Field of a Charged Ball

The Main Idea

In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells

A Mathematical Model

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of 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 of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for. 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, we again treat the sphere as a point charge located at the center of the sphere.

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

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{ \oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} }[/math]

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

[math]\displaystyle{ \vec E * 4\pi*r^2= \frac{1}{\epsilon_0}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^3}r }[/math]

A Computational Model

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

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

   # GlowScript 1.1 VPython
    ## constants
   oofpez = 9e9
   qproton = 1.6e-19
   scalefactor = 2e-20
   Radius =3.5
   r=3

   ## objects
   particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
   ## initial values
   obslocation = vector(r,0,0)
   ##calculate position vector
   rvector = obslocation - particle.pos
   arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
   rmag=mag(rvector)

   ##colored arrow

   if r<= Radius:
       E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
       E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
       E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
       E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
       E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
       E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
       ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
       ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
       ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
       ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
       ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
       ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
   if r>Radius:
       E1= oofpez*(qproton)/(rmag)**2  *vector(1,0,0)
       E2= oofpez*(qproton)/(rmag)**2  *vector(-1,0,0)
       E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
       E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
       E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
       E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
       ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
       ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
       ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
       ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
       ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
       ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

Examples

Simple

Figure 2: This is an example of applying the equation for finding the electric field of a point charge

Describe the pattern of the electric field of charged ball from far away.

Solution

If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by: [math]\displaystyle{ \vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r }[/math]

Middling

Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.

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.

Solution

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.)

Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha

Solution

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

It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

History

(should be completed by a student)

See also

Further reading

• Point Charge

• Electric Dipole

• Gauss's Flux Theorem

References

• 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/

• Matter and Interactions, 4th Edition: 1-2