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

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

https://trinket.io/glowscript/48f6efd07a

use this code to see how it looks, change the r value.

Examples

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]

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.)

Useful equations:

[math]\displaystyle{ \vec p = \alpha *E }[/math]

[math]\displaystyle{ \vec p = Q*s }[/math]

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find [math]\displaystyle{ \alpha }[/math]

[math]\displaystyle{ \vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s }[/math]

[math]\displaystyle{ \alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 }[/math]

Connectedness

1. How is this topic connected to something that you are interested in?
2. How is it connected to your major?
3. Is there an interesting industrial application?

Looking Ahead, Gauss's Law

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

[math]\displaystyle{ }[/math]

See also

Further reading

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

References

This section contains the the references you used while writing this page

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/