Superposition and Continuous Charge Distributions
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You should be comfortable with vector addition, basic integration techniques, and the concept of a vector field. Familiarity with Coulomb's law for point charges is assumed.
Why this matters
Real-world charges are rarely isolated points; they form lines, sheets, or volumes.
The ability to calculate the electric field from arbitrary charge distributions is the fundamental skill of electrostatics. While Coulomb's law gives the force between two point charges, nature presents us with continuous distributions of charge. By leveraging the principle of superposition, we can break down these complex distributions into infinitesimal point charges, calculate the field contribution from each, and sum them up using integration. This lecture bridges the gap between the discrete point-charge model and the continuous field description, providing the mathematical tools necessary to solve real-world electrostatic problems.
The Principle of Superposition
Imagine you have a complex object made of many small pieces of charge. The total electric field at any point in space is simply the vector sum of the electric fields produced by each individual piece. This is the principle of superposition. It relies on the linearity of Coulomb's law: the force (and thus the field) produced by a collection of charges is the sum of the forces produced by each charge acting alone. There are no "interference" terms where charges interact with each other to change the field they produce at a distant point; each source charge contributes independently to the total field. This allows us to treat a continuous distribution as a sum of infinitesimal point charges, each contributing a small vector field \(d\vec{E}\). The total field is the integral of these contributions over the entire volume, surface, or line containing the charge.
From Point Charges to Continuous Distributions
To apply superposition to a continuous distribution, we first define the charge density. For a volume distribution, the charge density \(\rho(\vec{r}')\) is the charge per unit volume at a source point \(\vec{r}'\). An infinitesimal volume element \(d^3r'\) contains an infinitesimal charge \(dq = \rho(\vec{r}') \, d^3r'\). Similarly, for a surface distribution, we use surface charge density \(\sigma(\vec{r}')\) with \(dq = \sigma(\vec{r}') \, dA'\), and for a line distribution, linear charge density \(\lambda(\vec{r}')\) with \(dq = \lambda(\vec{r}') \, dl'\).
According to Coulomb's law, the electric field \(d\vec{E}\) at an observation point \(\vec{r}\) due to this infinitesimal charge \(dq\) is:
where \(\hat{r}\) is the unit vector pointing from the source point \(\vec{r}'\) to the observation point \(\vec{r}\). We can express \(\hat{r}\) in terms of the displacement vector \(\vec{R} = \vec{r} - \vec{r}'\). Thus, \(\hat{r} = \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|}\). Substituting this into the expression for \(d\vec{E}\) gives:
The total electric field is obtained by integrating this expression over the entire source region. This leads to the general integral form for the electric field due to a continuous volume charge distribution, as derived in D-039.
Vector Integration and Symmetry
When performing the integration \(\vec{E}(\vec{r}) = \int d\vec{E}\), it is crucial to remember that \(\vec{E}\) is a vector. We cannot simply integrate the magnitude; we must integrate the vector components. Often, symmetry arguments allow us to simplify the calculation significantly. For example, if the charge distribution is symmetric about an axis, the components of the electric field perpendicular to that axis may cancel out due to symmetry, leaving only the component along the axis to be calculated.
Consider a ring of charge with total charge \(Q\) and radius \(R\), lying in the \(xy\)-plane. By symmetry, the electric field at any point on the \(z\)-axis must point along the \(z\)-axis. The horizontal components from opposite sides of the ring cancel. We only need to integrate the \(z\)-component of \(d\vec{E}\). The distance from any point on the ring to the observation point \((0,0,z)\) is constant, \(r = \sqrt{R^2 + z^2}\). The \(z\)-component of the unit vector is \(\cos\theta = \frac{z}{\sqrt{R^2+z^2}}\). The integral becomes a scalar integral over the charge \(dq\), which simplifies the vector integration considerably. This technique of exploiting symmetry to reduce vector integrals to scalar integrals is a powerful tool in electrostatics.
Coordinate Systems and Integration Limits
The choice of coordinate system is critical for setting up the integral efficiently. Cartesian coordinates \((x,y,z)\) are suitable for rectangular distributions. Cylindrical coordinates \((s,\phi,z)\) are ideal for problems with cylindrical symmetry, such as charged wires or rings. Spherical coordinates \((r,\theta,\phi)\) are best for spherical distributions. The volume element \(d^3r'\) changes form in each system: \(dx\,dy\,dz\) in Cartesian, \(s\,ds\,d\phi\,dz\) in cylindrical, and \(r'^2 \sin\theta' \, dr' \, d\theta' \, d\phi'\) in spherical.
When setting up the integral, one must carefully define the limits of integration to cover the entire charge distribution. For a finite line charge from \(z'=-L/2\) to \(z'=L/2\), the limits for \(z'\) are \(-L/2\) to \(L/2\). For an infinite line, the limits are \(-\infty\) to \(+\infty\). The integrand must be expressed in terms of the chosen coordinates. For instance, in cylindrical coordinates, the vector \(\vec{r}-\vec{r}'\) must be decomposed into its \(s\), \(\phi\), and \(z\) components. This requires careful vector algebra to ensure that the direction of the field contribution is correctly accounted for at each source point.
The derivation
Electric Field from Coulomb's Law
The electrostatic field \(\vec{E}(\vec{r})\) is defined as the electric force per unit stationary test charge in the limit that the test charge is vanishingly small, \(\vec{E}(\vec{r}) = \lim_{q\to 0}\vec{F}(\vec{r})/q\). Combining this definition with Coulomb's law and the superposition of forces yields, for a source charge density \(\rho(\vec{r}')\) at rest, \(\displaystyle \vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \rho(\vec{r}')\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^{3}r'\).
A common error is to integrate the magnitude of the electric field \(|d\vec{E}|\) instead of the vector \(d\vec{E}\). The magnitude is a scalar and does not account for direction. If the field contributions point in different directions, their vector sum is not equal to the sum of their magnitudes. Always resolve the field into components before integrating, or use symmetry to justify integrating a single component. C
Worked example
Problem. A thin, non-conducting ring of radius \(R = 0.50\ \mathrm{m}\) carries a total charge \(Q = 2.0\ \mu\mathrm{C}\) uniformly distributed along its circumference. Calculate the magnitude of the electric field at a point on the axis of the ring, a distance \(z = 1.0\ \mathrm{m}\) from the center.
- Identify the charge element. For a ring, we use linear charge density \(\lambda = Q / (2\pi R)\). The charge element is \(dq = \lambda \, d\ell\), where \(d\ell = R \, d\phi\). Thus, \(dq = \frac{Q}{2\pi} d\phi\).
- Determine the distance and direction. The distance from any point on the ring to the observation point on the axis is \(r = \sqrt{R^2 + z^2}\). The electric field \(d\vec{E}\) from \(dq\) points away from the charge element. By symmetry, the transverse components cancel, and only the axial component \(dE_z\) remains.
- Express the axial component. The angle \(\theta\) between \(d\vec{E}\) and the \(z\)-axis satisfies \(\cos\theta = \frac{z}{r} = \frac{z}{\sqrt{R^2+z^2}}\). Therefore, \(dE_z = dE \cos\theta = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} \frac{z}{r} = \frac{1}{4\pi\varepsilon_0} \frac{z \, dq}{(R^2+z^2)^{3/2}}\).
- Integrate over the ring. Since \(z\), \(R\), and \(\varepsilon_0\) are constant, we integrate \(dq\) over the full circle: \(\int dq = Q\). Thus, \(E_z = \frac{1}{4\pi\varepsilon_0} \frac{z Q}{(R^2+z^2)^{3/2}}\).
- Substitute numerical values. \(\varepsilon_0 \approx 8.854 \times 10^{-12}\ \mathrm{F/m}\). \(R = 0.50\ \mathrm{m}\), \(z = 1.0\ \mathrm{m}\), \(Q = 2.0 \times 10^{-6}\ \mathrm{C}\). \(R^2 + z^2 = 0.25 + 1.00 = 1.25\ \mathrm{m^2}\). \((R^2+z^2)^{3/2} = (1.25)^{1.5} \approx 1.3975\ \mathrm{m^3}\). \(k_e = \frac{1}{4\pi\varepsilon_0} \approx 8.988 \times 10^9\ \mathrm{N\,m^2/C^2}\). \(E = (8.988 \times 10^9) \frac{(1.0)(2.0 \times 10^{-6})}{1.3975}\). \(E \approx (8.988 \times 10^9) \frac{2.0 \times 10^{-6}}{1.3975} \approx 1.286 \times 10^4\ \mathrm{N/C}\).
Units check. \(\frac{\mathrm{N\,m^2/C^2} \cdot \mathrm{m} \cdot \mathrm{C}}{\mathrm{m^3}} = \frac{\mathrm{N}}{\mathrm{C}}\). The units are consistent.
Three quick tests
- Why does the transverse component of the electric field from a charged ring cancel out on the axis?
Answer
Due to the rotational symmetry of the ring about the axis, for every charge element producing a transverse field component in one direction, there is a corresponding element on the opposite side producing an equal and opposite transverse component. - If the distance \(z\) from the center of the ring is much larger than the radius \(R\) (\(z \gg R\)), what does the electric field approach?
Answer
It approaches the field of a point charge \(Q\) located at the center, \(E \approx \frac{1}{4\pi\varepsilon_0} \frac{Q}{z^2}\). - In the integral \(\vec{E} = \int \frac{1}{4\pi\varepsilon_0} \frac{\rho(\vec{r}')(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3} d^3r'\), what is the role of the vector \(\vec{r}-\vec{r}'\)?
Answer
It provides the direction of the electric field contribution from the source element \(dq\), pointing from the source to the observation point.
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.