Comprehensive Study Notes on Electrostatics

Fundamental Principles of Electrostatics and Electric Charge

• Definition of Electrostatics:

  • Electrostatics is the branch of physics that investigates static (stationary) electric charges, the forces exerted between them, and the resulting physical phenomena manifested as electric fields and electric potentials.
  • It contrasts with current electricity, which originates from the coordinated macroscopic movement of electric charges.

• Manifestations in Everyday Phenomena:

  • Frictional Crackling in Textiles: Removing synthetic garments or woolen sweaters in dry winter conditions generates audible crackling and causes the fabric to cling to the skin. Mechanical friction transfers electrons between the body and fabric, producing localized static charges.
  • Contact Shocks: Rising from a polymer or plastic chair and contacting another individual or metallic conductor produces a brief electrostatic discharge shock.
  • Atmospheric Lightning: Massive electrostatic charges accumulate on cloud layers via aerodynamic friction and convection currents during turbulent pre-monsoon weather, discharging as lightning strikes.

• Historical Context and the Two-Charge Model:

  • Documented observations trace back to ancient Greece around $600\text{ BC}$, where friction between amber and animal fur produced an attraction toward light particles.
  • Electric charge is an intrinsic fundamental property of subatomic matter. Matter is constructed of atoms containing three primary elementary particles:
  • Protons: Positively charged particles ($+e$) bound within the atomic nucleus.
  • Electrons: Negatively charged particles ($-e$) occupying orbitals surrounding the nucleus.
  • Neutrons: Electrically neutral particles possessing zero net charge ($0\,C$).
  • In a normal atom, the number of orbital electrons equals the number of nuclear protons, rendering the overall atomic system electrically neutral.

Mechanisms of Charging and Electrostatic Induction

• Charging by Conduction and Friction:

  • Dissimilar insulating substances rubbed together experience mechanical transfer of orbital electrons across their contact boundaries.
  • The material accepting excess electrons acquires a net negative charge, while the donor material develops an identical magnitude of positive charge.
  • Examples include rubbing a glass rod with silk or drawing a polymer comb through dry human hair.
  • Other physical and biological processes separating charge include:
  • Electrochemical reactions inside galvanic and electrolytic cells.
  • Atmospheric thermal convection in thunderclouds.
  • Selective ionic diffusion across living cell membranes.

• Electrostatic Induction:

  • Electrostatic induction is the redistribution of electrical charge in an insulated conductor caused by the proximity of a nearby charged body without physical contact.
  • Free conduction electrons inside the metallic lattice migrate under the Coulomb force exerted by the external charged body.
  • The near surface acquires induced charges of opposite polarity, whereas the far surface exhibits an accumulation of identical polarity.
  • In pure induction, no net charge is exchanged between the inductor and the conductor; withdrawing the inducing body permits free electrons to redistribute uniformly, restoring electrical neutrality.

• Step-by-Step Charging of an Insulated Conductor by Induction:

  • Step 1: An uncharged, insulated spherical conductor is approached by a positively charged rod. Free electrons are attracted toward the rod, creating an accumulation of negative charge on the proximal side and leaving an equal positive charge on the distal hemisphere.   
  • Step 2: The distal side is connected to electrical ground (earthed). Conduction electrons flow upward from the Earth into the conductor, neutralizing the positive charge.   
  • Step 3: The ground path is severed while the external positively charged rod remains held in position, trapping the excess electrons on the conductor.   
  • Step 4: The inducing rod is withdrawn. Repulsive forces among the excess trapped electrons cause them to disperse symmetrically over the outer metallic surface, leaving a permanent negative net charge.   

The Gold Leaf Electroscope

• Apparatus and Structural Design:

  • A classic diagnostic instrument engineered for the qualitative detection and quantitative estimation of electric charge.
  • Consists of an external top horizontal metallic disc attached to a vertical conducting stem/rod, terminating in an ultra-thin gold leaf suspended adjacent to a fixed vertical metal plate.
  • The central stem is electrically insulated from the surrounding chassis by an insulating collar.
  • A protective glass window shields the leaf assembly from ambient air currents while permitting clear visual observation.
  • An earth terminal on the metallic casing prevents the accumulation of stray electrostatic charges.   

• Operating Principles and Methods of Charging:

  • Detection: Applying charge to the disc conducts charge down the central stem to both the fixed plate and the hinged gold leaf. Because like charges repel, the gold leaf diverges outward from the plate; greater divergence indicates higher charge magnitude.
  • Discharge: Touching the top disc with a finger grounds the electroscope, allowing excess charge to bleed to Earth and causing the leaf to collapse.
  • Charging by Contact: Bringing a charged conductor into direct mechanical contact with the disc transfers identical charge to the electroscope. This is inefficient and leaves the leaf charged with the same polarity as the source.
  • Charging by Induction: Bringing a charged rod near (without touching) the disc polarizes the instrument. Momentarily earthing the disc and then removing the ground before withdrawing the rod charges the electroscope with the opposite sign of charge.

Basic Properties of Electric Charge

• Additive Nature of Charge:

  • Electric charge behaves as a scalar quantity. The total electric charge on any composite body equals the algebraic sum of all point charges distributed throughout its constituent parts:     $q_{\text{total}} = \sum_{i=1}^N q_i = q_1 + q_2 + q_3 + \dots + q_N$
  • Must be calculated taking operational signs ($+$ or $-$) into account.
  • Comparison Between Mass and Charge:
  • Particle masses are strictly non-negative ($m > 0$); consequently, total mass is unconditionally positive.
  • Electric charges can be positive, negative, or zero; a system possessing equal numbers of positive and negative charges exhibits a net charge of exactly zero.

• Quantization of Charge:

  • Charge is discrete. The minimum unit of charge capable of independent existence is the elementary charge $e$, determined precisely by Robert Millikan's oil-drop experiment:     $e = 1.6 \times 10^{-19}\,C$
  • The net charge $q$ on any macroscopic or microscopic entity is an exact integer multiple of $e$:     $q = \pm n e \quad (n \in \mathbb{Z}^+)$
  • Charge modifications occur exclusively via integral transfers of electrons ($n = 1, 2, 3, \dots$).
  • Macroscopic Continuity: In macroscopic systems, working charges are typically measured on the order of microcoulombs ($1\,\mu C = 10^{-6}\,C$). The number of involved electrons is:     $n = \frac{10^{-6}\,C}{1.6 \times 10^{-19}\,C} = 6.25 \times 10^{12}$
  • Because this number is extremely large, the underlying granular quantum steps are imperceptible, giving the illusion of a continuous charge distribution.

• Subatomic Structure and the Quark Hypothesis:

  • Modern high-energy particle physics establishes that nucleons (protons and neutrons) are composed of sub-elementary entities called quarks.
  • Quarks occur in six distinct flavors carrying fractional elementary charges of either $+\frac{2}{3}e$ or $-\frac{1}{3}e$.
  • Nucleons consist of specific tri-quark combinations:
  • Proton: Composed of two up quarks and one down quark ($uud$):       $q_p = \left(+\frac{2}{3}e\right) + \left(+\frac{2}{3}e\right) + \left(-\frac{1}{3}e\right) = +1e$
  • Neutron: Composed of one up quark and two down quarks ($udd$):       $q_n = \left(+\frac{2}{3}e\right) + \left(-\frac{1}{3}e\right) + \left(-\frac{1}{3}e\right) = 0$
  • Quarks are permanently confined within composite hadrons by the strong color interaction; free, isolated quarks do not exist in nature.
  • Charge quantization remains completely intact under the quark model, with the fundamental step size decreasing from $e$ to $\frac{e}{3}$.
  • Highly sensitive cryogenic devices such as single-electron transistors (SETs) can experimentally measure discrete electron transfers at sub-attoampere current levels.

• Conservation of Electric Charge:

  • In any closed, isolated physical or chemical process, electric charge cannot be created or destroyed; it can only transfer from one region of matter to another.
  • The net algebraic sum of electric charge in an isolated thermodynamic system remains strictly invariant over time.
  • When an uncharged glass rod is rubbed with dry silk cloth, the positive charge acquired by the glass matches the negative charge accumulated on the silk, keeping the net charge of the two-body system zero.

• Forces Between Charges:

  • Charged bodies exert mutual mechanical forces across space without physical contact.
  • Like charges (positive-positive or negative-negative) mutually repel; unlike charges (positive-negative) mutually attract.   

Coulomb’s Law of Electrostatics

• Definition of a Point Charge:

  • A point charge is an idealized electrostatic model where the spatial physical dimensions of the charged object are negligible compared to the distance of separation $r$ between interacting bodies.

• Scalar Formulation:

  • Formulated by Charles-Augustin de Coulomb (1736–1806) using a sensitive torsion balance.
  • Statement: The magnitude of the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the scalar product of their charge magnitudes and inversely proportional to the square of the distance separating their centers, acting along the straight line connecting them.
  • Mathematical expression:     $F \propto \frac{q_1 q_2}{r^2}$$F = K \frac{q_1 q_2}{r^2}$     where $K$ represents an electrostatic proportionality constant governed by the measurement system and the physical medium.   

• Electrostatic Constants and Permittivity in Free Space:

  • In SI units, for charges situated in vacuum (free space):     $K = \frac{1}{4\pi\varepsilon_0} \approx 9.0 \times 10^9\,N\,m^2\,C^{-2}$
  • Here, $\varepsilon_0$ denotes the absolute electrical permittivity of free space:     $\varepsilon_0 = 8.85 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}$
  • The force in vacuum ($F_{\text{vac}}$) is given by:     $F_{\text{vac}} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$

• Dielectric Constant and Relative Permittivity:

  • When charges are placed in a material medium (e.g., purified water, mica, or paraffin oil), the medium attenuates the net electrostatic force:     $F_{\text{med}} = \frac{1}{4\pi\varepsilon} \frac{q_1 q_2}{r^2}$     where $\varepsilon$ represents the absolute permittivity of that material.
  • Taking the ratio of force in vacuum to force in the medium defines the relative permittivity $\varepsilon_r$ (also designated as dielectric constant $K$ or specific inductive capacity):     $\varepsilon_r = K = \frac{\varepsilon}{\varepsilon_0} = \frac{F_{\text{vac}}}{F_{\text{med}}}$
  • Characteristics of $\varepsilon_r$:
  • Dimensionless, unitless scalar quantity.
  • Always $\ge 1$ ($\varepsilon_r = 1.0$ for vacuum; $\approx 1.0006$ for air).
  • General expression for force within a medium:       $F_{\text{med}} = \frac{1}{4\pi\varepsilon_0\varepsilon_r} \frac{q_1 q_2}{r^2}$
  • Case of Liquid Water: For pure water, $\varepsilon_r = 80$:       $F_{\text{water}} = \frac{F_{\text{vac}}}{80} = \left(\frac{1}{80}\right) F_{\text{vac}}$       The electrostatic force between identical charges in water is reduced to $\frac{1}{80}$th of its value in a vacuum.
  • Derivation assumes the medium is linear, homogeneous, isotropic, and effectively infinite.

• Definition of the SI Unit of Charge (Coulomb):

  • Substituting $q_1 = q_2 = 1.0\,C$ and $r = 1.0\,m$ into Coulomb's vacuum formula:     $F = \left(9.0 \times 10^9\right) \frac{(1.0)(1.0)}{(1.0)^2} = 9.0 \times 10^9\,N$
  • Definition: One Coulomb ($1\,C$) is that quantity of electric charge which, when situated at a separation of one meter from an identical charge in vacuum, experiences a mutual repulsive electrostatic force of $9.0 \times 10^9\,N$.
  • Physical Scale: A force of $9.0 \times 10^9\,N$ corresponds to approximately $10\,\text{million metric tonnes}$. Because standard commercial haulage trucks carry roughly $10\,\text{metric tonnes}$, this force is equivalent to the static weight of about one million truckloads, making one Coulomb an exceptionally large quantity of charge.
  • Practical Sub-Units:
  • Microcoulomb: $1\,\mu C = 10^{-6}\,C$
  • Nanocoulomb: $1\,nC = 10^{-9}\,C$
  • Picocoulomb: $1\,pC = 10^{-12}\,C$

Vector Form of Coulomb’s Law and Comparison with Gravitation

• Vector Formulation:

  • Let point charges $q_1$ and $q_2$ be located at position vectors $\mathbf{r}_1$ and $\mathbf{r}_2$ at spatial coordinates $A$ and $B$.   
  • The displacement vector directed from $q_1$ to $q_2$ is $\mathbf{r}_{12}$, with unit vector $\hat{\mathbf{r}}_{12}$ pointing along the line $\vec{AB}$.
  • The force $\mathbf{F}_{21}$ exerted on charge $q_2$ by charge $q_1$ is:     $\mathbf{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{|\mathbf{r}_{21}|^2} \hat{\mathbf{r}}_{21}$
  • The force $\mathbf{F}_{12}$ exerted on charge $q_1$ by charge $q_2$ is:     $\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{|\mathbf{r}_{12}|^2} \hat{\mathbf{r}}_{12}$
  • Because $\hat{\mathbf{r}}_{12} = -\hat{\mathbf{r}}_{21}$, it follows directly that:     $\mathbf{F}_{21} = -\mathbf{F}_{12}$
  • Significance:
  • Obeys Newton's Third Law of Motion, forming an action-reaction pair.
  • Electrostatic force is a central force: it acts strictly along the radius vector linking the centers of the two charges.

• Comprehensive Comparison with Newton's Universal Gravitation:

  • Similarities:
  • Both fundamental interactions adhere to an inverse-square distance dependence ($F \propto \frac{1}{r^2}$).
  • Both forces are central forces acting along the line of centers.
  • Both forces operate across vacuum without requiring a physical medium.
  • Differences:
  • Gravitational force is unconditionally attractive; electrostatic force is either attractive (unlike charges) or repulsive (like charges).
  • Gravitational interaction is independent of the intervening medium; electrostatic force is strongly mediated by the surrounding medium's dielectric permittivity.
  • Electrostatic force is vastly stronger than gravitational force.

• Quantitative Force Ratio Between Protons:

  • Consider two isolated protons separated by a typical nuclear distance $r = 10^{-15}\,m$:
  • Charge on proton: $q = +1.6 \times 10^{-19}\,C$
  • Mass of proton: $m_p = 1.67 \times 10^{-27}\,kg$
  • Universal gravitational constant: $G = 6.674 \times 10^{-11}\,m^3\,kg^{-1}\,s^{-2}$
  • Repulsive Electrostatic Force:     $F_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2} = \frac{(9.0 \times 10^9) \times (1.6 \times 10^{-19})^2}{(10^{-15})^2} = 2.3 \times 10^2\,N$
  • Attractive Gravitational Force:     $F_g = G \frac{m_p^2}{r^2} = \frac{(6.674 \times 10^{-11}) \times (1.67 \times 10^{-27})^2}{(10^{-15})^2} = 1.86 \times 10^{-34}\,N$
  • Magnitude Comparison:     $\frac{F_e}{F_g} = \frac{2.30 \times 10^2\,N}{1.86 \times 10^{-34}\,N} = 1.23 \times 10^{36}$
  • The electrostatic force between protons is roughly $36\text{ orders of magnitude}$ stronger than their mutual gravitational attraction.

Principle of Superposition of Electrostatic Forces

• Formal Principle:

  • When multiple point charges interact simultaneously, the net resultant electrostatic force acting on any single selected charge is equal to the vector sum of the independent forces exerted on it by every other charge in the system.
  • The interaction between any specific pair of charges is entirely unaffected by the presence of surrounding charges.   

• Mathematical Formulation:

  • For a discrete configuration of charges $q_1, q_2, q_3, \dots, q_N$, the total force $\mathbf{F}_1$ on charge $q_1$ is:     $\mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \mathbf{F}_{14} + \dots + \mathbf{F}_{1N}$$\mathbf{F}_1 = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1 q_2}{|\mathbf{r}_{12}|^2} \hat{\mathbf{r}}_{12} + \frac{q_1 q_3}{|\mathbf{r}_{13}|^2} \hat{\mathbf{r}}_{13} + \dots + \frac{q_1 q_N}{|\mathbf{r}_{1N}|^2} \hat{\mathbf{r}}_{1N} \right]$
  • For a test charge $q_0$ acted upon by an assembly of $N$ source charges:     $\mathbf{F}_{\text{test}} = \sum_{n=1}^N \mathbf{F}_n = \frac{1}{4\pi\varepsilon_0} \sum_{n=1}^N \frac{q_0 q_n}{r_{0n}^2} \hat{\mathbf{r}}_{0n}$     where $\hat{\mathbf{r}}_{0n}$ is the unit vector oriented from source charge $q_n$ toward the test charge $q_0$, and $r_{0n}$ represents their spatial separation.

Electric Field and Field Intensity

• Concept and Definition:

  • An electric charge alters the physical properties of the space surrounding it, creating an electric field.
  • Any secondary charge introduced into this modified region experiences a Coulomb force.
  • Mathematical Definition: Electric field intensity $\mathbf{E}$ at any point is defined as the electrostatic force experienced per unit positive test charge placed at that location:     $\mathbf{E} = \frac{\mathbf{F}}{q_0}$
  • Limiting Formal Definition:$\mathbf{E} = \lim_{q \to 0} \frac{\mathbf{F}}{q}$
  • Test Charge Characterization: The test charge $q_0$ must be an infinitesimally small positive charge such that its own field does not disturb or modify the source charge distribution being evaluated.

• Field of an Isolated Point Charge:

  • Coulomb force on test charge $q$ from primary source charge $Q$ separated by distance $r$ in vacuum:     $\mathbf{F} = \frac{1}{4\pi\varepsilon_0} \frac{Q q}{r^2} \hat{\mathbf{r}}$
  • Electric field intensity:     $\mathbf{E} = \frac{\mathbf{F}}{q} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}$
  • The electric field has spherical symmetry; its magnitude is constant over any concentric sphere of radius $r$ centered on $Q$.
  • Directed radially outward if $Q > 0$, and radially inward if $Q < 0$.
  • Inside a material dielectric medium with relative permittivity $K$:     $\mathbf{E} = \frac{1}{4\pi\varepsilon_0 K} \frac{q}{r^2} \hat{\mathbf{r}}$

• Units and Dimensional Formula:

  • SI Unit: Newton per Coulomb ($N\,C^{-1}$) or Volt per meter ($V\,m^{-1}$).
  • Dimensional Derivation:$[E] = \frac{[F]}{[q_0]} = \frac{[L M T^{-2}]}{[I T]} = [L M T^{-3} I^{-1}]$

• Classification of Electric Fields:

  • Uniform Electric Field: Magnitude and vector direction remain identical at all points in space. Represented by parallel, equidistant field lines (e.g., between oppositely charged wide parallel plates).
  • Non-Uniform Electric Field: Magnitude, direction, or both vary across spatial coordinates (e.g., the radial inverse-square field of a single point charge).

• Potential-Gradient Formulation ($E = V/d$):

  • Consider two parallel conducting plates separated by distance $d$ across which an electrostatic potential difference $V$ is maintained.
  • The work $W$ required to transport an exploratory positive charge $+q$ from negative plate $B$ to positive plate $A$ against the uniform electric force is:     $W = F d$
  • By the thermodynamic definition of electrostatic potential difference:     $W = V q$
  • Equating both mechanical energy formulations:     $V q = F d \implies \frac{F}{q} = \frac{V}{d}$$E = \frac{V}{d}$
  • This yields the practical unit Volt per meter ($V\,m^{-1}$), directly equivalent to Newton per Coulomb ($N\,C^{-1}$).

Electric Lines of Force

• Conceptual Origin:

  • Introduced by Michael Faraday (1791–1867) as a geometric tool to visualize the spatial distribution and strength of electric fields without direct algebraic calculation.

• Definition:

  • An electric line of force is an imaginary curve drawn through a region of space such that the tangent constructed at any point on the curve indicates the vector direction of the resultant electric field at that position.
  • Represents the geometric path along which a massless, free positive unit test charge would accelerate.   

• Fundamental Properties and Characteristics:

  • Lines of force originate on positive charges and terminate on negative charges.
  • Lines of force cannot intersect or cross. If two lines intersected, two tangents could be drawn at the intersection point, which would imply two different directions of the net electric field at that single coordinate—a physical impossibility.
  • Field lines intersect conducting surfaces at perpendicular ($90^\circ$) angles in static equilibrium.
  • Field lines never pass through the interior of a static conductor; the electrostatic field inside a conductor is zero ($E = 0$). Field lines do pass through dielectric insulators.
  • The relative spatial density of lines (number of lines passing perpendicularly through a unit surface area) is directly proportional to field intensity ($E$):
  • Crowded field lines correspond to regions of strong electric field.
  • Widely spaced field lines denote regions of weak electric field.   
  • Field lines for a perfectly uniform electric field are straight, mutually parallel, and equally spaced.
  • Lines of force are purely geometric constructions with no independent physical existence.

• Field Line Topologies for Standard Charge Geometries:

  • Isolated Positive Charge ($+q$): Lines emerge radially outward into infinity.   
  • Isolated Negative Charge ($-q$): Lines converge radially inward from infinity.   
  • Electric Dipole ($+q$ and $-q$): Continuous curved paths loop outward from the positive charge and terminate on the negative charge.   
  • Two Like Charges ($+q$ and $+q$): Lines curve laterally away from each other, creating a central null point ($E = 0$) where no field lines pass.   
  • Conducting Body Shielding: Lines terminate normally at the outer boundary, leaving the hollow interior completely free of field lines ($E = 0$).   

Electric Flux and Gauss’ Law

• Definition of Electric Flux ($\Phi$):

  • Electric flux measures the total number of electric field lines penetrating a given surface area.
  • For a planar area element $d\mathbf{S}$ inclined at an angle $\theta$ relative to a local electric field $\mathbf{E}$:     $d\phi = E (dS \cos(\theta)) = \mathbf{E} \cdot d\mathbf{S}$
  • Integrating across an extended macroscopic surface $S$ yields the total electric flux:     $\Phi = \int_S d\phi = \int_S \mathbf{E} \cdot d\mathbf{S} = \mathbf{E} \cdot \mathbf{S}$
  • Units of Flux:$[\Phi] = (V/m) \times m^2 = V\,m = N\,m^2\,C^{-1}$

• Gauss' Law Statement:

  • Formulated by Karl Friedrich Gauss (1777–1855).
  • Theorem: The total electric flux $\Phi_E$ of the net electric field passing through any arbitrary closed mathematical surface equals the net enclosed algebraic charge $Q_{\text{enclosed}}$ divided by the vacuum permittivity $\varepsilon_0$:     $\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

• Mathematical Derivation Using Solid Angles:

  • Consider an isolated positive point charge $+q$ placed at origin $O$ inside an irregular closed boundary.
  • At an arbitrary surface element $d\mathbf{S}$ at distance $r$, the radial electric field is:     $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$
  • The differential electric flux through area element $dS$ is:     $d\phi = E \cos(\theta)\,dS = \left( \frac{q}{4\pi\varepsilon_0 r^2} \right) \cos(\theta)\,dS = \left( \frac{q}{4\pi\varepsilon_0} \right) \left( \frac{dS \cos(\theta)}{r^2} \right)$
  • By geometric definition, the solid angle element $d\omega$ subtended by area $dS$ at point $O$ is:     $d\omega = \frac{dS \cos(\theta)}{r^2}$$d\phi = \left( \frac{q}{4\pi\varepsilon_0} \right) d\omega$
  • Integrating over the entire closed enclosing surface:     $\Phi_E = \oint_S d\phi = \frac{q}{4\pi\varepsilon_0} \oint_S d\omega$
  • Because the total solid angle subtended by any complete closed surface surrounding an interior point is identically $4\pi\text{ steradians}$ ($\oint_S d\omega = 4\pi$):     $\Phi_E = \frac{q}{4\pi\varepsilon_0} (4\pi) = \frac{q}{\varepsilon_0}$

• Physical Implications and Surface Characteristics:

  • Charge Outside the Enclosing Surface: If a source charge lies outside the closed boundary, any field line entering the volume through one surface must exit through another, producing a net flux of zero.
  • Superposition of Flux: For a collection of charges $q_1, q_2, \dots, q_n$ distributed within the enclosed volume:     $\Phi_E = \frac{q_1}{\varepsilon_0} + \frac{q_2}{\varepsilon_0} + \dots + \frac{q_n}{\varepsilon_0} = \frac{1}{\varepsilon_0} \sum_{i=1}^n q_i = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$
  • Flux Sign Conventions:
  • Positive enclosed charge yields net outward flux ($\Phi > 0$).
  • Negative enclosed charge yields net inward flux ($\Phi < 0$).
  • Invariance to Shape and Scale: Electric flux depends solely on the magnitude of the enclosed charge and is independent of the size, radius, or geometry of the closed surface.   

• The Gaussian Surface:

  • An imaginary closed surface chosen to exploit coordinate symmetry (e.g., spherical, cylindrical, or planar) so that the electric field magnitude is constant over the surface and oriented either parallel or normal to the area vectors.

Electric Dipole Dynamics

• Dipole Definitions and Terminology:

  • Electric Dipole: A system composed of two equal and opposite point charges ($+q$ and $-q$) separated by a finite spatial distance $2l$.
  • Dipole Axis: The straight line passing through the centers of both charges.
  • Axial Line: The infinite collinear extension passing along the dipole axis.
  • Equatorial Line: The perpendicular bisector passing through the geometric midpoint of the dipole axis.

• Electric Dipole Moment Vector ($\mathbf{p}$):

  • The vector measure of dipole strength, defined as the product of the charge magnitude $q$ and the vector separation $2\mathbf{l}$ directed from the negative charge to the positive charge:     $\mathbf{p} = q(2\mathbf{l})$
  • SI Unit: Coulomb-meter ($C\,m$).
  • Debye Unit ($D$): A conventional molecular unit:     $1\,D = 3.33 \times 10^{-30}\,C\,m$
  • Reference Value: Two elementary charges ($\pm e$) separated by $1.0\,\text{Å} = 1.0 \times 10^{-10}\,m$ yield:     $p = (1.6 \times 10^{-19}\,C) \times (1.0 \times 10^{-10}\,m) = 1.6 \times 10^{-29}\,C\,m \approx 4.8\,D$

• Polar vs. Non-Polar Molecular Dipoles:

  • Polar Molecules: Chemical bonds have an asymmetric charge distribution, creating an inherent separation between centers of positive and negative charge and giving the molecule a permanent dipole moment.
  • Examples: $H_2O$, $NH_3$, $SO_2$, $NaCl$.
  • Water ($H_2O$): Displays a non-linear bent structure with a bond angle of $105^\circ$. Oxygen draws electron density away from hydrogen, creating a permanent dipole moment of $p = 6.172 \times 10^{-30}\,C\,m = 1.85\,D$ oriented from oxygen toward the hydrogen atoms.
  • Non-Polar Molecules: Highly symmetric structures where the centers of positive and negative charge coincide, producing zero permanent dipole moment in the ground state.
  • Examples: $H_2$, $Cl_2$, $CO_2$, $CH_4$.
  • An external electric field displaces the positive nuclear charges and negative electron clouds in opposite directions, inducing a temporary dipole moment.

• Torque on an Electric Dipole in a Uniform Field:

  • Consider a dipole with charges $\pm q$ separated by distance $2l$ placed in a uniform field $\mathbf{E}$ at angle $\theta$:
  • Force on positive charge $+q$ at $B$: $\mathbf{F}_B = +q\mathbf{E}$ (directed along $\mathbf{E}$.
  • Force on negative charge $-q$ at $A$: $\mathbf{F}_A = -q\mathbf{E}$ (directed opposite to $\mathbf{E}$.
  • The net translational force is zero ($\mathbf{F}_{\text{net}} = \mathbf{F}_A + \mathbf{F}_B = 0$), meaning the center of mass remains stationary.
  • Because the equal and opposite forces act along different lines of action separated by perpendicular distance $d = 2l\sin(\theta)$, they form a couple.
  • The magnitude of the resulting torque $\tau$ is:     $\tau = \text{Magnitude of force} \times \text{Perpendicular distance} = (qE)(2l\sin(\theta)) = (q \cdot 2l) E \sin(\theta)$$\tau = p E \sin(\theta)$
  • Vector formulation:     $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$
  • Special Cases:
  • Maximum Torque ($\theta = 90^\circ$): $\sin(90^\circ) = 1 \implies \tau_{\text{max}} = pE$.
  • Stable Equilibrium ($\theta = 0^\circ$): $\sin(0^\circ) = 0 \implies \tau = 0$. The dipole moment is aligned with the external electric field.
  • Unstable Equilibrium ($\theta = 180^\circ$): $\sin(180^\circ) = 0 \implies \tau = 0$. The dipole moment is anti-parallel to the electric field.

Electric Field Intensity Due to an Electric Dipole

• Case 1: Field at an Axial Point:

  • Consider a dipole consisting of charges $-q$ at point $A$ and $+q$ at point $B$, separated by distance $2l$ with center $C$.
  • Let point $P$ lie on the dipole axis at distance $r$ from the center $C$.
  • Electric field component at $P$ due to $-q$ at $A$:     $\mathbf{E}_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r + l)^2} \hat{\mathbf{PD}}$     (directed along $\vec{PD}$, pointing toward $A$)
  • Electric field component at $P$ due to $+q$ at $B$:     $\mathbf{E}_B = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r - l)^2} \hat{\mathbf{PQ}}$     (directed along $\vec{PQ}$, pointing away from $B$)
  • Because $(r - l) < (r + l)$, the magnitude of $\mathbf{E}_B$ exceeds $\mathbf{E}_A$. The net axial field $\mathbf{E}_a = \mathbf{E}_B + \mathbf{E}_A$ has magnitude:     $E_a = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(r - l)^2} - \frac{1}{(r + l)^2} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{(r + l)^2 - (r - l)^2}{(r^2 - l^2)^2} \right]$$E_a = \frac{q}{4\pi\varepsilon_0} \left[ \frac{4lr}{(r^2 - l^2)^2} \right] = \frac{1}{4\pi\varepsilon_0} \frac{2(q \cdot 2l)r}{(r^2 - l^2)^2}$$E_a = \frac{1}{4\pi\varepsilon_0} \frac{2pr}{(r^2 - l^2)^2}$
  • Short Dipole Approximation ($r \gg l$):
  • Neglecting $l^2$ relative to $r^2$:       $E_a = \frac{1}{4\pi\varepsilon_0} \frac{2pr}{r^4} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}$
  • Vector direction: Points along the dipole moment vector $\mathbf{p}$ (from $-q$ to $+q$).

• Case 2: Field at an Equatorial Point:

  • Let point $P$ lie on the equatorial line at distance $r$ from the center $C$.
  • From right triangle $\Delta ACP$, the distance to each charge is:     $AP = BP = \sqrt{r^2 + l^2}$
  • Field magnitudes due to each individual charge:     $E_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}$$E_B = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}$
  • Component Resolution at Point $P$:
  • Vertical Components (along equatorial line): $E_A \sin(\theta)$ acts toward $-q$, while $E_B \sin(\theta)$ acts away from $+q$. Being equal in magnitude and opposite in direction, they cancel completely:       $E_y = E_B \sin(\theta) - E_A \sin(\theta) = 0$
  • Horizontal Components (parallel to dipole axis): Both $E_A \cos(\theta)$ and $E_B \cos(\theta)$ point in the same direction, opposite to $\mathbf{p}$. They add directly:       $E_{\text{eq}} = E_A \cos(\theta) + E_B \cos(\theta) = 2E_A \cos(\theta)$
  • From the geometry of $\Delta ACP$:     $\cos(\theta) = \frac{l}{\sqrt{r^2 + l^2}}$
  • Substituting gives:     $E_{\text{eq}} = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2} \right) \left( \frac{l}{\sqrt{r^2 + l^2}} \right) = \frac{1}{4\pi\varepsilon_0} \frac{q(2l)}{(r^2 + l^2)^{3/2}}$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + l^2)^{3/2}}$
  • Short Dipole Approximation ($r \gg l$):
  • Neglecting $l^2$ relative to $r^2$:       $E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}$
  • Vector direction: Points anti-parallel to the dipole moment vector $\mathbf{p}$.

• Comparison of Axial and Equatorial Intensities:

  • For identical distances $r$ from the center of a short dipole:     $E_{\text{axial}} = 2 E_{\text{equatorial}}$
  • Both field regimes decay with an inverse-cube distance dependence ($E \propto \frac{1}{r^3}$), which drops off significantly faster than the inverse-square field of a single point charge ($E \propto \frac{1}{r^2}$).

Continuous Charge Distributions

• Concept of Continuous Distribution:

  • When charges are packed together with microscopic spacing much smaller than the distance to the observation point, the discrete charge array is modeled as a continuous charge distribution.   

• Linear Charge Density ($\lambda$):

  • Used for charges distributed along a one-dimensional curve or thin wire of length $l$:     $\lambda = \frac{q}{l}$
  • For non-uniform distributions, the charge on a differential element $dl$ is:     $dq = \lambda\,dl$
  • SI Unit: Coulomb per meter ($C\,m^{-1}$).
  • Examples: Thin uniformly charged metallic rod, fine nylon fiber.

• Surface Charge Density ($\sigma$):

  • Used for charges distributed across a two-dimensional surface of area $A$:     $\sigma = \frac{q}{A}$
  • For non-uniform distributions, the charge on a differential element $dA$ is:     $dq = \sigma\,dA$
  • SI Unit: Coulomb per square meter ($C\,m^{-2}$).
  • Examples: Charged metallic plates, planar conducting sheets, dielectric discs.

• Volume Charge Density ($\rho$):

  • Used for charges distributed through a three-dimensional volume $V$:     $\rho = \frac{q}{V}$
  • For non-uniform distributions, the charge on a differential element $dV$ is:     $dq = \rho\,dV$
  • SI Unit: Coulomb per cubic meter ($C\,m^{-3}$).
  • Examples: Non-conducting solid spheres, uniformly charged polymer cubes.

• Total Field Calculation:

  • The total electric field is obtained by integrating the field contributions from all infinitesimal charge elements:     $\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{r^2} \hat{\mathbf{r}}$

Practical Applications, Hazards, and Safety Precautions

• Industrial and Technological Applications:

  • Electrophotography (Photocopiers): High-voltage corona wires deposit uniform electrostatic charges on a selenium drum. Light projected from an original document discharges specific regions, allowing negatively charged toner powder to adhere electrostatically to the remaining charged pattern and transfer onto paper.
  • Inkjet Printing: Microscopic ink droplets expelled from nozzles acquire directed electrostatic charges, guiding them between charged deflection plates to print precise patterns.
  • Electrostatic Powder Coating: Atomized paint or polymer powder is electrically charged at an applicator nozzle while the grounded metal workpiece is kept at opposite polarity, ensuring uniform, wraparound coating with minimal overspray.
  • Electrostatic Precipitators: Industrial exhaust filtration units that pass combustion smoke through high-voltage ionizing wire grids. Particulate matter acquires an electrical charge and collects on oppositely charged plates, preventing environmental release.

• Electrostatic Hazards:

  • Fuel Handling Sparks: Liquid petroleum or diesel flowing through rubber hoses develops dangerous static charges through pipe friction. If not safely discharged, a spark can ignite fuel vapors, causing explosions.
  • High-Altitude Lightning: Massive static buildup on storm cloud bases can initiate energetic dielectric breakdown of air, destroying trees, structures, and unshielded electrical installations.
  • Equipment Damage and Shock: Large static charges accumulated on human operators handling delicate complementary metal-oxide-semiconductor (CMOS) components can puncture gate oxides, permanently destroying microelectronics.
  • Display Dust Attraction: High electrostatic potentials generated on television and computer monitor screens attract airborne dust and particulate matter, accelerating optical fouling.

• Safety Precautions:

  • Earth Grounding: Electrical appliances and industrial fuel transport tankers must be grounded through heavy-duty copper lines to discharge static charges safely into the Earth.
  • Static Dissipative Footwear: Operators should avoid thick, highly insulating rubber-soled shoes in explosive or microelectronic assembly environments, using dissipative leather or conductive shoes instead.
  • Humidity Control: Maintaining ambient relative humidity ($> 50\%$) forms a microscopic, conductive water film on surfaces that bleeds off static charges through atmospheric ions.

Questions and Discussion

• Questions from Can You Recall?

  • Prompt: Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
  • Response: Yes. Friction between the person's clothing and the plastic chair transfers electrons, building a net electrostatic charge on the body. Approaching or shaking hands with a friend creates a localized, rapid electrostatic discharge through the air gap, causing a mild electric shock.
  • Prompt: Ever heard a crackling sound while taking out your sweater in winter?
  • Response: Yes. Moving woolen or synthetic fabric across clothing or skin separates positive and negative charges through friction. The dry winter air has low conductivity, so charge accumulates until miniature, audible electric sparks crackle across the fibers.
  • Prompt: Have you seen the lightning striking during pre-monsoon weather?
  • Response: Yes. Turbulence and convective circulation in pre-monsoon storm clouds cause vigorous collisions between ice crystals and water droplets, polarizing the cloud layers. The resulting potential difference breaks down the surrounding air, producing a massive lightning discharge.

• Questions from Can You Tell?

  • Prompt: When a petrol or a diesel tanker is emptied in a tank, it is grounded. Why?
  • Response: Liquid hydrocarbons flowing through delivery pipes generate significant static charge via fluid friction. Grounding the chassis and pipes with a conductive copper lead provides a safe, low-resistance path to Earth, preventing sparks that could ignite flammable fuel vapors.
  • Prompt: A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving. Why?
  • Response: Driving causes tire-road friction, engine air turbulence, and internal liquid sloshing, continuously building up static electricity. A trailing conductive chain maintains contact with the road, safely bleeding charge to Earth to prevent electrical sparking during transit.
  • Prompt: Three charges, $q$ each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge $q$ placed at the centroid of the triangle?
  • Response: The resultant force is zero ($0\,N$). By the three-fold rotational symmetry of an equilateral triangle, the centroid is equidistant from all three vertices. The three charges exert repulsive forces of equal magnitude oriented at mutual angles of $120^\circ$. Vector addition yields: $\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$.
  • Prompt: Why can a small voltage produce a reasonably large electric field?
  • Response: Because the electric field is governed by $E = \frac{V}{d}$. When the physical distance $d$ between two electrodes is made exceptionally small (such as an automotive spark plug gap of $d = 1.25 \times 10^{-3}\,m$ or a microscopic semiconductor layer of $10^{-9}\,m$), dividing even a modest potential difference ($V$) by a tiny denominator yields an exceptionally high electric field gradient ($E$).
  • Prompt: Lines of force are imaginary; can they have any practical use?
  • Response: Yes. Although field lines are conceptual geometric aids, their spatial distribution provides an intuitive, reliable map of field directions and gradient strengths. They simplify the design of high-voltage insulation, particle accelerators, electron microscopes, electrostatic precipitators, and Faraday cages.
  • Prompt: The surface charge density of Earth is $\sigma = -1.33\,nC/m^2$, which equals about $8.3 \times 10^9\text{ electrons}$ per square meter. Why don't we feel it?
  • Response: Human bodies are moderately conductive and, while standing on Earth, reside at the exact same electrostatic potential as the ground. Because there is no potential difference across our bodies to drive current, and ambient atmospheric ions neutralize localized field variations, we perceive no electric shock or force.

Detailed Mathematical Derivations and Worked Examples

• Example 10.1: Number of Electrons and Net Charge in Liquid Water:

  • Problem: Determine the total positive charge, total negative charge, and total number of electrons contained within $1.0\,g$ of pure water ($H_2O$). The molecular mass of water is $18.0\,g/mol$.
  • Step-by-step Solution:
  • The number of water molecules in one mole ($18.0\,g$) is given by Avogadro's constant, $N_A = 6.02 \times 10^{23}\text{ molecules/mol}$.
  • The number of molecules in $1.0\,g$ of water is:       $N = \frac{6.02 \times 10^{23}}{18.0} = 3.34 \times 10^{22}\text{ molecules}$
  • Each neutral water molecule ($H_2O$) contains two hydrogen atoms ($2 \times 1 = 2\text{ electrons}$) and one oxygen atom ($8\text{ electrons}$), giving a total of:       $Z = 2 + 8 = 10\text{ electrons per molecule}$
  • The total number of electrons (and protons) in $1.0\,g$ of water is:       $n = \left( \frac{6.02 \times 10^{23}}{18.0} \right) \times 10 = 3.34 \times 10^{23}\text{ electrons}$
  • The total positive nuclear charge is:       $Q_+ = n \times e = (3.34 \times 10^{23}) \times (1.6 \times 10^{-19}\,C) = 5.35 \times 10^4\,C$
  • Because water is electrically neutral, this is balanced by an identical negative electronic charge ($Q_- = -5.35 \times 10^4\,C$), yielding a net charge of zero.

• Example 10.2: Electron Count for One Coulomb:

  • Problem: How many electrons must be collected to accumulate a net charge of one Coulomb ($1.0\,C$)?
  • Step-by-step Solution:
  • Elementary charge magnitude: $e = 1.6 \times 10^{-19}\,C$.
  • Using the quantization relation $q = n e$:       $n = \frac{q}{e} = \frac{1.0\,C}{1.6 \times 10^{-19}\,C} = 0.625 \times 10^{19} = 6.25 \times 10^{18}\text{ electrons}$
  • Thus, accumulating one Coulomb of charge requires assembling $6.25 \times 10^{18}\text{ electrons}$.

• Example 10.4: Superposition of Electrostatic Forces on a Right-Angled Triangle:

  • Problem: Three charges ($q_A = 2\,\mu C$, $q_B = 3\,\mu C$, and $q_C = 4\,\mu C$) are positioned at vertices $A$, $B$, and $C$ of a right-angled triangle where $AB = 4.0\,cm$, $BC = 3.0\,cm$, and $\angle B = 90^\circ$. Find the magnitude and direction of the net resultant force on charge $A$.   
  • Step-by-step Solution:
  • Calculate the hypotenuse distance $AC$ using the Pythagorean theorem:       $AC = \sqrt{AB^2 + BC^2} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{16 + 9} = 5.0\,cm = 0.05\,m$
  • Calculate the repulsive force exerted on $A$ by $B$ ($\mathbf{F}_{AB}$):       $F_{AB} = \frac{1}{4\pi\varepsilon_0} \frac{q_A q_B}{r_{AB}^2} = (9 \times 10^9) \frac{(2 \times 10^{-6}) \times (3 \times 10^{-6})}{(4 \times 10^{-2})^2}$$F_{AB} = \frac{9 \times 10^9 \times 6 \times 10^{-12}}{16 \times 10^{-4}} = \frac{54 \times 10^{-3}}{16 \times 10^{-4}} = 3.375 \times 10 = 33.7\,N$       (Directed along line $\vec{BA}$, pointing horizontally westward)
  • Calculate the repulsive force exerted on $A$ by $C$ ($\mathbf{F}_{AC}$):       $F_{AC} = \frac{1}{4\pi\varepsilon_0} \frac{q_A q_C}{r_{AC}^2} = (9 \times 10^9) \frac{(2 \times 10^{-6}) \times (4 \times 10^{-6})}{(5 \times 10^{-2})^2}$$F_{AC} = \frac{9 \times 10^9 \times 8.0 \times 10^{-12}}{25 \times 10^{-4}} = \frac{72}{25} \times 10 = 28.8\,N$       (Directed along line $\vec{CA}$, pointing upward and westward)     
  • The angle $\theta$ between the vector directions of $\mathbf{F}_{AB}$ and $\mathbf{F}_{AC}$ matches the interior angle $\angle BAC$:       $\cos(\theta) = \frac{AB}{AC} = \frac{4.0}{5.0} = 0.8$
  • Using the vector law of cosines for resultant force magnitude $F$:       $F = \sqrt{F_{AC}^2 + F_{AB}^2 + 2 F_{AC} F_{AB} \cos(\theta)}$$F = \left[ (28.8)^2 + (33.7)^2 + 2(28.8)(33.7)(0.8) \right]^{1/2}$$F = \left[ 829.44 + 1135.69 + 1552.896 \right]^{1/2} = \sqrt{3518.026} = 59.3\,N$
  • Direction determination: Using the component angle formula:       $\tan(\alpha) = \frac{F_{AC} \sin(\theta)}{F_{AB} + F_{AC} \cos(\theta)}$       Since $\sin(\theta) = \frac{3}{5} = 0.6$ and $\cos(\theta) = 0.8$:       $\tan(\alpha) = \frac{(28.8)(0.6)}{33.7 + (28.8)(0.8)} = \frac{17.28}{33.7 + 23.04} = \frac{17.28}{56.74} \approx 0.3045$$\alpha = \arctan(0.3045) \approx 16.93^\circ$
  • The net force on charge $A$ is $59.3\,N$ directed $16.9^\circ\text{ north of west}$.     

• Example 10.5: Electric Field in an Automobile Spark Plug Gap:

  • Problem: The gap between the firing electrodes of an automotive spark plug is $d = 1.25\,mm$. Applying a potential difference of $V = 20\,V$ across the gap, calculate the resulting electric field intensity.
  • Step-by-step Solution:
  • Electrode separation: $d = 1.25\,mm = 1.25 \times 10^{-3}\,m$.
  • Potential difference: $V = 20\,V$.
  • Using the uniform field-potential relation:       $E = \frac{V}{d} = \frac{20\,V}{1.25 \times 10^{-3}\,m} = 16 \times 10^3\,V/m = 1.6 \times 10^4\,V/m$
  • This high field gradient is sufficient to ionize fuel-air vapor mixtures within the combustion chamber, initiating ignition.

• Example 10.6: Electric Field at the Midpoint of a Triangle's Hypotenuse:

  • Problem: Three charges are fixed at the corners of an isosceles right triangle: $q_A = +10\,\mu C$ at $A$, $q_B = +2\,\mu C$ at right-angled apex $B$, and $q_C = +10\,\mu C$ at $C$, where $AB = BC = 5\,cm$. Calculate the net electric field vector at point $P$, the midpoint of hypotenuse $AC$.   
  • Step-by-step Solution:
  • Hypotenuse length: $AC = \sqrt{5^2 + 5^2} = 5\sqrt{2}\,cm$.
  • Distance from midpoint $P$ to vertices $A$ and $C$:       $AP = PC = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}\,cm$
  • Because $q_A = q_C = +10\,\mu C$ and point $P$ is equidistant from $A$ and $C$, the fields generated by $A$ ($\mathbf{E}_A$) and $C$ ($\mathbf{E}_C$) are equal in magnitude and point in opposite directions along line $AC$:       $\mathbf{E}_A = -\mathbf{E}_C \implies \mathbf{E}_A + \mathbf{E}_C = 0$
  • The net field at $P$ is determined entirely by the apex charge $q_B = +2\,\mu C$:       $\mathbf{E}_P = \mathbf{E}_B$
  • The distance $BP$ is the median to the hypotenuse of an isosceles right triangle:       $BP = AB \cos(45^\circ) = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}\,cm = \frac{5}{\sqrt{2}} \times 10^{-2}\,m$
  • Calculating the field magnitude:       $E_P = \frac{1}{4\pi\varepsilon_0} \frac{q_B}{(BP)^2} = \frac{(9 \times 10^9) \times (2 \times 10^{-6})}{\left( \frac{5}{\sqrt{2}} \times 10^{-2} \right)^2}$$E_P = \frac{18 \times 10^3}{\frac{25}{2} \times 10^{-4}} = \frac{36 \times 10^7}{25} = 1.44 \times 10^7\,N\,C^{-1}$       (Directed along vector $\vec{BP}$, radially away from apex $B$)

• Example 10.7: Bohr Hydrogen Atom Electric Field and Force:

  • Problem: In a classical Bohr model of the hydrogen atom, an electron revolves around a central proton at a radius of $r = 5.3 \times 10^{-11}\,m$. Given the proton charge is $q = +1.6 \times 10^{-19}\,C$, calculate the electric field intensity produced by the proton at the electron's orbit, and find the resulting electrostatic attractive force.
  • Step-by-step Solution:
  • Electric field due to the proton:       $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} = \frac{(9.0 \times 10^9) \times (1.6 \times 10^{-19})}{(5.3 \times 10^{-11})^2}$$E = \frac{1.44 \times 10^{-9}}{28.09 \times 10^{-22}} = 5.1 \times 10^{11}\,N\,C^{-1}$
  • Force exerted on the orbiting electron ($q_e = -1.6 \times 10^{-19}\,C$):       $F = q_e E = (-1.6 \times 10^{-19}\,C) \times (5.1 \times 10^{11}\,N\,C^{-1}) = -8.16 \times 10^{-8}\,N$
  • Using direct Coulomb calculation:       $F = \frac{(9.0 \times 10^9) \times (-1.6 \times 10^{-19}) \times (+1.6 \times 10^{-19})}{(5.3 \times 10^{-11})^2} = -8.2 \times 10^{-8}\,N$
  • The negative sign indicates an attractive centripetal force directed toward the nuclear proton.

• Example 10.8: Gauss' Law Flux Through Concentric Spheres:

  • Problem: A point charge of $Q = 5.0\,C$ is fixed at the center of an imaginary sphere of radius $r = 1.0\,m$. Find the total electric flux passing through the sphere. How does this flux change if the radius is doubled to $r = 2.0\,m$?
  • Step-by-step Solution:
  • By Gauss' Law, total flux depends solely on enclosed charge:       $\Phi_E = \frac{Q}{\varepsilon_0} = \frac{5.0\,C}{8.85 \times 10^{-12}\,C^2\,N^{-1}\,m^{-2}} = 5.65 \times 10^{11}\,N\,m^2\,C^{-1} = 5.65 \times 10^{11}\,V\,m$
  • If the sphere's radius is doubled ($r \to 2r$), the surface area increases by a factor of four ($A \propto r^2$), while the electric field decreases by a factor of four ($E \propto \frac{1}{r^2}$):       $\Phi = E \times A = \left(\frac{E_0}{4}\right) \times (4 A_0) = E_0 A_0 = \Phi_0$
  • Because the enclosed charge remains $5.0\,C$, the total electric flux is unchanged at $5.65 \times 10^{11}\,V\,m$.

• Example 10.9: Charge Calculation on an Electric Dipole Experiencing Torque:

  • Problem: An electric dipole of length $2l = 2.0\,cm$ is placed with its axis inclined at an angle of $30^\circ$ relative to a uniform electric field of $E = 10^5\,N/C$. If the dipole experiences a torque of $\tau = 10\sqrt{3}\,N\,m$, determine the magnitude of the charges $q$.
  • Step-by-step Solution:
  • Given data: $\tau = 10\sqrt{3}\,N\,m$, $E = 10^5\,N/C$, $2l = 2.0 \times 10^{-2}\,m$, $\theta = 30^\circ$, $\sin(30^\circ) = 0.5$.
  • Using the torque formula:       $\tau = p E \sin(\theta) = q(2l)E \sin(\theta)$$10\sqrt{3} = q \times (2.0 \times 10^{-2}) \times (10^5) \times (0.5)$$10\sqrt{3} = q \times 10^3$$q = \frac{10\sqrt{3}}{10^3} = \sqrt{3} \times 10^{-2}\,C \approx 1.73 \times 10^{-2}\,C$
  • The charge magnitude on the dipole is $1.73 \times 10^{-2}\,C$.