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Calling all Oxbridge Physicists!
Right, am battling with EM and I have a query (thought this was best place to ask and get your attention!!!) I'm trying to find the electric field due to a line charge with uniform density λ coulombs/m. I want to solve it using Gauss' Law (it's much easier than using superposition principle). I know I need to draw a cylindrical Gaussian Surface surrounding the line charge and then use cylindrical coordinates. My question is, why do we not need to consider the electric field out of the ends of the cylinder? I know (according to Feynman 
) that the flux through the two ends are zero - because they're in opposite directions and cancel? However, surely if you draw the coordinate system so that z is along the axis, then there will be an electric field out of the ends? If someone could explain this to me, it would really help as I know how to do the rest of the calculation (as we should only have a radial electric field ie. Er, since Ez=0 & Eθ=0).
Thanks guys
) that the flux through the two ends are zero - because they're in opposite directions and cancel? However, surely if you draw the coordinate system so that z is along the axis, then there will be an electric field out of the ends? If someone could explain this to me, it would really help as I know how to do the rest of the calculation (as we should only have a radial electric field ie. Er, since Ez=0 & Eθ=0).Thanks guys
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I think you're trying to ask "Why is the electric field radial about the line of charge?".
If that's your question, then you simply remember we're working with electrostatics, and so none of the charges should be moving. Now let us suppose we have a non-radial electric field....that would mean the charges experience a sideways force along the wire (or whatever) and so would move. They would actually move until they are aligned such that there is only radial electric field...there can only be radial field because otherwise the charges would move (we neglect the ends at infinity)
if you want to look at it in terms of gaussian pillboxes (just a fancy name for the cylinders you draw)....well because I've said the electric field is always radial, you wont have ANY field out of the ends.
unless we're talking about a length of chrage which isnt infinite in length that is! But in that case you could (as you correctly say) arrange a pillbox so that the fields dont cancel.
does that make any sense or have i misinterpreted your question?
If that's your question, then you simply remember we're working with electrostatics, and so none of the charges should be moving. Now let us suppose we have a non-radial electric field....that would mean the charges experience a sideways force along the wire (or whatever) and so would move. They would actually move until they are aligned such that there is only radial electric field...there can only be radial field because otherwise the charges would move (we neglect the ends at infinity)
if you want to look at it in terms of gaussian pillboxes (just a fancy name for the cylinders you draw)....well because I've said the electric field is always radial, you wont have ANY field out of the ends.
unless we're talking about a length of chrage which isnt infinite in length that is! But in that case you could (as you correctly say) arrange a pillbox so that the fields dont cancel.
does that make any sense or have i misinterpreted your question?
well you could be sneaky and throw the superposition thing at him....just say "For every point in space, there will be a field from one point charge on line, and a field symmetrically positioned somewhere else on the line to cancel out any non-radial field. This means there must be only radial field left. (Edit: This is exactly what feynman the man himself says in Vol 2....the charges by symmetry cancel axial field!)
But personally I still use the idea that if they werent radial, the charges would surely experience sideways force, and so would move....this would then lead to a radial field. It explains nicely way why placing a metal ball in an electric field causes the field lines to point into the ball radially, and then point out of the ball radially.....if they didnt point radially, the charges would move. They keep moving until all field points radially, and so the charges stop moving: equilibrium
It's important to be able to define the vector direction as well, otherwise you can get in a pickle with signs. For example, Potential is MINUS the integral of F.dl, this is because the vector dl points in the direction of decreasing F (towards infinity), and so if you move in direction of increasing F, you are moving against dl, the minus sign is there to compensate. I think of it like the integral of F.-dl from one point to another....where -dl is now the vector in the direction of increasing potential! All fixed
But personally I still use the idea that if they werent radial, the charges would surely experience sideways force, and so would move....this would then lead to a radial field. It explains nicely way why placing a metal ball in an electric field causes the field lines to point into the ball radially, and then point out of the ball radially.....if they didnt point radially, the charges would move. They keep moving until all field points radially, and so the charges stop moving: equilibrium
It's important to be able to define the vector direction as well, otherwise you can get in a pickle with signs. For example, Potential is MINUS the integral of F.dl, this is because the vector dl points in the direction of decreasing F (towards infinity), and so if you move in direction of increasing F, you are moving against dl, the minus sign is there to compensate. I think of it like the integral of F.-dl from one point to another....where -dl is now the vector in the direction of increasing potential! All fixed
thanks. I have another question actually (think this is going to become the official Oxbridge Physics Revision thread )
If we have a very long neutral wire carrying a steady current I at distances much less than the length of the wire, then we can find the tangential component Bφ of the magentic field quite easily (by drawing a loop and using Ampere's Law). How can i PROVE that the radial component Br is zero? Do I basically draw a surface and then use the No Free Poles Law (Div.B = 0)? Similarly, for Bz? Thanks. These are questions my tutor has set and he specifically wants us to prove they're zero, rather than just assuming.
(think this is going to become the official Oxbridge Physics Revision thread )If we have a very long neutral wire carrying a steady current I at distances much less than the length of the wire, then we can find the tangential component Bφ of the magentic field quite easily (by drawing a loop and using Ampere's Law). How can i PROVE that the radial component Br is zero? Do I basically draw a surface and then use the No Free Poles Law (Div.B = 0)? Similarly, for Bz? Thanks. These are questions my tutor has set and he specifically wants us to prove they're zero, rather than just assuming.
Yup I think you've got it there, using the idea of No Free Poles. Br and Bz must be 0 because if you drew a gaussian pillbox centered on the wire, with the axis pointing along the wire, then you know there can be no outflow through the pillbox surface. The pillbox has a curved surface and two flat surfaces....outflow can only occur from Br and Bz components....which hence must both be 0.
If you want to get technical (your tutor might like) then you would say: From Divergence theorem, Integral of DivB over the pillbox volume = Integral of Vector B over the surface of the PillBox. But Integral of DivB = 0...so Integral of Vector B over the surface of the pillbox must be 0.
So we have a surface integral....now lets define the surface in terms of surface vectors. We have 3 surfaces: 1 curved surface with dS = dS(dr,0,0) (radial only) and then 2 flat surfaces with dS = dS(0,0,dz) and (0,0,-dz) (axis only). Since these integrals MUST = 0, the radial component of vector B = 0, and so does the z component. The only one which doesnt have to be 0 is the angular component cicrulating round the wire.
now excuse me while i attend a lonely erg
If you want to get technical (your tutor might like) then you would say: From Divergence theorem, Integral of DivB over the pillbox volume = Integral of Vector B over the surface of the PillBox. But Integral of DivB = 0...so Integral of Vector B over the surface of the pillbox must be 0.
So we have a surface integral....now lets define the surface in terms of surface vectors. We have 3 surfaces: 1 curved surface with dS = dS(dr,0,0) (radial only) and then 2 flat surfaces with dS = dS(0,0,dz) and (0,0,-dz) (axis only). Since these integrals MUST = 0, the radial component of vector B = 0, and so does the z component. The only one which doesnt have to be 0 is the angular component cicrulating round the wire.
now excuse me while i attend a lonely erg
lol, right just finished bathing the dog so I've bene enjoying sunshine, now back to EM! Have a quick question...
When we have to parallel sheets with the same, but opposite charge, I know that the electric field outside of them is zero. Is this because if you drew a Gaussian Surface covering them both, you do not have any net charge inside?
When we have to parallel sheets with the same, but opposite charge, I know that the electric field outside of them is zero. Is this because if you drew a Gaussian Surface covering them both, you do not have any net charge inside?
Hoofbeat
lol, right just finished bathing the dog so I've bene enjoying sunshine, now back to EM! Have a quick question...
When we have to parallel sheets with the same, but opposite charge, I know that the electric field outside of them is zero. Is this because if you drew a Gaussian Surface covering them both, you do not have any net charge inside?
When we have to parallel sheets with the same, but opposite charge, I know that the electric field outside of them is zero. Is this because if you drew a Gaussian Surface covering them both, you do not have any net charge inside?
yep!
cheers Nick! Glad to see that someone else is around to answer my questions! Willa is probably going to get very frustrated with me 
Here's another one: (you should all thank me that you're sort of revising too by having to explain it to me )
Q.Charges +q, +2q, -5q & +2q are placed at the four corners ABCD of a square of side a respectively, taken in cyclic order. Find the electric field strength at centre of square.
Right I know we use the field at each of the different charges and then superimpose. I've done this question before, but looking back I'm stuck with one bit. Say we're working out the field due to QA:
E = 2q/ka² (as the diagonal displacement from A -> O is a
If I want to resolve this into the x and y components so I can later work out the resultant easier, why do I get:
Ex = 2q/ka²√2 and Ey=2q/ka²√2
Where did the √2 come from?!
Here's another one: (you should all thank me that you're sort of revising too by having to explain it to me
)Q.Charges +q, +2q, -5q & +2q are placed at the four corners ABCD of a square of side a respectively, taken in cyclic order. Find the electric field strength at centre of square.
Right I know we use the field at each of the different charges and then superimpose. I've done this question before, but looking back I'm stuck with one bit. Say we're working out the field due to QA:
E = 2q/ka² (as the diagonal displacement from A -> O is a
If I want to resolve this into the x and y components so I can later work out the resultant easier, why do I get:
Ex = 2q/ka²√2 and Ey=2q/ka²√2
Where did the √2 come from?!
if the square is of side "a" then the distance to the centre of the square is: root(a^2 + a^2) / 2 = root(2)*a / 2 = a / Root(2)
so using that for a (with a-hat being the unit directional vector from the charge to the centre
E = 2q/ka^2 = 2q/k(a/root(2))^2 = 2q/ka^2 * 2 * a-hat
but a-hat is just (1/root(2),1/root(2)) so for the components
Ex = 2q/ka^2 * 2/root(2) = 2q/ka^2 * Root(2)
and same for Ey
Edit: I've just seen you were contented....oh well like you said it's good revision for me as well!
so using that for a (with a-hat being the unit directional vector from the charge to the centre
E = 2q/ka^2 = 2q/k(a/root(2))^2 = 2q/ka^2 * 2 * a-hat
but a-hat is just (1/root(2),1/root(2)) so for the components
Ex = 2q/ka^2 * 2/root(2) = 2q/ka^2 * Root(2)
and same for Ey
Edit: I've just seen you were contented....oh well like you said it's good revision for me as well!
Well if it's any consolation, by the sounds of things you're doing quite a bit more revision than a lot of people i know...namely the people i just spent the day with playing football in the park and drinking Pimms (4 of us are oxbridge students, and seem to have acquired a taste for that overly priced beverage!). Tomorrow I SHALL revise!
Willa
Well if it's any consolation, by the sounds of things you're doing quite a bit more revision than a lot of people i know...namely the people i just spent the day with playing football in the park and drinking Pimms (4 of us are oxbridge students, and seem to have acquired a taste for that overly priced beverage!). Tomorrow I SHALL revise!
Lol. I wish I wasn't though....especially when I think of all the lovely things I could be doing (and that even includes ergs!)...Shiny it doesn't help when you tell me you've been lazying around in ChCh Meadow all day
Sounds like you had a great day Will, wish I could have joined you (ummmm...Pimms 
). Fishpaste - I hate vector calculus too, but it does appear it has numerous applications in Physics <grumbles>. Anyways, after my cousing coming round yesterday I didn't finish the section on EM that I wanted to, so need to get that done today as well....back to the books <once I've checked a few more forums of course 
)
Related discussions
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- would a masters in Law in Oxbridge more appealing to top law firms
- Should I reapply to Cambridge?
- The experience I wanted
- Cancelling UCAS early application and re-applying
- GCSE results too low for Oxbridge?
- Oxford admissions mentor
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- Contemplating changing supervisor
- Comparing elastic and inelastic collision unknown mass
- Physics at oxbridge
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