Well I know that the point charge has a small dE towards the negative side of the wire while the positive side repels from the charge. Breaking it into component forms of both dE experienced by the point charge, I realized that the dEx would simply cancel each other out. From there I just took the integral of dEy. Was this wrong?gneill said:
If the wire WAS all positive, then in that case the y component would cancel out resulting in both x components of the dE in the positive x direction. From that only one integration was done to find a small section of dy. Are you implying that I have to add the integral of both dEs of the y-components?gneill said:
Oh wow, I really appreciate your help. Your questions got me thinking! Makes much more sense now when you think about a positive wire compared to one with half positive and negative.gneill said:
Aristotle said:
Glad I could help!So you're saying in this case it's best to "assume" that both dE are the same and by symmetric have E= 2 ∫(from 0 to L/2) for dE1 ? Hate me for this, but I'm a little confused on what you're inferring :/ if that's the case, the result from it is -2k*lamda[-1/ sqrt((a)^2 + (L^2)/4) + 1/a ].gneill said:
Our professor hasn't covered electric dipole potential yet so I don't think that will be useful to me. My main concern is to why we have to 2*∫(from 0 to L/2) for the dE of the positive side when they aren't really symmetric in terms of the dE of the negative.BvU said:You want to check the integrals table. I miss a factor y in the numerator.
[edit] Sorry, wrong component.
You can check your result -2k*lamda[-1/ sqrt((a)^2 + (L^2)/4) + 1/a ] here .Your q is ##\ {\lambda L\over 2}\ ## , your d is ##\ {L\over 2}\ ## so your ##\ p = {\lambda L^2\over 4}##
With ##\ {1\over\sqrt{1+({L\over 2a})^2} }\approx 1 - {L^2\over 8 a^2}\ ## you should get the ##\ E = {kp\over a^3} \ ## (and I think you do).
You'd be surprised ! Part 2 of your original post is asking precisely what the field far away is. So by the time he does mention a dipole field you have already calculated one !Aristotle said:
Ah I see, makes a little more sense nowgneill said:
...so since both the dEs are going in the negative y direction with the same magnitudes, its safe to say that the positive side & negative side of the wire can be found through integration by symmetry. Thank you for the visual! Just a curious thought though...when would it be the case to "integrate seperately"? I feel like with a linear wire, at least one of the component cancels each other out and symmetry seems to be always applicable.BvU said:
You can't apply symmetry if you cannot find a symmetry to exploit. So for example if the charge distributions above and below the x-axis were described by dissimilar functions you'd have to handle them separately.Aristotle said:
Hm, by dissimilar functions would you be referring to if the wire itself was non-uniform charge distributions--such as some positive & negative charges above x-axis and below? Another thing, it's been a while but...when taking symmetry in account: 2* ∫(from 0 to L/2), why does it not equal to when ∫(from -L/2 to L/2)? --Shouldnt they equal similar answers regardless?gneill said:
Well they too could exhibit symmetry if they were symmetrically arranged and sized. But if, for example, one followed a cubic density distribution while the other a square, that would throw a wrench into the symmetry. Or if one end was longer than the other, or,...Aristotle said:
Well, they were similar. It just happened to end up equal and opposite for the two halves resulting in a sum of zero.
gneill said:
A uniformly charged wire is a wire that has an equal amount of charge distributed along its entire length. This means that the charge per unit length of the wire is constant, resulting in a uniform electric field around the wire.
The electric field around a uniformly charged wire can be calculated using the equation E = λ/2πε*r, where λ is the charge per unit length of the wire, ε is the permittivity of the surrounding medium, and r is the distance from the wire to the point where the electric field is being measured.
The electric field around a uniformly charged wire is radial, meaning it points directly away from or towards the wire in all directions. This is because the wire has a cylindrical symmetry, and the electric field lines must be perpendicular to the wire's surface.
The electric field around a uniformly charged wire decreases with distance from the wire according to the inverse square law. This means that the electric field strength is inversely proportional to the square of the distance from the wire.
If the charge on the wire is doubled, the electric field around the wire will also double. This is because the electric field is directly proportional to the charge per unit length of the wire. Therefore, as the charge increases, the electric field strength also increases proportionally.