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Electric Field of Sphere 
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#1
Sep409, 04:21 PM

P: 142

We have a uniformly charged sphere (charge is all over, not just on surface) and want to determine the electric field at a point that is distance X from the center of the sphere. The radius of the sphere is known.
I first derived the electric field from a disk to a point that is distance x away from the center of the disk. s = surface density = Q/A ep = permitivity of free space constant r = radius of sphere R = radius of disk I found this to be [s/(2ep)]*[1  x/sqrt(x^2 + R^2)] Now I have to integrate this from X  R to X + R. But I cannot figure out how to express the radius of the disks in terms of the sphere radius without introducing more variables (ie angle variable). I am also confused as to how exactly to correlate surface density with volume density. I wish to use Coulomb's Law and not Gauss'. 


#2
Sep409, 04:46 PM

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P: 41,323

Consider a right triangle with hypotenuse equal to the radius of the sphere and one side equal to y. Express the radius of individual disks in terms of the variable y. 


#3
Sep509, 11:13 AM

P: 142

X = Given distance from center of sphere to point R^2 = r^2  (x_0  x)^2 Plugging this in gives: [s/(2ep)]*[1  x/sqrt(R^2 + 2*x*X  X^2 )]. I want to integrate this function from X  r to X + r. Did I perform this step correctly? If so, how do I integrate this function? Do I have to make some approximation? 


#4
Sep509, 03:00 PM

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P: 41,323

Electric Field of Sphere



#5
Sep509, 04:31 PM

P: 142

R^2 = r^2  (X  x)^2 Plugging this in gives: [s/(2ep)]*[1  x/sqrt(r^2 + 2*x*X  X^2 )]. I want to integrate this function from X  r to X + r. surface charge density = s = Q/(pi * R^2) volume charge density = p = Q/(4/3 * pi * r^3) pA = p * pi * R^2 = p * pi * (r^2  X^2  x^2 + 2*x*X) Plugging this in gives: [(p * pi * (r^2  X^2  x^2 + 2*x*X))/(2ep)]*[1  x/sqrt(r^2 + 2*x*X  X^2 )]. I want to integrate this function from X  r to X + r. It almost cancels out nicely... And I have absolutely no idea what to do with this gigantic integral. Is there another math approach I should use that would make this easier? My teacher said doing it by disks was simplest.. 


#6
Sep609, 08:27 AM

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P: 41,323

When I set it up, using my own notation, I get:
[tex]E = \int_{R}^{R} 2\pi k \rho (1  \frac{zx}{\sqrt{(zx)^2 + R^2  x^2}}) dx [/tex] Here I use z as the distance between the center of the sphere and the point in question, and x as the position of a particular disk from the origin. Assuming I didn't mess it up, I would then simplify a bit and start checking the integration tables (or Mathematica). It will be a mess, but when all is said and done it should simplify nicely. 


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