@Michel van Biezen if we draw a gaussian cylinder enclosing both the cylinders then net charge enclosed will be zero as there are equal and opposite charge densities.Then the electric field outside the cylinders should be zero according to gauss's law right?
I"m having trouble with a problem, infinite coaxial cylinders the inner cylinder has a charge density rho(non-conducting) and radius a. the outer cylinder has radius b and is a grounded conductor. I'm being asked to solve for the electric potential everywhere in space. I have a hard time figuring out how the grounding effects the system.
How did you get the electric field formula that is on the top? Isn't the denominator 4*pi*epsilon naught (because k sub e is 1/4*pi*epsilon naught)? I have never seen that formula before... Anyway, thx for all these videos...My AP Physics C grade is a 99 because of your videos!!!
Wait a minute... I figured it out... You applied Gauss' Law beforehand to derive that formula for the electric field using a cylindrical Gaussian surface. Is that right?
If the outside cylinder had a thickness to it, would a negative charge be induced on the inner wall of it and a positive charge be induced on the outside surface?
Think of charging a capacitor. It takes almost no work to add the first charge. But to add the last charge it takes a lot of work. To place the average charge on a capacitor, it takes 1/2 the work of the last one.
It depends on the application. Here the difference in potential depends on the strength of the electric field between the two cylinders. The electric field can be found using Gauss's law which shows that the electric field depends on the total charge inside the surface and it doesn't matter if it is along a line of charge or the surface of a small cylinder (as long as the charge in uniformly distributed), thus we can use line charge density just as well as surface charge density,
It is by convention. Since the potential is greater near the positive charge, the equation requires a negative sign to ensure that the potential increases as you get closer to the positive charge.
@@MichelvanBiezen sir understanding of sign it is so emberassing for me. If the potential is greater when it is toward to the positive why we put negative sign is it going to lower to bigger?
Michel van Biezen Hello sir, thank you for the help I would like to ask whether an electric field exists inside the larger cylinder. Since the larger cylinder is a conductor shouldn't the electric field inside it be 0. Moreover does the electric potential change as we move between r2 and r1 because if there's no electric field there should be no change in potential. Once again. thank you for untying the knot in my head All the best
Since there is charge on both the inner and outer cylinder, there will be an electric field between them. (No electric field inside the inner cylinder).
Excuse me, but what I mean to ask is, is it hollow?as of no presence of metal around the area I got mistaken thinking that it is two solid cylinder, that kept me wondering why there is electric field in between.... but at last i figure that maybe it is hollow.....if it is hollow i can understand why there is electric field in between....
It is wonderful that I can return to these lectures and listen again for more clarity. Thanks Professor for making these available.
You are welcome. Glad you are enjoying them.
So we changed the d from E=Vd into r when we did the integration when finding the voltage? It was a bit confusing...
Both d and r are variables representing linear distance.
Why do we go from the outer surface to the inner when integrating? Can we go the opposite direction
You can. There will be a sign difference.
what's the different about the result if the inner plate is given a negative charge and the outer plate is given a positive charge?
The direction of the electric field would be opposite, and the sign of the change in voltage would be opposite as well.
The negative charges on the outer cylinder will accumulate on the inner surface of that cylinder.
You are correct. They will migrate to the inner surface of the outer cylinder.
@Michel van Biezen
if we draw a gaussian cylinder enclosing both the cylinders then net charge enclosed will be zero as there are equal and opposite charge densities.Then the electric field outside the cylinders should be zero according to gauss's law right?
That is correct.
I"m having trouble with a problem, infinite coaxial cylinders the inner cylinder has a charge density rho(non-conducting) and radius a. the outer cylinder has radius b and is a grounded conductor. I'm being asked to solve for the electric potential everywhere in space. I have a hard time figuring out how the grounding effects the system.
Thank you sir! Wonderful and clear explanation!
You're welcome!
Wonderful. Thanks for an excellent lecture.
You're most welcome! 🙂
How did you get the electric field formula that is on the top? Isn't the denominator 4*pi*epsilon naught (because k sub e is 1/4*pi*epsilon naught)? I have never seen that formula before... Anyway, thx for all these videos...My AP Physics C grade is a 99 because of your videos!!!
Wait a minute... I figured it out... You applied Gauss' Law beforehand to derive that formula for the electric field using a cylindrical Gaussian surface. Is that right?
That is correct.
If the outside cylinder had a thickness to it, would a negative charge be induced on the inner wall of it and a positive charge be induced on the outside surface?
That is correct.
@@MichelvanBiezen thank you sir.
@@MichelvanBiezen It's awesome how you take the time to answer these questions. All the respect I can give for you dear sir!
Now if we calculate yhe electrostatic energy per unit length,we get a factor of 1/2...why is that sir?
Think of charging a capacitor. It takes almost no work to add the first charge. But to add the last charge it takes a lot of work. To place the average charge on a capacitor, it takes 1/2 the work of the last one.
hello! why are we using lamda and not sigma? isnt the charge spread out on the lateral surface area of the cylinder?
It depends on the application. Here the difference in potential depends on the strength of the electric field between the two cylinders. The electric field can be found using Gauss's law which shows that the electric field depends on the total charge inside the surface and it doesn't matter if it is along a line of charge or the surface of a small cylinder (as long as the charge in uniformly distributed), thus we can use line charge density just as well as surface charge density,
@@MichelvanBiezen So here we could also have used surface area charge density and would still get the same answer?
Hi Pr. Could you please explain to me why you always put the negative sign when you integrate. Thank you.
It is by convention. Since the potential is greater near the positive charge, the equation requires a negative sign to ensure that the potential increases as you get closer to the positive charge.
@@MichelvanBiezen sir understanding of sign it is so emberassing for me. If the potential is greater when it is toward to the positive why we put negative sign is it going to lower to bigger?
Thanks professor Michel.
sir do you have a video about finding v for concentric sphere/shell and one is earthed
This video may help: Physics - E&M: Electric Potential (20 of 22) Equipotential Surfaces in a Varied E-Field
man u the best
Ur a legend
so in this case we are assuming that the outside cylinder has no thickness - almost zero thickness-?
If you are referring to the Gaussian surface, you are correct.
Thank you so much sir from India,awesome teaching sir, with line by line explanation
Welcome to the channel
Michel van Biezen
Hello sir, thank you for the help
I would like to ask whether an electric field exists inside the larger cylinder. Since the larger cylinder is a conductor shouldn't the electric field inside it be 0. Moreover does the electric potential change as we move between r2 and r1 because if there's no electric field there should be no change in potential.
Once again. thank you for untying the knot in my head
All the best
Since there is charge on both the inner and outer cylinder, there will be an electric field between them. (No electric field inside the inner cylinder).
Thank you
+Michel van Biezen
Mr.Michel, is it empty between the two cylinder?
+Michel van Biezen
Mr.Michel, is it empty between the two cylinder?
Yes, there is no charge between the two cylinders.
Excuse me, but what I mean to ask is, is it hollow?as of no presence of metal around the area
I got mistaken thinking that it is two solid cylinder, that kept me wondering why there is electric field in between....
but at last i figure that maybe it is hollow.....if it is hollow i can understand why there is electric field in between....
Could we do it by the gaussian surface
It depends on what you mean by "do it".
thank you
Helpful. Thank you!