This is a great video but I do have a question... Since the electric field outside the coaxial cable is 0, does that mean that the electric field for the distance between a and b (a < r < b) is also 0?
No. Absolutely not. There will always be a field going from inner cable to the inner surface of the outer cable ( from + to - charge ) P.S : I know it's way too late for the answer, probably you don't care about all this now. School/college time over. 😂🙏
There is a problem when you calculate the charge density of the wires at the minute 2:30. There are 2 problems with that, first the charge density will be by surface, not by linear units, second you must equal charges, not densities so the total flux can be 0, in that case it would be something more like lambda*L=-sigma*2*pi*L*a where a is the radius of the inner surface and sigma is the density of the inner surface.
Charge density of the inner surface must be expressed as sigma not lambda and it will be equal = - delta/(2pi*radius of inner surface), also outer sigma = delta/(2pi*radius of outer surface).
No, the E between r=a and r=b will be (lambda)/(2*pi*r) where lambda is the linear charge density of the inner wire and r is the distance from the center of the wire to the place you want to know the field.
@@lasseviren1 So quick reply though it's 11years ago video... Thank you very much. It helped me with the homework I have to do by tomorrow. Best wishes to you forever, from korea.
In the first part Lambda outer=0, but later you showed that Lambda outer= Lambda not. It is confusing to me. Also, in part I video there exist E on the outer area of the shell, but here on outside the shell E=0. That's also confusing. What are the difference between two's? Nice video.
May you please explain what is the significant difference as to why E is 0 on the outside in this video but E = (Lambda) / 2*(Pi)*r * (Permittivity of free space)
Electric flux through a closed surface equals charge enclosed within that surface divided by permittivity ofbfree space (epsilon not). You'll get this in any standard book bro 🙄
Sorry, I'm kinda stuck. What if I wanted to find the electric field between two current carrying, nested, hollow cylinders? I can't seem to find anything on that anywhere.
I'm not sure what you mean by "nested" but almost certainly you would use the principle of superposition. For superposition you would find the E field for just one of the cylinders imagining the other cylinder was not even there, then reverse the process for the other cylinder, then add up up both vectors to get the combined effect.
PSA: In part one the outer cylinder had a net charge of zero, and in this video it has the opposite charge of the inner wire
@Julie Navia yours too
after spending 2 hours on my hw, I finally came across your video.....20 mins later, this is easy stuff! Thanks!
This is a great video but I do have a question...
Since the electric field outside the coaxial cable is 0, does that mean that the electric field for the distance between a and b (a < r < b) is also 0?
No. Absolutely not. There will always be a field going from inner cable to the inner surface of the outer cable ( from + to - charge )
P.S : I know it's way too late for the answer, probably you don't care about all this now. School/college time over. 😂🙏
There is a problem when you calculate the charge density of the wires at the minute 2:30. There are 2 problems with that, first the charge density will be by surface, not by linear units, second you must equal charges, not densities so the total flux can be 0, in that case it would be something more like lambda*L=-sigma*2*pi*L*a where a is the radius of the inner surface and sigma is the density of the inner surface.
Charge density of the inner surface must be expressed as sigma not lambda and it will be equal = - delta/(2pi*radius of inner surface), also outer sigma = delta/(2pi*radius of outer surface).
No, the E between r=a and r=b will be (lambda)/(2*pi*r) where lambda is the linear charge density of the inner wire and r is the distance from the center of the wire to the place you want to know the field.
But wouldn't the fact that it is within the metal make it equal to 0?
@@NumberFivezer it is not within the metal. It is hollow area, filled with air or any other medium.
Electric field intensity = 0 for ("rb") Do I understand well?
Yes! exactly right.
@@lasseviren1 So quick reply though it's 11years ago video... Thank you very much. It helped me with the homework I have to do by tomorrow. Best wishes to you forever, from korea.
@@Xhakabooom All the best!
this lamda inner and outer and lamda zero stuff is spinning me out.. I'm not sure what is what...
Great video, thanks!
In the first part Lambda outer=0, but later you showed that Lambda outer= Lambda not. It is confusing to me. Also, in part I video there exist E on the outer area of the shell, but here on outside the shell E=0. That's also confusing. What are the difference between two's? Nice video.
lambda naught..
this is a great video.:)...
I like it, ..
May you please explain what is the significant difference as to why E is 0 on the outside in this video but E = (Lambda) / 2*(Pi)*r * (Permittivity of free space)
Pravin Anand in part I **
anyway the video is great :) Nice explanation
this guy is bussin
hello what is gauss theorem
Electric flux through a closed surface equals charge enclosed within that surface divided by permittivity ofbfree space (epsilon not). You'll get this in any standard book bro 🙄
where did the third lambda came from? u trippin? or am i?
i like this.
i don't like you
Sorry, I'm kinda stuck. What if I wanted to find the electric field between two current carrying, nested, hollow cylinders? I can't seem to find anything on that anywhere.
I'm not sure what you mean by "nested" but almost certainly you would use the principle of superposition. For superposition you would find the E field for just one of the cylinders imagining the other cylinder was not even there, then reverse the process for the other cylinder, then add up up both vectors to get the combined effect.
i didnt understand why lambda egal -lambda
Mmkay
the comments are old 💀