We can calculate the electric field at the corner point y by first calculating the contribution of a square of side length a at a point d/2 in the vertical dimension, where d is the distance between the plates. If we do this you get E_z = (Q/(4*pi*eps0*a^2))*arctan( 2*(a/d)^2 / sqrt( 2*(a/d)^2 + 0.25 ) ). Using the principle of superposition, the field at the corner is double this. In the limit of a capacitor with side length much larger than plate spacing a >> d, the arctan() expression approaches a limiting value of pi/2, and so the field strength agrees with your analysis.
Sir, my concepts are extremely poor in electromagnetism, so pardon the silliness of the question- in both the cases when we equate 2E_x and 4E_x to E_i, why are we not increasing the value of A in the denominator? Shouldnt the denominators have 2A and 4A as the composite capacitors have much bigger areas?
It's a good question, and I probably should have said something about this in the video! You're right that the areas of the composite capacitors are larger, but their stored charges are also larger by the same proportion since we're just combining the individual charges on the smaller capacitors. So, the value of Q/A doesn't change and the same expression for E_i still applies.
@@DrBenYelverton So obvious! I am sorry, I had a terrible teacher who taught us EM and I have no sense or understanding of the subject- to the extent that I am terrified of even thinking about it. Thank you for clearing my doubt. Great video as usual!
Thanks! It's a great subject (and one that I'll definitely be doing more videos on) but can be a little intimidating and it takes time to build up intuition.
Hi! I am a physics enthusiast and I have to admit that your channel is way over the top: you explain in an excellent way! You did really a good job and you'd deserve more subscribers! I'd have a curiosity: in addition to the numerically computed field you mentioned, do you know if it is possible to write an analytical expression to describe those curved field lines? I mean, can we write the "edge" electric field as a precise function of cartesian coordinates x, y and z? Basically, all I have been able to find are papers about fringing effect capacitance, but nothing about the field itself..
Thanks for your kind words! In principle this could be done by splitting each plate of the capacitor into infinitesimal charge elements and integrating, then adding the fields due to each of the plates. I've done the integration to get the on-axis field of a single plate in the following video - ruclips.net/video/EJ0m5KASeh8/видео.html - but haven't tried doing this for arbitrary positions. There's a lot of work involved in the on-axis case already and it will be even worse for arbitrary positions due to the lack of symmetry! I'm not sure whether the resulting integrals can be done analytically - I suspect it's possible that they can, but certainly won't come out as neat expressions. I'd be interested to hear about it if you decide to give it a try though!
The vertical component of the field at the point where the three capacitors meet would just be three times as big as at the corner of a single capacitor, so 3Q/4ε₀A - this follows from the principle of superposition.
We can calculate the electric field at the corner point y by first calculating the contribution of a square of side length a at a point d/2 in the vertical dimension, where d is the distance between the plates. If we do this you get E_z = (Q/(4*pi*eps0*a^2))*arctan( 2*(a/d)^2 / sqrt( 2*(a/d)^2 + 0.25 ) ). Using the principle of superposition, the field at the corner is double this. In the limit of a capacitor with side length much larger than plate spacing a >> d, the arctan() expression approaches a limiting value of pi/2, and so the field strength agrees with your analysis.
Nice! I hadn't checked against the more general formula, good to know that both approaches agree.
really interesting topic :) - had fun learning it :D
Very lnteresting topic on e.m.!
Sir, my concepts are extremely poor in electromagnetism, so pardon the silliness of the question- in both the cases when we equate 2E_x and 4E_x to E_i, why are we not increasing the value of A in the denominator? Shouldnt the denominators have 2A and 4A as the composite capacitors have much bigger areas?
It's a good question, and I probably should have said something about this in the video! You're right that the areas of the composite capacitors are larger, but their stored charges are also larger by the same proportion since we're just combining the individual charges on the smaller capacitors. So, the value of Q/A doesn't change and the same expression for E_i still applies.
@@DrBenYelverton So obvious! I am sorry, I had a terrible teacher who taught us EM and I have no sense or understanding of the subject- to the extent that I am terrified of even thinking about it. Thank you for clearing my doubt. Great video as usual!
Thanks! It's a great subject (and one that I'll definitely be doing more videos on) but can be a little intimidating and it takes time to build up intuition.
Hi!
I am a physics enthusiast and I have to admit that your channel is way over the top: you explain in an excellent way! You did really a good job and you'd deserve more subscribers!
I'd have a curiosity: in addition to the numerically computed field you mentioned, do you know if it is possible to write an analytical expression to describe those curved field lines? I mean, can we write the "edge" electric field as a precise function of cartesian coordinates x, y and z?
Basically, all I have been able to find are papers about fringing effect capacitance, but nothing about the field itself..
Thanks for your kind words! In principle this could be done by splitting each plate of the capacitor into infinitesimal charge elements and integrating, then adding the fields due to each of the plates. I've done the integration to get the on-axis field of a single plate in the following video - ruclips.net/video/EJ0m5KASeh8/видео.html - but haven't tried doing this for arbitrary positions. There's a lot of work involved in the on-axis case already and it will be even worse for arbitrary positions due to the lack of symmetry! I'm not sure whether the resulting integrals can be done analytically - I suspect it's possible that they can, but certainly won't come out as neat expressions. I'd be interested to hear about it if you decide to give it a try though!
You should upload solutions to pathfinder physics book!
What would the Ey be if we just join 3 capacitors?
The vertical component of the field at the point where the three capacitors meet would just be three times as big as at the corner of a single capacitor, so 3Q/4ε₀A - this follows from the principle of superposition.