How Do You Calculate the Electric Field of a Uniformly Charged Sphere?

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Homework Help Overview

The discussion revolves around calculating the electric field of a uniformly charged sphere, specifically at a point a distance X from its center. Participants are exploring the relationship between surface charge density and volume charge density while attempting to derive the electric field using Coulomb's Law rather than Gauss's Law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss deriving the electric field from a disk model, integrating over the sphere's radius, and expressing variables in terms of each other. Questions arise regarding the correlation between surface and volume charge densities, as well as the setup of the integration limits.

Discussion Status

There is an ongoing exploration of the mathematical setup, with participants providing insights into expressing the integral in terms of volume charge density. Some guidance is offered regarding the notation and the potential complexity of the integral, but no consensus on a complete method has been reached.

Contextual Notes

Participants express confusion over the integration process and the relationship between different charge densities. There is a mention of homework constraints that may limit the methods available for solving the problem.

PhDorBust
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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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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.
 
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?
 
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 :rolleyes:). I would just look it up or plug it into Mathematica. (Thank goodness for Gauss's law!)
 
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 :rolleyes:). 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..
 
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.
 

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