Field of a Charged Ball: Difference between revisions
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<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math> | <math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math> | ||
We find that: | |||
<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math> | |||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 11:53, 1 December 2015
Claimed by Eric Erwood
In this section, the electric field due of sphere charged throughout its volume will be discussed.
The Main Idea
State, in your own words, the main idea for this topic
In this section, we will focus on a scenario where a sphere has charge distributed throughout the entire object. Calculating the electric field both outside and inside the sphere will be addressed.
A Mathematical Model
What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.
Step 1: Given a solid charged sphere throughout its volume, the first step is to cut up the the sphere into pieces. As a result, the solid sphere will appear as a series of spherical shells.
Step 2: Relationship between r and R. Next, it is necessary to determine whether the observation point is outside or inside the sphere.
If r>R, then we are outside the sphere. All the spherical shells appear as point charges at the center of the sphere. As a result, the electric field outside the sphere is a point charge:
[math]\displaystyle{ \vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r }[/math] when r>R, and R is the radius of the sphere.
However, when r<R, the observation location is inside some of the shells but outside others. To find [math]\displaystyle{ \vec E_{net} }[/math], add the contributions to the electric field from the inner shells. After adding the contributions of each inner shell, you should have an electric field equal to:
[math]\displaystyle{ \vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2} }[/math]
We find [math]\displaystyle{ \Delta Q }[/math]:
[math]\displaystyle{ \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} }[/math]
We find that:
[math]\displaystyle{ \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r }[/math]
A Computational Model
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
Examples
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Simple
Middling
Difficult
Connectedness
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History
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See also
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Further reading
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External links
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References
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