Field of a Charged Ball
Claimed by Eric Erwood
In this section, the electric field due of sphere charged throughout its volume will be discussed.
The Main Idea
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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
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]
The charge inside the sphere is proportional to r. When r=R,
[math]\displaystyle{ \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} }[/math]
A Computational Model
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Examples
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Simple
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.
Step 1: cut up the sphere into shells
step 2: we know that r<R
Next 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]
[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]
[math]\displaystyle{ \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C }[/math]
Middling
Difficult
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)
For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud
(s « R in the diagram). (Use the following as necessary: R and ε0.)
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Looking Ahead, Gauss's Law
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See also
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References
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http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html