Field of a Charged Ball

page edited by Jesse N. Maina (Spring 2026)

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

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Figure 2: An uncharged sphere produces no electric field at any distance.

Consider a solid sphere of radius 1.5 meters that carries no net charge. Determine the electric field at an observation point located 3.2 m from the center of the sphere.

Solution

Because the sphere is completely uncharged, there is no source of electric field anywhere in space. Using the point‑charge form of Coulomb’s law:

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

and substituting [math]\displaystyle{ Q }[/math] for 0 gives

[math]\displaystyle{ \vec{E} = \frac{1}{4\pi\epsilon_0}\frac{0}{r^2}\,\hat{r} = \vec{0}. }[/math]

Thus, the electric field is zero everywhere, regardless of the observation distance.

Middling

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Figure 3: A charged sphere behaves like a point charge when observed from outside the sphere.

A solid sphere of radius 2.0 meters carries a total charge of +5 C. A point of observation is located 6.0 meters from the center of the sphere. Because the observation point lies outside the sphere, the entire charged ball behaves exactly like a point charge. Use Coulomb’s law to determine the magnitude and direction of the electric field at this location.

Solution

Since the observation point is at [math]\displaystyle{ r = 6\ \text{m} \gt R = 2\ \text{m} }[/math], the sphere acts like a point charge of total charge [math]\displaystyle{ Q = +5\ \text{C} }[/math]. Thus, the electric field is given by Coulomb’s law:

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

Substitute the values:

[math]\displaystyle{ E = (8.99\times 10^9)\frac{5}{(6)^2} = 8.99\times 10^9 \cdot \frac{5}{36} \approx 1.73\times 10^9\ \text{N/C} }[/math]

Because the charge is positive, and due to symmetry, the electric field points radially outward from the center of the sphere.

Difficult

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Figure 3: A non‑uniformly charged ball requires integrating the charge density to find the field.

A solid sphere of radius 0.40 m has a non‑uniform charge density

[math]\displaystyle{ \rho(r) = \rho_0 \left(\frac{r}{R}\right) }[/math]

with [math]\displaystyle{ \rho_0 = 6.0\times 10^{-6}\ \text{C/m}^3 }[/math]. A point P is located 0.25 m from the center of the sphere. Find the total charge of the sphere by integrating the charge density over the volume. Then, use Gauss’s law to find the magnitude of the electric field at point P.

Solution

Step 1: Compute the total charge.

Cut the sphere into thin spherical shells. Each shell has charge

[math]\displaystyle{ dQ = \rho(r)\,4\pi r^2\,dr = \rho_0\left(\frac{r}{R}\right)4\pi r^2\,dr. }[/math]

Then you will want to integrate from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ R }[/math]:

[math]\displaystyle{ Q = \int_0^R \rho_0\frac{r}{R}4\pi r^2\,dr = \frac{4\pi\rho_0}{R}\int_0^R r^3\,dr = \pi\rho_0 R^3. }[/math]

Plugging in numbers gives you

[math]\displaystyle{ Q \approx 1.2\times 10^{-6}\ \text{C}. }[/math]

---

Step 2: Find the enclosed charge at [math]\displaystyle{ r = 0.25\ \text{m} }[/math].

[math]\displaystyle{ Q_{\text{enc}}(r) = \int_0^r \rho_0\frac{r'}{R}4\pi r'^2\,dr' = \frac{4\pi\rho_0}{R}\int_0^r r'^3\,dr' = \pi\rho_0\frac{r^4}{R}. }[/math]

---

Step 3: Apply Gauss’s law.

For a point inside a spherically symmetric charge distribution:

[math]\displaystyle{ E(r) = \frac{1}{4\pi\epsilon_0}\frac{Q_{\text{enc}}(r)}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{\pi\rho_0 r^2}{R}. }[/math]

Evaluating at [math]\displaystyle{ r = 0.25\ \text{m} }[/math] gives

[math]\displaystyle{ E \approx 2.6\times 10^{4}\ \text{N/C}. }[/math]

The direction is radially outward. Or, in other words, the only direction the electric field is allowed to point towards is straight along the radius (due to symmetry).

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