Electric Potential and Field of Continuous Charge Distributions

Hello! Welcome to your next lesson in electrostatics.

In our previous lesson, we established that the electrostatic field $E$ is conservative ( $\nabla \times E = 0$ ). This profound property allowed us to define a scalar electric potential, $V$ , from which the field can be recovered using the relation $E = - \nabla V$ . We applied this to systems of point charges.

Today, we will extend this framework to more realistic scenarios involving continuous distributions of charge. The learning outcome for this lesson is to solve for the electric potential and field of continuous charge distributions using integration. This is a fundamental skill that bridges the gap between abstract field theory and concrete problem-solving. We will master the techniques for setting up and solving the necessary integrals for various charge geometries.

1. The Core Principle: From Sums to Integrals

For a collection of discrete point charges, we found the total potential and field by simple (scalar or vector) summation.
$V = \sum_{i} \frac{k q _{i}}{r _{i}} and E = \sum_{i} \frac{k q _{i}}{r _{i}^{2}} r^_{i}$
When charges are so numerous and close together that they form a continuous distribution, this sum becomes an integral. We consider the object to be made of an infinite number of infinitesimal charge elements, $d q$ . Each $d q$ acts like a point charge, creating a differential potential $d V$ and a differential field $d E$ .

Electric Field Due to Continuous Charge Distributions — This diagram illustrates the conceptual leap. We replace the sum over discrete charges $\Sigma q_i$ with an integral over a continuous charge element $dq$. The field $\mathbf{E}$ at a point P is the vector integral of all the differential fields $d\mathbf{E}$ produced by each $dq$ in the source distribution.

The master equations we will use are:

Electric Potential: $V = \int d V = \int \frac{k d q}{r}$
Electric Field: $E = \int d E = \int \frac{k d q}{r ^{2}} r^$

Here, $r$ is the distance from the charge element $d q$ to the point of interest, and $r^$ is the unit vector pointing from $d q$ to that point.

The charge element $d q$ is expressed using charge density:

Linear charge density $λ$ : $d q = λ d l$ (for lines/rods)
Surface charge density $σ$ : $d q = σ d A$ (for surfaces/disks)
Volume charge density $ρ$ : $d q = ρ d V$ (for solids/spheres)

The Electric Field II: Continuous Charge Distributions

To formalize these definitions, please review the introductory section of the following chapter from Tipler's "Physics for Scientists and Engineers".

Read the first page of the chapter (p. 682). Focus on the definitions of volume, surface, and linear charge densities (Equations 22-1, 22-2, 22-3) and the general integral formulation for the electric field (Equation 22-4).

2. Calculating Electric Potential by Direct Integration

Since potential is a scalar, its calculation involves a scalar integral, which is generally simpler than the vector integral for the electric field. Let's start with a classic one-dimensional example.

Example 1: Uniformly Charged Rod (Axial Point)

Consider a thin rod of length $L$ with a total charge $Q$ distributed uniformly, giving it a linear charge density $λ = Q / L$ . We want to find the potential at a point P on the axis of the rod, at a distance $r$ from its nearest end.

Class 12 Physics | Electric Potential |#12 Electric Potential due to a Uniformly Charged rod on Axis

The channel Physics Galaxy provides a concise and clear derivation for this exact problem. It's a perfect demonstration of the direct integration method for potential.

Please watch the entire video (00:00 - 03:27). Follow these key steps: Setting up the coordinate system and identifying the charge element dq = \lambda dx. Writing the expression for the differential potential dV from this element. Defining the correct limits of integration and solving the integral.

The result is $V = \frac{k Q}{L} ln (1 + \frac{L}{r})$ . This process—defining $d q$ , writing $d V$ , and integrating over the geometry—is the fundamental workflow.

Example 2: Potential on the Rim of a Charged Disc

Let's tackle a more challenging problem that showcases the power of choosing a clever coordinate system. Consider a disc of radius $R$ with uniform surface charge $σ$ . What is the potential at a point on its rim?

Electromagnetism I: Electrostatics

This problem, from a collection of Physics Olympiad materials by Kevin Zhou, requires a more sophisticated integration setup. The solution is elegant and a good test of your mathematical skills.

Find Problem 20(a) on page 17. Study the problem and its solution. Notice how placing the origin of a polar coordinate system at the point on the rim simplifies the integral for the potential.

This example shows that with a well-chosen coordinate system, even seemingly complex 2D integrals can become manageable.

3. Calculating Electric Field by Direct Integration

Calculating the electric field involves a vector integral, which means we must handle components and pay close attention to symmetry. Often, symmetry arguments can drastically simplify the problem by showing that certain components of the field must cancel out.

Example 3: Uniformly Charged Ring (On-axis)

A circular ring of radius $R$ has a uniform charge $Q$ . We want to find the electric field at a point P on the axis of the ring, a distance $x$ from its center.

Electric Field from a Ring and a Disk

The Physics Ninja channel offers an excellent, step-by-step derivation for the ring and the disk. Let's start with the ring.

Watch from 00:44 to 10:12. The first part (until 07:11) covers the core derivation. Pay close attention to the symmetry argument: for every dq on top, there's another on the bottom whose perpendicular field component cancels out, leaving only the axial component. The second part (07:11 - 10:12) checks the result in two physical limits: at the center (x=0) and very far away (x >> R). This is a critical step in professional problem-solving to validate your results.

The key takeaway is that by exploiting symmetry, we reduced a 3D vector integration problem to a 1D scalar integration. The final field is purely along the x-axis: $E_{x} = \frac{k Q x}{( x ^{2} + R ^{2} ) ^{3/2}}$ .

Example 4: Uniformly Charged Disk (On-axis)

We can now find the field of a charged disk by treating it as a continuous set of concentric rings. The field of each ring is already known from the previous example.

Electric Field from a Ring and a Disk

Let's continue with the Physics Ninja video to see how the result for the ring is used to solve for the disk.

Watch from 10:12 to 20:36. The first part (until 17:37) shows how to set up the integral. The field of the disk, dE_disk, is the field from one ring, dE_ring. The total field is found by integrating this expression from r=0 to r=R. Note the use of a u-substitution to solve the integral. The second part (18:26 - 20:36) again checks a physical limit: what happens when you are very close to the disk (x -> 0)? The result, E = \sigma / (2\epsilon_0), should be familiar from Gauss's Law for an infinite plane.

This method of building up a 2D object from 1D slices is a powerful technique you will use in many areas of physics.

Test your understanding!

In the previous lesson, we established that $E = - \nabla V$ . Can you use the potential of the rod from Example 1, $V (r) = \frac{k Q}{L} ln (1 + \frac{L}{r})$ , to find the electric field $E$ at the same point P?
(Note: the point P is on the x-axis, at a distance $r$ from the end of the rod. So you need to calculate $E_{x} = - \frac{\partial V}{\partial r}$ using the chain rule).

Show answer

Yes, we can. The field will only have an x-component, which we'll call the r-component here.
$E_{r} = - \frac{d V}{d r} = - \frac{d}{d r} [\frac{k Q}{L} ln (1 + \frac{L}{r})]$
Using the chain rule, where $u = 1 + L / r$ :
$\frac{d}{d r} (ln u) = \frac{1}{u} \cdot \frac{d u}{d r} = \frac{1}{1 + L / r} \cdot (- \frac{L}{r ^{2}}) = \frac{r}{r + L} \cdot (- \frac{L}{r ^{2}}) = - \frac{L}{r ( r + L )}$
Substituting this back into the expression for $E_{r}$ :
$E_{r} = - \frac{k Q}{L} (- \frac{L}{r ( r + L )}) = \frac{k Q}{r ( r + L )}$
This is exactly the result one would get by integrating Coulomb's Law directly for the field, which is a more complex integral ( $\int d x / x^{2}$ ). This shows how calculating potential first can be a simpler path to the electric field.

4. Alternative Strategy: Field First, Then Potential

For highly symmetric charge distributions, we can often find the electric field $E$ easily using Gauss's Law. Once $E$ is known, we can find the potential $V$ by integrating the field, using the definition from our last lesson:
$V (B) - V (A) = - \int_{A}^{B} E \cdot d l$
This is often much easier than direct integration for the potential.

Example 5: Potential Inside and Outside a Uniformly Charged Sphere

Consider a solid sphere of radius $R$ and total charge $Q$ , with uniform volume charge density $ρ$ . From Gauss's law, we know the electric field is:
$E (r) = {\frac{k Q r}{R ^{3}} r^\frac{k Q}{r ^{2}} r^for r \leq R (inside) for r \geq R (outside)$
Let's find the potential everywhere, assuming $V (\infty) = 0$ .

Electromagnetism I: Electrostatics

The Kevin Zhou handout provides a concise solution to this exact problem, demonstrating the strategy of integrating a known E-field.

Please read the solution to Problem 3(a) on page 2. Follow the two-step integration: Find the potential at the surface, V(R), by integrating the outside field from \infty to R. Find the additional potential difference to get to the center, \Delta V, by integrating the inside field from R to 0. The potential at the center is then V(0) = V(R) + \Delta V.

This method is extremely powerful. The potential at the center, $V (0) = \frac{3}{2} \frac{k Q}{R}$ , is a classic result. It shows that the potential at the center is 1.5 times the potential at the surface.

Conclusion

You have now learned and practiced the essential techniques for analyzing the fields and potentials of continuous charge distributions. This is a workhorse skill in electromagnetism.

Key Takeaways:

Continuous charge distributions are analyzed by integrating the contributions of infinitesimal charge elements, $d q$ .
The fundamental formulas are $V = \int k d q / r$ and $E = \int k d q / r^{2} r^$ .
Potential is a scalar integral and is often easier to compute directly. The field can then be found via $E = - \nabla V$ .
Field calculations are vector integrals where exploiting symmetry is crucial for simplification.
For highly symmetric cases, it is often most efficient to find $E$ first using Gauss's Law, and then find $V$ by integrating the field via $V = - \int E \cdot d l$ .

Preview of the Next Lesson:
While direct integration is a powerful tool, it can be mathematically intensive. For a large class of problems known as boundary-value problems, there is an even more powerful approach. In the next lesson, we will combine $\nabla \cdot E = ρ / ϵ_{0}$ with $E = - \nabla V$ to derive Poisson's Equation, $\nabla^{2} V = - ρ / ϵ_{0}$ , and its source-free counterpart, Laplace's Equation. These partial differential equations are at the heart of advanced electrostatics.

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