Fields of continuous charge distributions Knowledge-Based Study

Study notes

1. Transition from Discrete to Continuous Charges

   In macroscopic systems, an enormous number of discrete elementary charges are often distributed over a region of space. Rather than summing the individual contributions of billions of discrete point charges, we model the charge as a smooth, continuous distribution.

   Depending on the geometry of the charged body, we define three fundamental charge densities:

   • Linear Charge Density ($\lambda$): Used for one-dimensional structures such as thin wires or rings. It is defined as $\lambda = \frac{dq}{dl}$ with SI units of C/m. For a uniformly charged object of total charge $Q$ and length $L$, $\lambda = \frac{Q}{L}$.
   • Surface Charge Density ($\sigma$): Used for two-dimensional structures such as flat sheets or curved surfaces. It is defined as $\sigma = \frac{dq}{dA}$ with SI units of C/m2. For a uniform distribution over area $A$, $\sigma = \frac{Q}{A}$.
   • Volume Charge Density ($\rho$): Used for charges distributed throughout a three-dimensional volume. It is defined as $\rho = \frac{dq}{dV}$ with SI units of C/m3. For a uniform distribution over volume $V$, $\rho = \frac{Q}{V}$.

   Formulating the Electric Field Integral

   The electric field of a continuous charge distribution is found by applying the principle of superposition. We divide the total charge into infinitesimal charge elements $dq$, treat each element as a point charge, and integrate over the entire distribution.

   An infinitesimal charge element $dq$ located at position vector $\vec{r}'$ creates an electric field $d\vec{E}$ at the field point $\vec{r}$:

   d\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{dq}{\mathcal{r}^2} \hat{\mathcal{r}}\

   where $\vec{\mathcal{r}} = \vec{r} - \vec{r}'$ is the displacement vector from the source element to the field point, $\mathcal{r} = |\vec{\mathcal{r}}|$ is its magnitude, and $\hat{\mathcal{r}} = \frac{\vec{\mathcal{r}}}{\mathcal{r}}$ is the unit vector pointing toward the field point.

   The net electric field is obtained by vector integration:

   \vec{E} = \int d\vec{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{\mathcal{r}^2} \hat{\mathcal{r}}\

   Because the integral is a vector operation, it is generally evaluated by separating the field into Cartesian components: $E_x = \int dE_x$, $E_y = \int dE_y$, and $E_z = \int dE_z$.

   Simplifying Integrals Using Symmetry

   Directly computing vector integrals can be challenging, but geometric symmetry often simplifies the calculation significantly.

   • Pairwise Cancellation: If the charge distribution is symmetric about a coordinate axis or plane, corresponding pairs of charge elements $dq$ produce field components that cancel in one or more directions.
   • Component Selection: By identifying which components vanish due to symmetry before integrating, we only need to set up and evaluate the integral for the non-zero component.
   • Coordinate Systems: Choosing a coordinate system that matches the boundary of the charge distribution (such as polar coordinates for circles/disks or cylindrical coordinates for long rods) ensures that limits of integration remain constant and integrals are straightforward.

   Electric Field of a Ring of Charge

   Consider a thin ring of radius $R$ carrying a uniform positive charge $Q$, centered at the origin in the $xy$-plane. We wish to determine the electric field at a point $P$ on the $z$-axis at distance $z$ from the center.

   1. Symmetry Analysis

   For any charge element $dq$ on the ring, there is a diametrically opposite element of equal charge. The components of $d\vec{E}$ perpendicular to the $z$-axis cancel pairwise, leaving only the component along the $z$-axis:

   $E_z = \int dE_z = \int dE \cos\theta$

   2. Expressing Geometric Factors

   Every charge element $dq$ is at the exact same distance from point $P$:

   $\mathcal{r} = \sqrt{R^2 + z^2}$

   The angle factor is:

   $\cos\theta = \frac{z}{\mathcal{r}} = \frac{z}{\sqrt{R^2 + z^2}}$

   3. Evaluating the Integral

   Substituting these into the component integral gives:

   $E_z = \int \frac{1}{4\pi\varepsilon_0} \frac{dq}{R^2 + z^2} \left( \frac{z}{\sqrt{R^2 + z^2}} \right) = \frac{z}{4\pi\varepsilon_0 (R^2 + z^2)^{3/2}} \int dq$

   Since $\int dq = Q$, the net electric field along the axis is:

   E_z = \frac{1}{4\pi\varepsilon_0} \frac{Qz}{(R^2 + z^2)^{3/2}}\

   Limiting Cases: At the center ($z = 0$), $E_z = 0$, as expected by symmetry. For points far from the ring ($z \gg R$), the denominator approaches $z^3$, yielding $E_z \approx \frac{Q}{4\pi\varepsilon_0 z^2}$, which reproduces Coulomb's law for a point charge.

   Electric Field of a Uniformly Charged Disk

   A uniformly charged flat disk of radius $R$ with surface charge density $\sigma$ can be treated as a collection of concentric rings of radius $r$ and infinitesimal radial width $dr$.

   1. Setting up the Element

   The area of each thin ring is $dA = 2\pi r\,dr$, so its charge is $dq = \sigma dA = 2\pi\sigma r\,dr$. Using our result for a single ring, the field contributed by this concentric ring at distance $z$ along the axis is:

   dE_z = \frac{1}{4\pi\varepsilon_0} \frac{(2\pi\sigma r\,dr)z}{(r^2 + z^2)^{3/2}}\

   2. Integration over Radius

   Integrating from $r = 0$ to $r = R$:

   $E_z = \frac{\sigma z}{4\varepsilon_0} \int_{0}^{R} \frac{2r\,dr}{(r^2 + z^2)^{3/2}} = \frac{\sigma z}{2\varepsilon_0} \left[ -\frac{1}{\sqrt{r^2 + z^2}} \right]_{0}^{R}$

   Evaluating the limits gives:

   $E_z = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right)$

   3. The Infinite Plane Limit

   As the radius becomes very large compared to the distance ($R \to \infty$), the term $\frac{z}{\sqrt{R^2 + z^2}} \to 0$, yielding the field of an infinite sheet of charge:

   $E = \frac{\sigma}{2\varepsilon_0}$

   This field is uniform in magnitude and directed perpendicular to the sheet.