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1 year ago with 300 notes
Electric Field Due to a Uniformly Charged Disk 
Perhaps one of the most difficult concepts in general physics is uniformly charged surfaces. Really, I can’t think of anything more rigorous than E&M and complex integration.
Usually with these problem sets we are given the task of finding the E field at a distance and/or the E field when two charges are in the system.
So when given a uniformly charged disk and asked to find the E field at a distant point here’s how we do it:
First we need to divide the disk into flat rings so we can calculate the E field at our point P by adding up all the rings. Each ring will be out at a distance y and will have a radial width of dy. 
Next we’ll set up the E field equation E=kQ/r2, but write it as dE=kdQ/√(x2 + y2).
Now we come to our first tricky part of E field integration; the dQ. When we do these problems we must remember our charge densities - linear, surface, volume; lambda, sigma, and rho respectively. 
Linear - lambda, 1 dimension : dQ = λdx
Surface - sigma, 2 dimensions : dQ = σdA
Volume - rho, 3 dimensions : dQ = ρdV
Here, we use sigma and use the relationship dQ = σdA = σ2πydy. All that means is that our charge, dq, is equal to the surface charge density, σ, times the area of the charged concentric ring.  
Now that we have our charge set we can look at the x and y components of the E field. If you look at dEy you’ll see that as you go around the ring there is another y component that cancels each other out. Therefore we only focus on dEx since the vector sum of dEy will add to 0. 
Our dEx = kdQ cosθ /(x2 + y2). We use cosθ because on our free-body diagram dEx is the adjacent-hypotenuse component of dE. Further substitute the dEx equation by using x/√(x2 + y2) for cosθ  and 1/4πε for k.
Only thing left to worry about is your trig identities of integration. Take the integral of dEx from 0 to R and you will find the E field at point P. 

Electric Field Due to a Uniformly Charged Disk 

Perhaps one of the most difficult concepts in general physics is uniformly charged surfaces. Really, I can’t think of anything more rigorous than E&M and complex integration.

Usually with these problem sets we are given the task of finding the E field at a distance and/or the E field when two charges are in the system.

So when given a uniformly charged disk and asked to find the E field at a distant point here’s how we do it:

First we need to divide the disk into flat rings so we can calculate the E field at our point P by adding up all the rings. Each ring will be out at a distance y and will have a radial width of dy

Next we’ll set up the E field equation E=kQ/r2, but write it as dE=kdQ/√(x2 + y2).

Now we come to our first tricky part of E field integration; the dQ. When we do these problems we must remember our charge densities - linear, surface, volume; lambda, sigma, and rho respectively. 

Here, we use sigma and use the relationship dQ = σdA = σ2πydy. All that means is that our charge, dq, is equal to the surface charge density, σ, times the area of the charged concentric ring.  

Now that we have our charge set we can look at the x and y components of the E field. If you look at dEy you’ll see that as you go around the ring there is another y component that cancels each other out. Therefore we only focus on dEx since the vector sum of dEy will add to 0. 

Our dEx = kdQ cosθ /(x2 + y2). We use cosθ because on our free-body diagram dEx is the adjacent-hypotenuse component of dE. Further substitute the dEx equation by using x/√(x2 + y2) for cosθ  and 1/4πε for k.

Only thing left to worry about is your trig identities of integration. Take the integral of dEx from 0 to R and you will find the E field at point P. 



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    Jajaja que era entrete y a la vez frustrante hacer estos ejercicios.
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    This may be relevant to my life right now.