Electric Field of Sphere

In summary: X - r to X + r is not an exact procedure. A better approximation could be: [(p * pi * (r^2 - X^2 - x^2 + 2*x*X))/(4*ep)]*[1 - x/sqrt(r^2 + 2*x*X - X^2 )]
  • #1
PhDorBust
143
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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'.
 
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  • #2
PhDorBust said:
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.
 
  • #3
Doc Al said:
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?
 
  • #4
PhDorBust said:
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 :uhh:). I would just look it up or plug it into Mathematica. (Thank goodness for Gauss's law!)
 
  • #5
Doc Al said:
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 :uhh:). 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)

Doc Al said:
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..
 
  • #6
When I set it up, using my own notation, I get:

[tex]E = \int_{-R}^{R} 2\pi k \rho (1 - \frac{z-x}{\sqrt{(z-x)^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.
 

1. What is the electric field of a sphere?

The electric field of a sphere refers to the force exerted by a charged sphere on another charged particle at any point in space around it. It is a measure of the strength and direction of the electric force experienced by a particle due to the presence of the charged sphere.

2. How is the electric field of a sphere calculated?

The electric field of a sphere can be calculated using Coulomb's law, which states that the electric field at a point is directly proportional to the magnitude of the charge on the sphere and inversely proportional to the square of the distance from the center of the sphere.

3. What is the direction of the electric field of a sphere?

The direction of the electric field of a sphere is always radially outward or inward, depending on the type of charge on the sphere. If the sphere has a positive charge, the electric field will point radially outward, and if it has a negative charge, the electric field will point radially inward.

4. How does the electric field of a sphere change with distance?

The electric field of a sphere follows an inverse-square law, meaning that it decreases with the square of the distance from the center of the sphere. This means that as the distance from the sphere increases, the electric field strength decreases.

5. Can the electric field of a sphere be uniform?

Yes, the electric field of a sphere can be uniform if the charge distribution on the sphere is also uniform. This means that the electric field strength will be the same at all points on the surface of the sphere, and the direction will also be the same. However, this is only possible for a perfectly symmetric and uniformly charged sphere.

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