Your video helped me see I was on the right track and even an easier way of doing it because I brute forced it by doing all the cartesian algebra you mentioned. How do you start part B though? I thought maybe you differentiate the potential and set it =0 but then i get lost.
I have a few doubts. 1) Is it necessary as till now in Griffiths, I didn't deal with two systems and finding its potential value, must have the same reference points. What if they are not the same. Then would the superposition principle fail and we can't add the two potentials. 2) I am seriously not getting how you chose "a" to be the reference point as the reference point can be anything, say point P in yz plane (as finding potential in x-axis would be meaningless) except infinity for long wire?
Yes, I too had the same question as you asked in the second point. I have an answer. I don't know whether it's fully correct. But I think it makes sense. So, the reference point can be anything. We generally assume it to be a point far away. There may exist two possibilities. 1. They chose 'a' as an arbitrary point. It has nothing to do with the distance of the wires from the origin. 2. They chose 'a', as the reference point can be anything. And at the end it doesn't affect the result. Any other reference point would have led us to the same result!
So, assuming you're talking about the calculation of the potential, the reason is that the lower limit of integration in the potential formula is whatever point of reference we're choosing for the potential (if you have the Griffiths textbook on hand, see eq. 2.21 on p. 79). The problem statement asks us to set the origin as our reference, and in the cylindrical coordinate system we initially set up for each wire, the origin would be at s=a. The picture I draw around 14:00 shows this a little better. Hope this helps!
@@gregdoesphysics7619 thanks, so (1) when they say origin they mean origin wrt the x-y coord frame (ie not the coordinate frame of the single cylinder where origin (x=a in xy frame) would be 0)? (2) Also for part (b) I found R_c = 2a*sqrt(k)/(1-k) where k = exp(4*Pi*epsilon0*V0/lambda0) -- I know you haven't solved (b) publicly but is it the answer you got? thanks
@@queseraseraballdance For (1), yes exactly! And for (2), yes that is what I got. In fact, if you plug in that exponential you can simplify your answer a little further if you notice that it gives you the equation for a hyperbolic trig function. But you don't necessarily need to do that--your answer already is correct.
Your video helped me see I was on the right track and even an easier way of doing it because I brute forced it by doing all the cartesian algebra you mentioned. How do you start part B though? I thought maybe you differentiate the potential and set it =0 but then i get lost.
you are awsome greg
Great video! thanks a lot:)
I have a few doubts.
1) Is it necessary as till now in Griffiths, I didn't deal with two systems and finding its potential value, must have the same reference points. What if they are not the same. Then would the superposition principle fail and we can't add the two potentials.
2) I am seriously not getting how you chose "a" to be the reference point as the reference point can be anything, say point P in yz plane (as finding potential in x-axis would be meaningless) except infinity for long wire?
Yes, I too had the same question as you asked in the second point. I have an answer. I don't know whether it's fully correct. But I think it makes sense. So, the reference point can be anything. We generally assume it to be a point far away. There may exist two possibilities. 1. They chose 'a' as an arbitrary point. It has nothing to do with the distance of the wires from the origin. 2. They chose 'a', as the reference point can be anything. And at the end it doesn't affect the result.
Any other reference point would have led us to the same result!
why does the radius start from a, shouldn't it start from 0 ? thanks
So, assuming you're talking about the calculation of the potential, the reason is that the lower limit of integration in the potential formula is whatever point of reference we're choosing for the potential (if you have the Griffiths textbook on hand, see eq. 2.21 on p. 79). The problem statement asks us to set the origin as our reference, and in the cylindrical coordinate system we initially set up for each wire, the origin would be at s=a. The picture I draw around 14:00 shows this a little better. Hope this helps!
@@gregdoesphysics7619 thanks, so
(1) when they say origin they mean origin wrt the x-y coord frame (ie not the coordinate frame of the single cylinder where origin (x=a in xy frame) would be 0)?
(2) Also for part (b) I found R_c = 2a*sqrt(k)/(1-k) where k = exp(4*Pi*epsilon0*V0/lambda0) -- I know you haven't solved (b) publicly but is it the answer you got? thanks
@@queseraseraballdance For (1), yes exactly! And for (2), yes that is what I got. In fact, if you plug in that exponential you can simplify your answer a little further if you notice that it gives you the equation for a hyperbolic trig function. But you don't necessarily need to do that--your answer already is correct.