location of lower Potential where +q has low PE).

But these electrons are in the sheath at 2.1 mm,
and are trying to move TOWARD the ++Q on the wire,
(against E, to location of higher Potential where -q would have LOW PE).

They'll keep moving inward until:
a) there is zero TOTAL charge inside of their location
b) there is zero E-field at the place that they are at
c) the Potential is the same to their inside as to their outside.

All 3 are the same condition. Gauss relates a) to b) .
 
  • #40
so electrons move toward the wire. Does that make the charge on the inner surface negative?
 
  • #41
yes.

And you know how much neg. charge it will take (/meter length)
to offset the charge on the inner wire.
 
  • #42
I don't get what you mean by offset the charge on the inner wire. I just want to figure out the charge on the inner surface of the sheath..

Are they the same thing?
 
  • #43
Gauss says that the total E outward thru surface A
equals the total Qharge enclosed by the surface A.

for E to be 0, the total Q inside has to add up to zero.
positives and negatives offset ...
 
  • #44
t_n_p said:
I don't get what you mean by offset the charge on the inner wire. I just want to figure out the charge on the inner surface of the sheath..

Are they the same thing?

The charge on the inner surface of the sheath offsets the charge on the wire (due to Gauss law as lightgrav explained)... so Q/x for the sheath is the negative of Q/x for the wire... From that you can get the Q/area

Or here's a different way to do it. Take a tiny cylindrical volume (or any shape doesn't matter) that goes through the inner surface of the sheath. So one end is inside the sheath and the other end is outside... Do gauss' law over that volume... The field inside the sheath is 0. The field outside, you already calculated before at 2mm. That will also give the same answer...
 
  • #45
I am really, really confused now!

Q/x of sheath = -Q/x for wire = -1.112*10^-12 C/m

Then I use Gauss?
But I want to find Q?:confused::confused::confused:
 
  • #46
t_n_p said:
I am really, really confused now!

Q/x of sheath = -Q/x for wire = -1.112*10^-12 C/m

Then I use Gauss?
But I want to find Q?:confused::confused::confused:

Total Q? Are you sure they aren't asking for charge per unit length or charge per unit area?
 
  • #47
the exact question..

"A hollow cylindrical 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?"

I'm not sure if they are after a qualitative or quantitative answer, how do you interpret the question?
 
  • #48
so the rest of the electrons are not attracted toward the inside.
Q/x is as good as you can do.

(unless you want to find the charge per Area...)

Gauss is not a new foundation equation,
it deals with the same info as Coulomb's law,
so in situations like this one (symmetry!)
you often don't actually have to COMPUTE a new result.

it is sort of like weighing a mass on a spring using Newton's 3rd law;
once you know the Force by the spring, the Force by gravity is obvious.
 
  • #49
t_n_p said:
the exact question..

"A hollow cylindrical 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?"

I'm not sure if they are after a qualitative or quantitative answer, how do you interpret the question?

If they gave the length of the cylinder (or the initial wire) then you could give a total Q... I assumed that it was an infinitely long wire and cylinder... in which case charge per unit length, or charge per unit area is the only type of answer you can give.
 
  • #50
Yeah in the initial question the wire is described as "very long", however the question states nothing about area or length. :confused:
 
  • #51
Due to the wire being "very long" you can ignore the length component.
Refer to prescribed text, “Understanding Physics” Page 702.
 

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