Field of a Charged Ball

Page Claimed/Edited by Jesse N. Maina (Spring 2026)

Overview

In this section, we will discuss the electric field of a solid sphere. A solid 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 in the provided text.

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 [math]\displaystyle{ r }[/math], the distance of the observation point to the center of the sphere, and [math]\displaystyle{ R }[/math], the radius of the sphere itself.

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

If r is greater than R ([math]\displaystyle{ r \gt R }[/math]), 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]

where [math]\displaystyle{ r \gt R }[/math], and [math]\displaystyle{ R }[/math] is the radius of the sphere.

However, when r is less than R ([math]\displaystyle{ r \lt R }[/math]), 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. Note that 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 phenomenon is due to the electric field produced by these larger shells, which 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 the change of charge, [math]\displaystyle{ \Delta Q }[/math], through:

[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]

And, subsequently, 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 [math]\displaystyle{ r }[/math]. When [math]\displaystyle{ r = R }[/math], 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*[1].

For the time being, know that Gauss's law validates the follwing statement:

The total electric flux through any closed surface equals the enclosed charge divided by [math]\displaystyle{ \epsilon_0 }[/math].

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 {enc}} }[/math]

Using this, one can calculate the electric field when [math]\displaystyle{ r \lt R }[/math] throughout a charged solid sphere.

[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 VPython demonstrations provides a visualization of the of the electric field fo a charged solid sphere. Try adjusting the radius, [math]\displaystyle{ R }[/math], or the observation distance, [math]\displaystyle{ r }[/math], in the code to see how the field changes!

Link to Code: https://trinket.io/glowscript/48f6efd07a

   # GlowScript 1.1 VPython

   ##Constants
   oofpez = 9e9
   qproton = 1.6e-19
   scalefactor = 2e-20
   Radius = 3e-10
   r = 3e-10

   ##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)

   ##Offset for arrows (for visuals)
   offset = 1e-9

   ##Colored arrows (In Sphere; Integration)
   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)
       arrow(pos=vector(offset,0,0), axis=E1*scalefactor, color=color.orange)
       arrow(pos=vector(-offset,0,0), axis=E2*scalefactor, color=color.orange)
       arrow(pos=vector(0,offset,0), axis=E3*scalefactor, color=color.orange)
       arrow(pos=vector(0,-offset,0), axis=E4*scalefactor, color=color.orange)
       arrow(pos=vector(0,0,offset), axis=E5*scalefactor, color=color.orange)
       arrow(pos=vector(0,0,-offset), axis=E6*scalefactor, color=color.orange)

   ##Colored arrows (Out Sphere; Point Charge)
   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)
       arrow(pos=vector(offset,0,0), axis=E1*scalefactor, color=color.orange)
       arrow(pos=vector(-offset,0,0), axis=E2*scalefactor, color=color.orange)
       arrow(pos=vector(0,offset,0), axis=E3*scalefactor, color=color.orange)
       arrow(pos=vector(0,-offset,0), axis=E4*scalefactor, color=color.orange)
       arrow(pos=vector(0,0,offset), axis=E5*scalefactor, color=color.orange)
       arrow(pos=vector(0,0,-offset), axis=E6*scalefactor, color=color.orange)

Examples

Simple

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Figure 1: 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.

Normal

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Figure 2: 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 useful to discuss real-life applications of a charged solid sphere.

The foremost and probably most valuable example is to look at the structure of an atomic nucleus. An atom's nucleus is very dense, so much so that the positive charge can be treated as a uniformly charged sphere.

The surrounding electron cloud can also be approximated as a roughly spherical region of negative charge, though not perfectly uniform. Keep in mind that these structures are quite small, with the nuclear radii being about [math]\displaystyle{ 10e-15 }[/math] m and the electron cloud extending to about [math]\displaystyle{ 10e-10 }[/math]. Maybe a microscope could help?

Other real systems also behave like charged Spheres. For instances, charged droplets (such as tiny water droplets or dust particles in a thunderstorm) are nearly perfect spheres dues to surface tension, and their electric fields follow the same principles as a uniformly charged ball. The same could be applied to an electrolyte drink, though one might consider what such a beverage is doing in the middle of a thunderstorm.

Another example would be spherical plasmas created in laboratories, which hold charge throughout their volume. You may find an example experiment in the following link: https://plasma.physics.ucla.edu/

History

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Figure 4: A torsion balance. Pretty cool, right?

To understand the “history” of the Electric Field of a Charged Ball is really to understand the broader history of the electric field itself.

In 1785, physicist Charles‑Augustin de Coulomb performed the first precise measurements of electrical forces using a torsion balance (seen in Figure 4). His experiments established the inverse‑square law, which all later electric‑field theory depends on.

There were, of course, earlier thinkers such as Thales (c. 600 BC)*[2] and Benjamin Franklin (d. 1790)*[3] who noticed electrical attraction, but Coulomb’s work is what transformed those scattered observations into quantitative physics.

Now, over the next several decades, the modern idea of an electric field took shape. Physicists began to understand that a charged object affects the space around it, creating a region in which any other charge will experience a force... which is what we now call the electric field.

These developments eventually led to the modern derivations of the electric field we use today, including the field of a charged sphere. However, historical sources tend to focus on the evolution of electric forces and field concepts in general, rather than the specific spherical case.

See also

Read the Provided texts for more context regarding Point Charges, Electric Diploes, and Gauss's Flux Theorem.

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/

• https://www.aps.org/apsnews/2016/06/coulomb-measures-electric-force

• https://www.britannica.com/science/electromagnetism/Coulombs-law

• https://openstax.org/books/college-physics-2e/pages/18-5-electric-field-lines-multiple-charges

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