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Chapter 1 • Theory & Derivations

Electric Field and Gauss's Law

A rigorous foundation of electrostatics: charge quantization and conservation, Coulomb's inverse-square law, vector superposition, electric field lines, dipole dynamics and torque, electric flux, Gauss's law in integral and differential forms, and applications to spherical, cylindrical, and planar symmetries.

§1.1Electric Charge, Quantization, and Conservation Laws

Electrostatics investigates electric charges at rest and the static electric fields they establish in vacuum and material media.

1. The Fundamental Nature of Electric Charge

Electric charge is an intrinsic fundamental property of subatomic matter. Matter exhibits two complementary types of charge:
  • Positive Charge ($+q$): Borne by protons ($+e$).
  • Negative Charge ($-q$): Borne by electrons ($-e$).
Like charges repel one another with mutual electrostatic forces; unlike charges attract.

2. Quantization of Electric Charge

Robert A. Millikan's oil-drop experiment (1909) experimentally confirmed that electric charge is not continuous, but exists exclusively in discrete, integer multiples of the elementary charge $e$: $$q = \pm n e, \quad n = 0, 1, 2, 3, \dots$$ where the elementary quantum of charge is defined by the 2019 SI standard: $$e = 1.602176634 \times 10^{-19} \text{ Coulombs (exact)}$$ (Note: Although quarks carry fractional charges $\pm \frac{1}{3}e$ and $\pm \frac{2}{3}e$, quark confinement strictly forbids isolated fractional charges under ordinary conditions).

3. The Law of Conservation of Electric Charge

In any closed, isolated physical system, the algebraic sum of electric charges remains strictly constant over time: $$\sum q_i = \text{constant}, \quad \frac{d Q_{\text{total}}}{dt} = 0$$ In relativistic and high-energy particle physics, while particles can be created or annihilated (e.g., pair production $\gamma \to e^- + e^+$ or electron-positron annihilation $e^- + e^+ \to 2\gamma$), net electric charge is conserved in every known physical interaction.

4. Continuous Charge Distributions

On macroscopic scales where individual charges cannot be resolved, charge is modeled via continuous spatial density distributions:
  • Linear Charge Density ($\lambda$): $\lambda = \frac{dq}{dl}$ (Units: C/m)
  • Surface Charge Density ($\sigma$): $\sigma = \frac{dq}{dA}$ (Units: C/m²)
  • Volume Charge Density ($\rho$): $\rho = \frac{dq}{dV}$ (Units: C/m³)

§1.2Coulomb's Law and the Electrostatic Superposition Principle

Charles-Augustin de Coulomb (1785) established the fundamental quantitative law of electrostatic force using a precision torsion balance.

1. Coulomb's Law in Vector Form

The electrostatic force $\vec{F}_{12}$ exerted by a point charge $q_1$ located at position $\vec{r}_1$ on a second point charge $q_2$ located at $\vec{r}_2$ in vacuum is: $$\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^2} \hat{r}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1)$$ where:
  • $\epsilon_0$: Permittivity of Free Space (Vacuum Permittivity): $$\epsilon_0 = 8.8541878128 \times 10^{-12} \text{ F/m (or C}^2/(\text{N}\cdot\text{m}^2))$$
  • Coulomb's constant: $$k_e = \frac{1}{4\pi\epsilon_0} \approx 8.98755 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$$
  • By Newton's third law of action and reaction: $\vec{F}_{21} = -\vec{F}_{12}$.

2. Coulomb's Force in a Dielectric Medium

When point charges are embedded in a linear, isotropic, homogeneous dielectric medium of relative permittivity $\epsilon_r$ (dielectric constant $\kappa$): $$\vec{F} = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2} \hat{r} = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2} \hat{r} = \frac{\vec{F}_{\text{vacuum}}}{\epsilon_r}$$ Because $\epsilon_r > 1$ for all physical media (e.g., $\epsilon_r \approx 80$ for pure water at 20°C), polarization of the surrounding dielectric shields the charges, reducing the electrostatic force significantly.

3. The Superposition Principle

The net electrostatic force exerted on a test charge $q_0$ by an assembly of $N$ discrete point charges $q_1, q_2, \dots, q_N$ is the vector sum of the individual Coulomb forces: $$\vec{F}_{\text{net}} = \sum_{i=1}^N \vec{F}_i = \frac{q_0}{4\pi\epsilon_0} \sum_{i=1}^N \frac{q_i}{|\vec{r}_0 - \vec{r}_i|^3} (\vec{r}_0 - \vec{r}_i)$$ For a continuous volume charge distribution $\rho(\vec{r}')$: $$\vec{F}_{\text{net}} = \frac{q_0}{4\pi\epsilon_0} \int_V \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}') dV'$$

§1.3The Electric Field Vector and Field Line Topology

The concept of the electric field, introduced by Michael Faraday, replaces direct action-at-a-distance with a local physical intermediary.

1. Definition of the Electric Field Vector

The electric field $\vec{E}(\vec{r})$ at a point in space is defined as the electrostatic force experienced per unit positive test charge placed at that location, in the limit where the test charge $q_0$ approaches zero to prevent perturbing the source charge distribution: $$\vec{E}(\vec{r}) = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0}$$ SI Unit: $\text{N/C}$ (equivalent to $\text{Volts/meter}$, $\text{V/m}$).

2. Field of a Point Charge and Continuous Distribution

For an isolated point charge $q$ at the origin: $$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}$$ For an arbitrary continuous charge distribution occupying volume $V'$: $$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int_{V'} \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}') dV'$$

3. Motion of a Point Charge in an Electric Field

A particle of mass $m$ and charge $q$ placed in an electric field $\vec{E}$ experiences acceleration: $$\vec{a} = \frac{\vec{F}}{m} = \frac{q \vec{E}}{m}$$ In a uniform electric field $\vec{E} = E_0 \hat{j}$:
  • A charged particle launched perpendicular to $\vec{E}$ executes a parabolic trajectory, completely analogous to projectile motion under uniform gravity (the operational principle of cathode ray oscilloscopes and ink-jet printers).

4. Electric Field Lines (Lines of Force)

Electric field lines are imaginary curves whose tangent at any point indicates the direction of the local electric field vector $\vec{E}$:
  1. Field lines originate on positive charges and terminate on negative charges (or extend to infinity).
  2. The local spatial density of lines (number of lines per unit area normal to $\vec{E}$) is directly proportional to field magnitude $|\vec{E}|$.
  3. Field lines never intersect in free space, because the electric field vector is uniquely defined at every point.

§1.4The Electric Dipole in an External Electric Field

An electric dipole consists of two equal and opposite point charges $+q$ and $-q$ separated by a fixed distance $2a$.

1. The Electric Dipole Moment Vector

The electric dipole moment $\vec{p}$ is defined as: $$\vec{p} = q \vec{d}$$ where $\vec{d}$ is the displacement vector directed from the negative charge $-q$ toward the positive charge $+q$. SI Unit: $\text{Coulomb}\cdot\text{meter}$ (C·m). (In molecular physics, the Debye unit is commonly used: $1 \text{ D} = 3.33564 \times 10^{-30} \text{ C}\cdot\text{m}$).

2. Torque on a Dipole in a Uniform Electric Field

When placed in a uniform external field $\vec{E}$, the net translational force on the dipole vanishes: $$\vec{F}_{\text{net}} = (+q)\vec{E} + (-q)\vec{E} = 0$$ However, the forces act along different lines of action, producing a net mechanical restoring torque: $$\vec{\tau} = \vec{r}_+ \times (q\vec{E}) + \vec{r}_- \times (-q\vec{E}) = (\vec{r}_+ - \vec{r}_-) \times q\vec{E} = \vec{d} \times q\vec{E}$$ $$\vec{\tau} = \vec{p} \times \vec{E}$$ Magnitude: $\tau = p E \sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The torque acts to align the dipole moment parallel to the external field ($\theta = 0$).

3. Potential Energy of an Electric Dipole

The external work required to rotate the dipole from reference angle $\theta_0 = 90^{\circ}$ to angle $\theta$ is: $$U(\theta) = \int_{90^{\circ}}^\theta \tau_{\text{ext}} d\theta' = \int_{90^{\circ}}^\theta (p E \sin\theta') d\theta' = -p E \cos\theta$$ $$U = -\vec{p} \cdot \vec{E}$$
  • Stable Equilibrium ($\theta = 0^{\circ}$): Dipole aligned with $\vec{E}$; minimum potential energy $U_{\min} = -p E$.
  • Unstable Equilibrium ($\theta = 180^{\circ}$): Dipole antiparallel to $\vec{E}$; maximum potential energy $U_{\max} = +p E$.

4. Dipole in a Non-Uniform Electric Field

In an inhomogeneous field where $\vec{E}$ varies spatially, the forces on $+q$ and $-q$ do not cancel. The net translational force is: $$\vec{F} = (\vec{p} \cdot \nabla) \vec{E} = \nabla (\vec{p} \cdot \vec{E})$$ A neutral dipole is always pulled toward regions of stronger electric field strength.

§1.5Electric Flux and Gauss's Law

Carl Friedrich Gauss (1835) formulated Gauss's law, which relates the total electric flux passing through a closed geometric surface to the enclosed net electric charge.

1. Definition of Electric Flux

The electric flux $d\Phi_E$ through an infinitesimal oriented surface element $d\vec{A} = \hat{n} dA$ is: $$d\Phi_E = \vec{E} \cdot d\vec{A} = E \cos\theta \, dA$$ where $\hat{n}$ is the outward unit normal vector. For an arbitrary closed Gaussian surface $S$: $$\Phi_E = \oint_S \vec{E} \cdot d\vec{A}$$ SI Unit: $\text{N}\cdot\text{m}^2/\text{C} = \text{V}\cdot\text{m}$.

2. Gauss's Law in Integral Form

For an isolated point charge $q$ enclosed by a concentric sphere of radius $r$: $$\oint_S \vec{E} \cdot d\vec{A} = \oint_S \left( \frac{q}{4\pi\epsilon_0 r^2} \hat{r} \\right) \cdot (\hat{r} dA) = \frac{q}{4\pi\epsilon_0 r^2} (4\pi r^2) = \frac{q}{\epsilon_0}$$ By superposition, for any arbitrary closed surface enclosing net charge $Q_{\text{encl}}$: $$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$ This is **Gauss's Law** (the first of Maxwell's four foundational equations of electromagnetism).

3. Key Properties of Gauss's Law

  • Gauss's law holds for *any* closed surface of arbitrary shape (called a Gaussian surface).
  • Charges located *outside* the closed surface contribute zero net flux through the surface (any flux entering must leave).
  • Gauss's law is a direct mathematical consequence of the inverse-square nature of Coulomb's force ($F \propto 1/r^2$). If the force followed $1/r^{2+\delta}$, Gauss's law would fail.

4. Differential Form of Gauss's Law

Applying Gauss's Divergence Theorem: $$\oint_S \vec{E} \cdot d\vec{A} = \int_V (\nabla \cdot \vec{E}) dV = \frac{1}{\epsilon_0} \int_V \rho \, dV$$ Since this holds for any arbitrary volume $V$: $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ This is the differential form of Gauss's law, stating that positive charge density acts as a local source (divergence) of electric field lines, and negative charge acts as a sink.

§1.6Applications of Gauss's Law: Spherical, Cylindrical, and Planar Symmetries

Gauss's law enables direct calculation of electric field configurations when the charge distribution possesses high geometric symmetry.

1. Spherically Symmetric Charge Distribution

Consider a solid insulating sphere of radius $R$ with uniform volume charge density $\rho$ (total charge $Q = \frac{4}{3}\pi R^3 \rho$). By spherical symmetry, $\vec{E} = E(r) \hat{r}$. Construct a concentric spherical Gaussian surface of radius $r$: $$\oint_S \vec{E} \cdot d\vec{A} = E(r) \oint_S dA = E(r) (4\pi r^2)$$
  • Outside the Sphere ($r \ge R$): $$Q_{\text{encl}} = Q \implies E(r) (4\pi r^2) = \frac{Q}{\epsilon_0} \implies E(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$$ The external field is identical to that of a point charge $Q$ at the center.
  • Inside the Sphere ($r < R$): $$Q_{\text{encl}} = \rho \left( \frac{4}{3}\pi r^3 \\right) = Q \left( \frac{r^3}{R^3} \\right)$$ $$E(r) (4\pi r^2) = \frac{Q r^3}{\epsilon_0 R^3} \implies E(r) = \frac{Q}{4\pi\epsilon_0 R^3} r = \frac{\rho}{3\epsilon_0} r$$ Inside the sphere, the electric field increases linearly from zero at the center to maximum at the surface!

2. Infinitely Long Line of Charge (Cylindrical Symmetry)

Consider an infinite line carrying uniform linear charge density $\lambda$ (C/m). Construct a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. Flux through the two flat end caps is zero because $\vec{E} \perp \hat{n}_{\text{caps}}$. Flux through the curved cylindrical jacket of area $2\pi r L$: $$\oint_S \vec{E} \cdot d\vec{A} = E(r) (2\pi r L) = \frac{Q_{\text{encl}}}{\epsilon_0} = \frac{\lambda L}{\epsilon_0}$$ $$E(r) = \frac{\lambda}{2\pi\epsilon_0 r}$$ The field decays inversely with radial distance ($E \propto 1/r$).

3. Infinite Plane Sheet of Charge (Planar Symmetry)

Consider an infinite non-conducting flat plane with uniform surface charge density $\sigma$ (C/m²). Construct a cylindrical Gaussian pillbox of cross-sectional area $A$ piercing the sheet perpendicularly. Flux through the cylindrical mantle is zero ($\vec{E} \parallel \text{mantle}$). Flux passes equally out of both end faces: $$\oint_S \vec{E} \cdot d\vec{A} = E A + E A = 2 E A = \frac{Q_{\text{encl}}}{\epsilon_0} = \frac{\sigma A}{\epsilon_0}$$ $$E = \frac{\sigma}{2\epsilon_0}$$ Remarkably, the field is completely **uniform and independent of distance** from the sheet! For a conducting surface where all charge resides on the outer boundary and $\vec{E}_{\text{inside}} = 0$: $$E_{\text{conductor}} = \frac{\sigma}{\epsilon_0}$$
Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

StandardCoulomb Vector Superposition for Multiple Point Charges
Three point charges are positioned at the vertices of an equilateral triangle of side length $a = 20.0\text{ cm}$ in vacuum: $q_1 = +4.00\ \mu\text{C}$ at $(0, 0)$, $q_2 = +4.00\ \mu\text{C}$ at $(a, 0)$, and $q_3 = -2.00\ \mu\text{C}$ at the top vertex $(a/2, a\sqrt{3}/2)$.\n(a) Calculate the magnitude and direction of the net electrostatic force $\vec{F}_3$ exerted on charge $q_3$, and\n(b) Determine the electric field vector $\vec{E}$ at the centroid of the triangle.
Step 1: Compute forces from q1 and q2 on q3
$$F_{13} = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_3|}{a^2} = \frac{(8.988 \times 10^9) \times (4.00 \times 10^{-6}) \times (2.00 \times 10^{-6})}{(0.200)^2}$$ $$F_{13} = \frac{0.07190}{0.0400} = 1.798 \text{ N}$$ $$F_{23} = F_{13} = 1.798 \text{ N (by symmetry)}$$

Because $q_3$ is negative and $q_1, q_2$ are positive, both forces are attractive and point down toward the base vertices.

Step 2: Vector resolution of net force on q3
$$\text{Angle with vertical: } \theta = 30.0^{\circ}$$ $$F_{3x} = F_{13}\sin(30^{\circ}) - F_{23}\sin(30^{\circ}) = 0$$ $$F_{3y} = -F_{13}\cos(30^{\circ}) - F_{23}\cos(30^{\circ}) = -2 \times 1.798 \times \cos(30^{\circ})$$ $$F_{3y} = -2 \times 1.798 \times 0.8660 = -3.114 \text{ N}$$ $$\vec{F}_3 = -3.11 \hat{j} \text{ N}$$

Horizontal components cancel identically by symmetry, leaving a purely downward net force of 3.11 N.

Step 3: Electric field at the centroid
$$\text{Distance from vertex to centroid: } r_c = \frac{a}{\sqrt{3}} = \frac{0.200}{\sqrt{3}} = 0.1155 \text{ m}$$ $$\vec{E}_{1+2} = \text{sum of fields from } q_1, q_2 \text{ points straight up along } +\hat{j}:$$ $$E_y = \frac{1}{4\pi\epsilon_0 r_c^2} [q_1 \cos(30^{\circ}) + q_2 \cos(30^{\circ}) + |q_3|] = \frac{8.988 \times 10^9}{(0.1155)^2} \times [2(4.00\times 10^{-6})(0.866) + 2.00\times 10^{-6}]$$ $$E_y = \frac{8.988 \times 10^9}{0.01333} \times [6.928 + 2.00] \times 10^{-6} = (6.743 \times 10^{11}) \times (8.928 \times 10^{-6}) = 6.02 \times 10^6 \text{ V/m} \,\hat{j}$$

The net electric field at the centroid points vertically upward with magnitude $6.02 \times 10^6$ V/m.

AdvancedElectric Dipole Field and Torque in External Field
A water molecule ($H_2O$) possesses a permanent electric dipole moment $p = 6.17 \times 10^{-30}\text{ C}\cdot\text{m}$. It is placed in a uniform electric field $E = 4.50 \times 10^5\text{ N/C}$ directed along the $+x$-axis, with its dipole moment initially inclined at $\theta = 60.0^{\circ}$ to the field.\n(a) Calculate the magnitude of the torque $\vec{\tau}$ acting on the molecule,\n(b) Find the potential energy $U$ of the dipole in this orientation, and\n(c) Calculate the external work required to rotate the molecule from $\theta = 60.0^{\circ}$ to $\theta = 180.0^{\circ}$.
Step 1: Compute torque magnitude
$$\tau = p E \sin\theta = (6.17 \times 10^{-30} \text{ C}\cdot\text{m}) \times (4.50 \times 10^5 \text{ N/C}) \times \sin(60.0^{\circ})$$ $$\tau = (2.7765 \times 10^{-24}) \times 0.86603 = 2.404 \times 10^{-24} \text{ N}\cdot\text{m}$$

The torque acts clockwise to rotate the dipole into alignment with the positive x-axis.

Step 2: Potential energy in field
$$U = -\vec{p} \cdot \vec{E} = -p E \cos\theta = -(6.17 \times 10^{-30}) \times (4.50 \times 10^5) \times \cos(60.0^{\circ})$$ $$U = -2.7765 \times 10^{-24} \times 0.500 = -1.388 \times 10^{-24} \text{ Joules}$$

The negative potential energy confirms the configuration is bound relative to the 90° reference.

Step 3: Work required to rotate to antiparallel orientation
$$W_{\text{ext}} = \Delta U = U(180^{\circ}) - U(60^{\circ})$$ $$U(180^{\circ}) = -p E \cos(180^{\circ}) = +p E = +2.7765 \times 10^{-24} \text{ J}$$ $$W_{\text{ext}} = 2.7765 \times 10^{-24} - (-1.3882 \times 10^{-24}) = +4.165 \times 10^{-24} \text{ Joules}$$

An external agent must perform $4.17 \times 10^{-24}$ J of work against the electrostatic restoring torque.

StandardGauss's Law for Non-Uniform Spherical Charge Density
A solid insulating sphere of radius $R = 12.0\text{ cm}$ carries a spherically symmetric charge distribution whose volume charge density varies with radial distance $r$ according to $\rho(r) = \rho_0 (1 - r/R)$, where $\rho_0 = 3.50 \times 10^{-6}\text{ C/m}^3$.\n(a) Determine the total charge $Q$ of the sphere,\n(b) Derive the electric field $E(r)$ inside the sphere ($r \le R$) and find the radial position $r_{\max}$ where the electric field attains its maximum value, and\n(c) Calculate the magnitude of the electric field at $r = R$ and at $r = 2R$.
Step 1: Calculate total charge by volume integration
$$Q = \int_0^R \rho(r) (4\pi r^2 dr) = 4\pi \rho_0 \int_0^R \left( r^2 - \frac{r^3}{R} \\right) dr$$ $$Q = 4\pi \rho_0 \left[ \frac{R^3}{3} - \frac{R^4}{4R} \\right] = 4\pi \rho_0 R^3 \left( \frac{1}{3} - \frac{1}{4} \\right) = \frac{\pi \rho_0 R^3}{3}$$ $$Q = \frac{\pi \times (3.50 \times 10^{-6}) \times (0.120)^3}{3} = \frac{\pi \times 3.50 \times 10^{-6} \times 1.728 \times 10^{-3}}{3} = 6.333 \times 10^{-9} \text{ C} = 6.33 \text{ nC}$$

Integrating spherical shells of volume $4\pi r^2 dr$ yields a total enclosed charge of 6.33 nC.

Step 2: Derive electric field inside and find maximum
$$Q_{\text{encl}}(r) = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \\right]$$ $$\oint \vec{E} \cdot d\vec{A} = E(r) (4\pi r^2) = \frac{Q_{\text{encl}}(r)}{\epsilon_0} \implies E(r) = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \\right)$$ $$\text{For maximum } E: \frac{dE}{dr} = \frac{\rho_0}{\epsilon_0} \left( \frac{1}{3} - \frac{2r}{4R} \\right) = 0 \implies \frac{1}{3} = \frac{r}{2R} \implies r_{\max} = \frac{2}{3}R$$ $$r_{\max} = \frac{2}{3} \times 12.0 \text{ cm} = 8.00 \text{ cm}$$

The electric field reaches its peak at two-thirds of the sphere's radius.

Step 3: Evaluate electric fields at surface and r = 2R
$$E(R) = \frac{\rho_0}{\epsilon_0} \left( \frac{R}{3} - \frac{R}{4} \\right) = \frac{\rho_0 R}{12 \epsilon_0} = \frac{(3.50 \times 10^{-6}) \times 0.120}{12 \times (8.854 \times 10^{-12})} = \frac{4.20 \times 10^{-7}}{1.0625 \times 10^{-10}} = 3953 \text{ V/m}$$ $$E(2R) = \frac{1}{4\pi\epsilon_0} \frac{Q}{(2R)^2} = \frac{E(R)}{4} \times \left(\frac{R^2}{R^2}\\right) = \frac{8.988 \times 10^9 \times 6.333 \times 10^{-9}}{(0.240)^2} = \frac{56.92}{0.0576} = 988 \text{ V/m}$$

At the surface $E = 3.95$ kV/m; outside at $2R = 24$ cm, it drops to 988 V/m via the inverse-square law.