The electric field is the fundamental concept that explains how charged objects communicate forces across empty space. Instead of "action at a distance," we understand that a charge creates a property in space—a field—that exerts force on other charges. In this post, we explore the precise definitions of the electric field and perform a rigorous step-by-step derivation of the field produced by a finite line of charge.
Electric Field: Definitions, Distributions, and Line Charge Derivation
1. Electric Field: Definition and Meaning
The electric field at any point in space is defined operationally by the force it exerts on a hypothetical test charge.
Mathematical Definition
The electric field \(\vec E\) at a position \(\vec r\) is the electrostatic force \(\vec F\) per unit positive test charge \(q_0\), in the limit that the test charge becomes infinitesimally small (to avoid disturbing the source charges):
Physical Meaning: The field is a property of space itself, generated by source charges. Once the source charges are fixed, the vector field \(\vec E(\vec r)\) exists at every point, regardless of whether another charge is present to experience a force.
2. Electric Field Due to a Point Charge
For a single point source charge \(q\) located at position \(\vec r'\), the electric field at an observation point \(\vec r\) is given by Coulomb's Law:
\[ \vec E(\vec r) = \frac{1}{4\pi\varepsilon_0} \frac{q}{|\vec r - \vec r'|^2} \hat{R} \]Where the separation vector and unit vector are defined as:
\[ \vec R = \vec r - \vec r', \qquad \hat R = \frac{\vec R}{|\vec R|} \]Key Properties:
- Magnitude: Falls off as the inverse square of the distance (\(1/R^2\)).
- Direction: Radially outward if \(q > 0\), radially inward if \(q < 0\).
3. The Superposition Principle
For a system of \(n\) discrete point charges \(q_i\) located at positions \(\vec r_i\), the total electric field is the vector sum of the individual fields:
\[ \vec E(\vec r) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^n \frac{q_i}{|\vec r - \vec r_i|^2} \hat R_i \]This principle implies that electric fields satisfy linearity: the presence of one charge does not alter the field produced by another.
4. Continuous Charge Distributions
When charge is distributed continuously rather than in discrete points, the summation becomes an integral:
\[ \vec E(\vec r) = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{|\vec r - \vec r'|^2} \hat R \]Types of Charge Densities
Depending on the geometry of the source, we express the differential charge element \(dq\) using different densities:
- Line Charge Density (\(\lambda\)): For wires and rods. \[ dq = \lambda\, dl \implies \lambda = \frac{dq}{dl} \]
- Surface Charge Density (\(\sigma\)): For sheets and plates. \[ dq = \sigma\, dS \implies \sigma = \frac{dq}{dS} \]
- Volume Charge Density (\(\rho\)): For solid spheres or insulators. \[ dq = \rho\, d\tau \implies \rho = \frac{dq}{d\tau} \]
5. Derivation: Electric Field Due to a Finite Line Charge
Geometry Setup
Consider a straight wire carrying a uniform line charge density \(\lambda\).
- The wire lies along the \(z\)-axis, extending from \(z = -L_1\) to \(z = +L_2\).
- The observation point \(P\) is located at a perpendicular distance \(D\) from the wire.
- We place the origin at the foot of the perpendicular from \(P\) to the wire.
Consider a small element \(dz\) at position \(z\). The distance \(r\) from this element to point \(P\) is:
\[ r = \sqrt{z^2 + D^2} \]The Differential Electric Field
The magnitude of the field produced by the element \(dq = \lambda dz\) is:
\[ dE = \frac{1}{4\pi\varepsilon_0} \frac{\lambda\,dz}{z^2 + D^2} \]We resolve this vector into two perpendicular components:
- Radial Component (\(dE_r\)): Along the perpendicular \(D\). \(\cos\theta = D/r\).
- Axial Component (\(dE_z\)): Parallel to the wire. \(\sin\theta = z/r\).
A. Calculation of the Radial Component (\(E_r\))
The radial component is the projection of \(dE\) along \(D\):
\[ dE_r = dE \cos\theta = \frac{1}{4\pi\varepsilon_0} \frac{\lambda\,dz}{z^2 + D^2} \left(\frac{D}{\sqrt{z^2 + D^2}}\right) \] \[ E_r = \frac{\lambda D}{4\pi\varepsilon_0} \int_{-L_1}^{L_2} \frac{dz}{(z^2 + D^2)^{3/2}} \]Integration Technique: Use trigonometric substitution.
Let \(z = D\tan\theta\), then \(dz = D\sec^2\theta\,d\theta\). The term \((z^2+D^2)^{3/2}\) becomes \(D^3\sec^3\theta\).
The limits of integration change from lengths to angles \(-\beta\) (for \(-L_1\)) and \(+\alpha\) (for \(L_2\)).
\[ E_r = \frac{\lambda D}{4\pi\varepsilon_0} \int_{-\beta}^{\alpha} \frac{D\sec^2\theta\,d\theta}{D^3\sec^3\theta} \] \[ E_r = \frac{\lambda}{4\pi\varepsilon_0 D} \int_{-\beta}^{\alpha} \cos\theta\,d\theta \] \[ E_r = \frac{\lambda}{4\pi\varepsilon_0 D} [\sin\theta]_{-\beta}^{\alpha} \]Converting back to linear dimensions using the geometry of the triangle:
\[ \sin\alpha = \frac{L_2}{\sqrt{L_2^2 + D^2}}, \quad \sin\beta = \frac{L_1}{\sqrt{L_1^2 + D^2}} \] \[ E_r = \frac{\lambda}{4\pi\varepsilon_0 D} \left( \frac{L_1}{\sqrt{L_1^2 + D^2}} + \frac{L_2}{\sqrt{L_2^2 + D^2}} \right) \]B. Calculation of the Axial Component (\(E_z\))
The axial component is the projection along the wire:
\[ dE_z = dE \sin\theta = \frac{1}{4\pi\varepsilon_0} \frac{\lambda\,dz}{z^2 + D^2} \left(\frac{z}{\sqrt{z^2 + D^2}}\right) \] \[ E_z = \frac{\lambda}{4\pi\varepsilon_0} \int_{-L_1}^{L_2} \frac{z\,dz}{(z^2 + D^2)^{3/2}} \]Integration Technique: Standard substitution \(t = z^2 + D^2\), \(dt = 2z\,dz\).
\[ \int \frac{z\,dz}{(z^2+D^2)^{3/2}} = -\frac{1}{\sqrt{z^2+D^2}} \]Applying the limits:
6. Special Cases and Physical Interpretation
Case 1: The Symmetric Wire (\(L_1 = L_2 = L\))
If the observation point is at the midpoint of the wire:
- Axial Field (\(E_z\)): The terms cancel out perfectly. \(E_z = 0\). This makes physical sense as the "push" from the top half is canceled by the "push" from the bottom half.
- Radial Field (\(E_r\)): The components add up. \[ E_r = \frac{1}{4\pi\varepsilon_0} \frac{2\lambda L}{D\sqrt{L^2 + D^2}} \]
Case 2: The Infinite Line (\(D \ll L\))
If we are very close to the wire, or the wire is very long, \(L \to \infty\). In this limit, \(\sqrt{L^2 + D^2} \approx L\).
\[ E_r \approx \frac{1}{4\pi\varepsilon_0} \frac{2\lambda L}{D \cdot L} = \frac{\lambda}{2\pi\varepsilon_0 D} \]This matches the standard result derived from Gauss's Law.
Case 3: The Far Field (\(D \gg L\))
If we are very far away, the wire looks like a point. Here \(\sqrt{L^2 + D^2} \approx D\).
\[ E_r \approx \frac{1}{4\pi\varepsilon_0} \frac{2\lambda L}{D^2} \]Since the total charge \(Q = 2\lambda L\), this becomes:
\[ E \approx \frac{1}{4\pi\varepsilon_0} \frac{Q}{D^2} \]This confirms that from a large distance, any finite charge distribution behaves like a point charge.