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# Proof: Field from infinite plate (part 2)

## Video transcript

so where I left off we had this infinite this infinite plate it's just an infinite plane and it's a charged plate with a charge density Sigma and we what we did is we said okay well we have some you know we're taking this point up here that's H units above the surface of our charge plate and we wanted to figure out the the electric field at that point generated by a ring of radius R essentially centered you know at the base of above the base of where that point is above we want to figure out what is the electric field generated by this ring at that point and we figured out that well the electric field was this and then because we made a symmetry argument in the last video we only care about the Y component because we figured out if the electric field generated from any point the X components cancel out because if we have a point here it he'll have some X component field the fields X component might be in that direction to the right then you have another point out here and it's X component will just cancel it out so we only compare care about the Y component so at the end we meticulously calculated what the Y component of the electric field generated by the ring is at H units above the surface so with that out of the way let's see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total Y component or essentially the total electric field because we realize that all the X's cancel out anyway the total electric field at that point H units above the surface of the of the plane so let me erase a lot of this just so I can free it up for some hardcore math and this is pretty much all calculus at this point so let me erase all of this watch the previous video if you forgot how it was derived and let me even erase that because I think I will need a lot of space there you go okay so let me let me redraw a little bit just so we never forget what we what we're doing here because that's easy that that happens so that's my plane goes off in every direction I have my point above the plane where we're trying to figure out the electric field and we've come to the conclusion that the field is going to point upward so we only care about the Y component it's H units above the surface it's H units above the surface and we're figuring out the electric field generated by a ring around that that this point of radius R and what's it's white with the y component of that electric field we figured out it was this so now what we're going to do is take the integral so the total electric field from the plate so the total electric field is going to be the integral from our that's a really ugly looking integral from a radius of zero to a radius of infinity so we're going to take a sum of all of the rings starting with the radius of zero all the way to the ring that has a radius of infinity because it's an infinite plane so we're figuring out the impact of the entire plane tot going take the sum of every ring so at the charge of or the field generated by every ring and this is the field generated by each of the rings so let's let me do in a different color this light blue is getting a little not lit monotonous kh2 pi sigma r dr over H squared plus R squared to the three-halves now let's simplify this a little bit let's take some constants out of it just so this looks like a slightly simpler equation so this equals the integral from 0 to n hopes I didn't want to write that just yet so let's take the K I'm going to leave the two there and you'll see why in a second I'm gonna take all the other constants out that we're not integrating across so it's equal to K H Pi Sigma Pi Sigma times the integral from 0 to infinity of what is this so what did I what did I leave in there I left a 2r so we could rewrite this as well actually I'm running out of space but let me write to our D R over H squared plus R squared to the three-halves or we could think of it as the negative three-halves right and so what's the antiderivative of here well this is essentially the reverse chain rule right I could make a substitution here if you're more comfortable using the substitution rule but you might be able to eyeball this at this point we could make the substitution that let me pick a good color we could make the substation that U is equal to if we just want to figure out the antiderivative this if u is equal to H squared plus R squared H is just a constant right then D U is just equal to I mean D the the dudr right dudr is just this is a constant so it equals to R or we could say D U is equal to 2r dr and so if we're trying to take the antiderivative if we're trying to take the antiderivative of 2 r dr / h squared plus R squared to the three-halves this is the exact same thing as taking the antiderivative with this substitution to our dr we just show right here that the same thing is d u right so that's D u over and then this is just u right H squared plus R squared is you that we do that by definition so u to the three-halves which is equal to the antiderivative of we could write this as u to the minus three-halves d u and now that's easy this is just kind of reverse I guess you would call it what the exponent rule so that equals minus 2 u to the minus 1/2 and we can confirm right if we take the derivative this minus 1/2 times -2 is 1 and then subtract from 1 from here we get minus three-halves and then we could add plus c but since we're eventually going to do a definite integral note the C's all cancel out or we could say that this is equal to since we made that substitution this equals -2 over we could say you know 2 minus 1/2 that's the same thing as over the square root of H squared plus R squared right so all of this stuff I did in magenta was just to figure out the antiderivative of this and we figured it out to be this minus 2 over H squared + / the square root of H squared plus R squared so with that out of the way let's continue evaluating our definite integral so this expression simplifies to this is a marathon problem but satisfying K let's get all the constants K H Pi Sigma we can even take this minus 2 out let's take this x times minus 2 and all of that and we're going to evaluate the definite integral at the two boundaries 1 over the square root of H squared plus R squared evaluated at an infinity minus it evaluated at 0 right well what is what is this expression equal well when you what is 1 over the square root of 8 squared plus infinity right what happens when we evaluate R at infinity well the square root of an infinity is still infinity and 1 over infinity is 0 so this expression right here is becomes 0 when you evaluate it at infinity this becomes 0 minus this expression evaluated 0 so what happens when it's at 0 when R squared is 0 we get 1 over the square root of H squared right so let's let's write it all out this becomes minus 2 K H Pi Sigma times zero minus one over the square root of H squared well this equals minus 2 K H Pi Sigma times well 1 over the square root of H squared that's just 1 over H right and there's a minus x minus 1 over h well this - and that - cancel out fair enough and then this H and this 1 over H should cancel out fair enough and all we're left with after doing all of that work and I'll do it in a bright color because we've done a lot of work to get here is 2 K PI Sigma so that's needed on a lot of levels first of all what we even do here we might have gotten lost in the math this is the net electric field the total electric field at a point at height H above this infinite ly uniform to this infinite plate that it has a uniform charge and the charge density is Sigma but notice this this is the electric field at that point but there's no H in here so it essentially is telling us that there's no the the the strength of the field is in no way dependent on how high above the field we are which tells us is going to be a constant field we can be anywhere above the plate and the charge will be the same the only thing of the necessary not to charge the field will be the same the only thing that and if we have a test charge the force would be the same and the only thing that the strength of the field or the strength of the exerted electrostatic force is dependent on is the charge density right this is Coulomb's constant pi is PI 2 pi and I think it's kind of cool that it involves PI but that's something else to to ponder but all it mattered all that matters is the charge density so hopefully you found that reasonably satisfying and the big thing that we learned here is that if I have an infinite uniformly charged plate the field you know and I put a some distance H above that field you know above that I keep mixing upwards above that plate it doesn't matter what that H is be here I could be here I could be here at all of those points the field has the exact same strength or the net electrostatic force on a test charge at those points has the exact same strength and that's kind of a neat thing and now if you do believe everything that occurred in the last two videos you can now believe that we have that there are such things as uniform electric fields and they occur between parallel plates especially far away from the boundaries see you soon