# Electric Field of Sphere

by PhDorBust
Tags: electric, field, sphere
 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'.
Mentor
P: 41,568
 Quote by PhDorBust 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).
Imagine the disks to be stacked along the y-axis, with thickness dy. Put the center of the sphere at y = 0.

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.

 I am also confused as to how exactly to correlate surface density with volume density.
Consider each disk of thickness dy to have a charge equal to ρAdy, where A is the area of the disk (a function of y).

 I wish to use Coulomb's Law and not Gauss'.
It's a good exercise.
P: 142
 Quote by Doc Al Imagine the disks to be stacked along the y-axis, with thickness dy. Put the center of the sphere at y = 0. 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.
OK, translating that to the x-axis I did:

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?

Mentor
P: 41,568
Electric Field of Sphere

 Quote by PhDorBust OK, translating that to the x-axis I did: 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?
You seem to have the right idea, but I can't follow your notation easily. Express the integral in terms of volume charge density (instead of surface charge density), the radius of the sphere, and the distance from center to the point. The integral is not one I can solve off the top of my head anymore (assuming I was able to at one point ). I would just look it up or plug it into Mathematica. (Thank goodness for Gauss's law!)
P: 142
 Quote by Doc Al You seem to have the right idea, but I can't follow your notation easily. Express the integral in terms of volume charge density (instead of surface charge density), the radius of the sphere, and the distance from center to the point. The integral is not one I can solve off the top of my head anymore (assuming I was able to at one point ). I would just look it up or plug it into Mathematica. (Thank goodness for Gauss's law!)
Sorry about inconsistency.. didn't notice it until now. =/ Here is fixed:

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)

 Quote by Doc Al Consider each disk of thickness dy to have a charge equal to ρAdy, where A is the area of the disk (a function of y).
R^2 = r^2 - (X - x)^2
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..
 Mentor P: 41,568 When I set it up, using my own notation, I get: $$E = \int_{-R}^{R} 2\pi k \rho (1 - \frac{z-x}{\sqrt{(z-x)^2 + R^2 - x^2}}) dx$$ 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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