Electric Charges and Fields

Introduction to Electrostatics

• Everyday Phenomena of Static Electricity:

  • Experiencing a spark or hearing a crackle when removing synthetic clothes or sweaters, particularly in dry weather.

  • Atmospheric lightning observed during thunderstorms.

  • Experiencing an electric shock when touching a car door handle or holding the iron bar of a bus after sliding across a seat.

• Physical Cause: Electric charge accumulation resulting from rubbing insulating surfaces together, followed by discharge through the human body.

• Definition of Electrostatics: The branch of physics that studies forces, electric fields, and electric potentials arising from static (stationary) electric charges. Static denotes anything that does not move or change over time.

Electric Charge

• Historical Discovery:

  • Thales of Miletus, Greece (~600 BC) discovered that amber rubbed with wool or silk cloth acquires the property to attract light objects such as straw, bits of paper, and pith balls.

  • The term electricity originates from the Greek word elektron, meaning amber.

• Experimental Observations of Electrification:

  • Two glass rods rubbed with wool or silk cloth repel each other when brought close together.

  • The two pieces of silk or wool used to rub the glass rods also repel each other; however, a glass rod attracts the silk or wool cloth.

  • Two plastic rods rubbed with cat's fur repel each other, but each plastic rod attracts the cat's fur.

  • A plastic rod attracts a glass rod, while repelling the wool or silk used on the glass rod.   

    

• Fundamental Conclusions on Charge:

  • There exist only two kinds of electric charge in nature.

  • Like charges repel each other, and unlike charges attract each other.

  • Polarity of Charge: The fundamental property that distinguishes the two types of electric charges.

  • Neutralization: When two electrified bodies carrying opposite charges are brought into physical contact, they lose their electrification and nullify each other's effects.

• Naming Conventions (Benjamin Franklin):

  • By international convention, the charge acquired by a glass rod or cat's fur is termed positive.

  • The charge acquired by a plastic rod or silk cloth is termed negative.

  • An object possessing net electric charge is termed electrified or charged; an object with no net charge is electrically neutral.

• Detection of Charge (Gold-Leaf Electroscope):

  • Consists of a vertical metal rod housed inside a glass window box, with two thin gold leaves attached to its lower tip.

  • When a charged body touches the metal knob at the top, charge transfers down the rod to both gold leaves.

  • Repulsion between like charges on the leaves causes them to diverge; the degree of divergence indicates the amount of charge.

• Microscopic Origin of Electrification:

  • Matter consists of atoms containing positively charged nuclei and negatively charged electrons.

  • Intermolecular and atomic forces (solid binding, adhesive forces of glue, surface tension) are fundamentally electrical in origin.

  • Electrification occurs via the transfer of loosely bound valence electrons between rubbing bodies.

  • Positive Charging: Occurs when a body loses electrons.

  • Negative Charging: Occurs when a body gains electrons.

  • Rubbing transfers only a minuscule fraction of the total electrons; no new charges are created or destroyed.

Conductors and Insulators

• Conductors:

  • Materials that allow electricity and electric charges (free electrons) to move freely through their interior.

  • Examples: Metals, human and animal bodies, and the Earth.

  • Any charge placed on a conductor rapidly redistributes across its entire outer surface.

• Insulators:

  • Materials that offer high resistance to the flow of electricity and do not allow mobile charge transport.

  • Examples: Non-metals such as glass, porcelain, plastic, nylon, and dry wood.

  • Any charge placed on an insulator remains localized at the exact spot where it was introduced.

• Semiconductors:

  • A distinct third category possessing electrical resistance intermediate between conductors and insulators.

• Earthing / Grounding:

  • A metal object held directly in a human hand cannot retain charge because excess charge leaks through the human body (a conductor) into the Earth.

  • Electrification of a metal rod requires holding it by an insulating handle (e.g., plastic or wooden handle).

Basic Properties of Electric Charge

• Point Charge Approximation:

  • When the linear dimensions of charged bodies are much smaller than the distance separating them, all charge is treated as concentrated at a single spatial point.

• Additivity of Electric Charges:

  • Total charge of a system containing multiple point charges is the algebraic scalar sum of all individual charges.      \ntotal\,q = q_1 + q_2 + q_3 + \dots + q_n\n     

  • Unlike mass (which is strictly non-negative), electric charge can be positive or negative; algebraic signs must be explicitly included in calculations.

  • Example: A system with charges $+1\,\mu\text{C}$, $+2\,\mu\text{C}$, $-3\,\mu\text{C}$, $+4\,\mu\text{C}$, and $-5\,\mu\text{C}$ has a total net charge of $-1\,\mu\text{C}$.

• Conservation of Electric Charge:

  • Within an isolated system, the total net electric charge remains strictly constant over time.

  • Charges can redistribute between interacting bodies, but net charge cannot be created or destroyed.

  • Example (Pair creation/decay): A neutral neutron decays into a proton ($+e$) and an electron ($-e$); total net charge before and after the process is exactly zero.

• Quantization of Electric Charge:

  • All observable free electric charges are integral multiples of a basic elementary unit of charge, denoted by $e$      \n  q = n e \quad \text{where } n = 0, \pm 1, \pm 2, \pm 3, \dots\n     

  • The elementary charge $e$ corresponds to the magnitude of charge on an electron ($-e$) or proton ($+e$).

  • First suggested by Michael Faraday's laws of electrolysis; experimentally confirmed by Robert A. Millikan in 1912.

  • SI Unit of Charge: The Coulomb ($\text{C}$), defined as the charge transported by a current of $1\,\text{A}$ in $1\,\text{s}$ ($1\,\text{C} = 1\,\text{A}\cdot\text{s}$).

  • Elementary charge value:      \n  e = 1.602192 \times 10^{-19}\,\text{C}\n     

  • A charge of $-1\,\text{C}$ contains approximately $6 \times 10^{18}$ electrons.

  • Microscopic units: $1\,\mu\text{C} = 10^{-6}\,\text{C}$, $1\,\text{mC} = 10^{-3}\,\text{C}$.

  • Macroscopic vs. Microscopic Scale:

  • At the macroscopic level ($\sim 1\,\mu\text{C} \approx 10^{13} e$), the granular nature of charge is negligible, and charge behaves as a continuous fluid.

  • At the microscopic level (where charge involves tens or hundreds of $e$), charge quantization must be rigorously accounted for.

Coulomb's Law

• Definition: The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of their charge magnitudes, inversely proportional to the square of the distance separating them, and acts along the line joining their centers.      \n  F = k \frac{|q_1 q_2|}{r^2}\n     

• Historical Background:

  • Established by French physicist Charles Augustin de Coulomb (1736–1806) in 1785 using a sensitive torsion balance.

  • Coulomb established charge ratios ($q \rightarrow q/2 \rightarrow q/4$) by touching charged metallic spheres to identical uncharged spheres.

  • Valid from macroscopic distances down to subatomic separations ($r \sim 10^{-10}\,\text{m}$).

• SI Electrostatic Constants:

  • The proportionality constant $k$ is expressed as:      \n  k = \frac{1}{4 \pi \varepsilon_0} \approx 8.9875 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2 \approx 9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2\n     

  • Permittivity of free space $\varepsilon_0$:      \n  \varepsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}\n     

• Definition of $1\,\text{C}$: The quantity of charge that, when placed at a distance of $1\,\text{m}$ from an equal charge in vacuum, experiences an electrostatic repulsive force of $9 \times 10^9\,\text{N}$.

• Vector Form of Coulomb's Law:

  • Position vector leading from charge 1 ($q_1$) at $\mathbf{r}_1$ to charge 2 ($q_2$) at $\mathbf{r}_2$:      \n  \mathbf{r}_{21} = \mathbf{r}_2 - \mathbf{r}_1\n     

  • Unit vector in direction of $\mathbf{r}_{21}$:      \n  \mathbf{\hat{r}}_{21} = \frac{\mathbf{r}_{21}}{r_{21}}\n     

  • Force $\mathbf{F}_{21}$ exerted on charge $q_2$ by charge $q_1$:      \n  \mathbf{F}_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \mathbf{\hat{r}}_{21}\n     

  • Force $\mathbf{F}_{12}$ exerted on charge $q_1$ by charge $q_2$:      \n  \mathbf{F}_{12} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \mathbf{\hat{r}}_{12} = -\mathbf{F}_{21}\n     

  • Confirming complete agreement with Newton's Third Law of Motion.

Forces Between Multiple Charges (Superposition Principle)

• Principle of Superposition: The net electrostatic force exerted on a given point charge by a system of multiple charges is equal to the vector sum of the individual Coulomb forces exerted on it by each charge acting independently. The interaction between any pair of charges remains completely unaffected by the presence of surrounding charges.

• Mathematical Formulation:

  • For a system of $n$ point charges $q_1, q_2, q_3, \dots, q_n$, the total force $\mathbf{F}_1$ on charge $q_1$ is:      \n  \mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{13} + \dots + \mathbf{F}_{1n}\n        \n  \mathbf{F}_1 = \frac{q_1}{4 \pi \varepsilon_0} \sum_{i=2}^n \frac{q_i}{r_{1i}^2} \mathbf{\hat{r}}_{1i}\n     

  • Individual vector forces are added using the parallelogram law of vector addition.

Electric Field

• Conceptual Definition: A source charge $Q$ alters the surrounding space by generating an electric field $\mathbf{E}(\mathbf{r})$. When a test charge $q$ is introduced at position $\mathbf{r}$, the field exerts an electrostatic force $\mathbf{F}(\mathbf{r})$ on it.

• Mathematical Expression for Point Charge:      \n  \mathbf{E}(\mathbf{r}) = \frac{\mathbf{F}(\mathbf{r})}{q} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \mathbf{\hat{r}}\n     

• Operational Definition:      \n  \mathbf{E}(\mathbf{r}) = \lim_{q \to 0} \left( \frac{\mathbf{F}}{q} \right)\n     

  • Taking the limit $q \to 0$ ensures the test charge is vanishingly small so its presence does not displace the source charge $Q$

• Key Characteristics:

  • $\mathbf{E}(\mathbf{r})$ is independent of the magnitude or sign of the test charge $q$

  • Directed radially outward for positive source charges ($Q > 0$) and radially inward for negative source charges ($Q < 0$).

  • Possesses spherical symmetry around a point charge (magnitude depends strictly on radial distance $r$

  • SI Units: Newton per Coulomb ($\text{N/C}$) or Volt per meter ($\text{V/m}$).

• Electric Field of a System of Point Charges:      \n  \mathbf{E}(\mathbf{r}) = \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_{iP}^2} \mathbf{\hat{r}}_{iP}\n     

• Physical Significance and Electrodynamics:

  • While electrostatic interactions can be computed using Coulomb's law directly, the field concept is indispensable in electrodynamics.

  • Accelerated charges emit electromagnetic waves travelling at finite speed $c$ ($3 \times 10^8\,\text{m/s}$), causing a delayed force response on distant charges.

  • Electric and magnetic fields are physical entities that store and transport energy and momentum.

Electric Field Lines

• Definition: An electric field line is a continuous space curve drawn in an electric field such that the tangent at any point gives the direction of the net electric field vector $\mathbf{E}$ at that point.

• Field Intensity and Line Density:

  • Magnitude of the electric field is proportional to the relative density (closeness) of field lines crossing unit area normal to the lines.

  • Lines crowd tightly where the electric field is strong and spread apart where the field is weak.

• Solid Angle Basis ($1/r^2$ dependence):

  • Solid angle subtended by area element $\Delta S$ at distance $r$ is $\Delta \Omega = \frac{\Delta S}{r^2}$.

  • The number of radial field lines $n$ within a fixed solid angle $\Delta \Omega$ is constant.

  • Field line density equals $\frac{n}{\Delta S} = \frac{n}{r^2 \Delta \Omega} \propto \frac{1}{r^2}$, proving field strength decreases as $\frac{1}{r^2}$.

• Fundamental Properties of Field Lines:

  1. Field lines originate on positive charges and terminate on negative charges (or extend to/from infinity for isolated single charges).

  2. Field lines are continuous curves without any breaks in charge-free space.

  3. Two field lines can never intersect. If they crossed, two tangents could be drawn at the intersection point, implying two different net directions of $\mathbf{E}$ at one point, which is physically impossible.

  4. Electrostatic field lines never form closed loops. This property reflects the conservative nature of electrostatic forces.

Electric Flux

• Definition: Electric flux $\Phi$ quantifies the total number of electric field lines passing through a specified surface area.

• Flux Through Small Area Element:

  • For an area element $\Delta S$ represented by vector $\Delta \mathbf{S} = \Delta S \mathbf{\hat{n}}$ (where $\mathbf{\hat{n}}$ is the unit outward normal vector):      \n  \Delta \Phi = \mathbf{E} \cdot \Delta \mathbf{S} = E \Delta S \cos(\theta)\n     

  • $\theta$ is the angle between $\mathbf{E}$ and outward normal $\mathbf{\hat{n}}$.

  • When $\theta = 0^\circ$, flux is maximum ($\Delta \Phi = E \Delta S$).

  • When $\theta = 90^\circ$, field lines run parallel to the surface, and zero flux crosses it ($\Delta \Phi = 0$).

• Total Flux Over an Arbitrary Surface:      \n  \Phi = \int_S \mathbf{E} \cdot d\mathbf{S} \approx \sum \mathbf{E} \cdot \Delta \mathbf{S}\n     

• Closed Surface Normal Convention: By universal convention, the direction of the area vector $\Delta \mathbf{S}$ for a closed surface points along the outward normal.

• SI Unit of Flux: $\text{N}\cdot\text{m}^2/\text{C}$ or $\text{V}\cdot\text{m}$.

Electric Dipole

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

• Dipole Vector Moment ($\mathbf{p}$):      \n  \mathbf{p} = q (2a) \mathbf{\hat{p}}\n     

  • Directed along the dipole axis strictly from $-q$ to $+q$.

  • Magnitude $p = 2qa$. SI unit: Coulomb-meter ($\text{C}\cdot\text{m}$).

• Electric Field on Axial Line (at distance $r$ from midpoint, $r \gg a$):      \n  \mathbf{E}_{\text{axial}} = \frac{1}{4 \pi \varepsilon_0} \frac{2 \mathbf{p}}{r^3}\n     

  • Directed parallel to dipole moment $\mathbf{p}$.

• Electric Field on Equatorial Line (at distance $r$ on perpendicular bisector, $r \gg a$):      \n  \mathbf{E}_{\text{equatorial}} = -\frac{1}{4 \pi \varepsilon_0} \frac{\mathbf{p}}{r^3}\n     

  • Directed antiparallel to dipole moment $\mathbf{p}$.

  • Ratio of axial to equatorial field magnitude at equal large distance $r$ is exactly $2 : 1$

• Point Dipole: The theoretical limit as separation $2a \to 0$ and charge $q \to \infty$ such that product $p = 2qa$ remains finite. For point dipoles, $\frac{1}{r^3}$ formulas are exact at all distances $r > 0$

• Polar vs. Non-Polar Molecules:

  • Non-Polar Molecules (e.g., $\text{CO}_2, \text{CH}_4$): Centers of positive and negative charges coincide; net dipole moment is zero in the absence of external field.

  • Polar Molecules (e.g., $\text{H}_2\text{O}$): Centers of positive and negative charge do not coincide; possess permanent electric dipole moments.

Dipole in a Uniform External Field

• Net Force in Uniform Field:      \n  \mathbf{F}_{\text{net}} = q\mathbf{E} + (-q\mathbf{E}) = 0\n     

  • A dipole experiences zero net translational force in a uniform electric field.

• Net Torque ($\boldsymbol{\tau}$):

  • Opposite forces acting at different lines of action form a couple resulting in torque:      \n  \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}\n        \n  \tau = p E \sin(\theta)\n     

  • Torque acts to align dipole moment $\mathbf{p}$ parallel to $\mathbf{E}$.

  • Equilibrium is stable at $\theta = 0^\circ$ ($\tau = 0$) and unstable at $\theta = 180^\circ$ ($\tau = 0$).

• Dipole in Non-Uniform External Field:

  • Net force is non-zero.

  • If $\mathbf{p}$ is parallel to $\mathbf{E}$, the dipole experiences a net force pointing toward the region of increasing field strength.

  • If $\mathbf{p}$ is antiparallel to $\mathbf{E}$, net force points toward the region of decreasing field strength.

  • Practical Application: A charged comb attracting neutral pieces of paper. The comb's non-uniform field induces dipoles in paper and exerts a net attractive force pulling the paper toward the comb.

Continuous Charge Distribution

• Macroscopic Smoothing: Replaces microscopic discrete charges with continuous charge density functions averaged over macroscopically small but microscopically large volume/area/line elements.

• Linear Charge Density ($\lambda$):      \n  \lambda = \frac{\Delta Q}{\Delta l} \quad [\text{SI Unit: } \text{C/m}]\n     

• Surface Charge Density ($\sigma$):      \n  \sigma = \frac{\Delta Q}{\Delta S} \quad [\text{SI Unit: } \text{C/m}^2]\n     

• Volume Charge Density ($\rho$):      \n  \rho = \frac{\Delta Q}{\Delta V} \quad [\text{SI Unit: } \text{C/m}^3]\n     

• Total Electric Field from Continuous Volume Distribution:      \n  \mathbf{E}(\mathbf{r}) = \frac{1}{4 \pi \varepsilon_0} \int_V \frac{\rho(\mathbf{r}')}{r'^2} \mathbf{\hat{r}}' dV'\n     

Gauss's Law

• Statement: The total electric flux $\Phi$ passing through any closed surface $S$ is equal to $\frac{1}{\varepsilon_0}$ times the total net enclosed electric charge $q_{\text{enclosed}}$.      \n  \Phi = \oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0}\n     

• Cylindrical Example (Zero Enclosed Charge):

  • Uniform field $\mathbf{E}$ along cylinder axis.

  • Flux through flat face 1: $\Phi_1 = -E S$

  • Flux through flat face 2: $\Phi_2 = +E S$

  • Flux through curved side: $\Phi_3 = 0$

  • Total net flux: $\Phi = -E S + E S + 0 = 0$

• Critical Points for Application:

  1. Valid for closed surfaces of any arbitrary shape or size.

  2. $q_{\text{enclosed}}$ includes the algebraic sum of all charges enclosed inside the surface.

  3. The field $\mathbf{E}$ on the left-hand side is the resultant electric field due to all charges (both inside and outside the Gaussian surface).

  4. Gaussian Surface: The imaginary closed surface chosen to apply Gauss's law. It must not pass directly through discrete point charges (where $\mathbf{E}$ is undefined), but can pass through continuous charge distributions.

  5. Relying directly on the inverse-square dependence ($1/r^2$) of Coulomb's law; any deviation from Gauss's law implies departure from inverse-square distance scaling.

Applications of Gauss's Law

• 1. Field Due to an Infinitely Long Straight Uniformly Charged Wire:

  • Linear charge density $\lambda$

  • Gaussian Surface: Coaxial cylinder of radius $r$ and length $l$

  • Flux through flat circular end caps is zero ($\mathbf{E} \perp d\mathbf{S}$).

  • Flux through curved cylindrical surface equals $E (2 \pi r l)$.

  • Enclosed charge: $q_{\text{enclosed}} = \lambda l$

  • Applying Gauss's Law: $E (2 \pi r l) = \frac{\lambda l}{\varepsilon_0}$      \n  \mathbf{E} = \frac{\lambda}{2 \pi \varepsilon_0 r} \mathbf{\hat{n}}\n     

  • Directed radially outward if $\lambda > 0$ and radially inward if $\lambda < 0$

• 2. Field Due to a Uniformly Charged Infinite Plane Sheet:

  • Surface charge density $\sigma$

  • Gaussian Surface: Cylindrical or rectangular pillbox of cross-sectional area $A$ extending normally through the sheet.

  • Flux through curved sides is zero ($\mathbf{E} \parallel$ surface).

  • Flux through two flat end faces equals $E A + E A = 2 E A$

  • Enclosed charge: $q_{\text{enclosed}} = \sigma A$

  • Applying Gauss's Law: $2 E A = \frac{\sigma A}{\varepsilon_0}$      \n  \mathbf{E} = \frac{\sigma}{2 \varepsilon_0} \mathbf{\hat{n}}\n     

  • Electric field is completely independent of distance $x$ from the sheet.

• 3. Field Due to a Uniformly Charged Thin Spherical Shell:

  • Radius $R$, total charge $q = 4 \pi R^2 \sigma$

  • Case 1: Field Outside the Shell ($r \ge R$):

  • Gaussian sphere of radius $r \ge R$ concentric with shell.

  • Total flux: $E (4 \pi r^2)$

  • Applying Gauss's Law: $E (4 \pi r^2) = \frac{q}{\varepsilon_0}$          \n    \mathbf{E} = \frac{q}{4 \pi \varepsilon_0 r^2} \mathbf{\hat{r}} \quad (r \ge R)\n         

  • The shell acts as if its entire charge $q$ were concentrated at its geometric center O.

  • Case 2: Field Inside the Shell ($r < R$):

  • Gaussian sphere of radius $r < R$ concentric with shell.

  • Enclosed charge $q_{\text{enclosed}} = 0$

  • Applying Gauss's Law: $E (4 \pi r^2) = 0$          \n    \mathbf{E} = 0 \quad (r < R)\n         

  • Electrostatic field inside a uniformly charged thin spherical shell is identically zero everywhere.

Worked Examples and Detailed Solutions

• Example 1.1:

  • Problem: If $10^9$ electrons move out of a body per second, calculate time required to accumulate $1\,\text{C}$ charge on another body.

  • Solution:

  • Charge transferred per second: $10^9 \times 1.6 \times 10^{-19}\,\text{C} = 1.6 \times 10^{-10}\,\text{C/s}$.

  • Time required $t = \frac{1\,\text{C}}{1.6 \times 10^{-10}\,\text{C/s}} = 6.25 \times 10^9\,\text{s}$.

  • In years: $t = \frac{6.25 \times 10^9}{365 \times 24 \times 3600} \approx 198\,\text{years} \approx 200\,\text{years}$.

  • Demonstrates that $1\,\text{C}$ is an immense practical unit of charge. Note: $1\,\text{cm}^3$ of copper contains $\approx 2.5 \times 10^{24}$ electrons.

• Example 1.2:

  • Problem: Calculate amount of positive and negative charge in a cup of water ($250\,\text{g}$).

  • Solution:

  • Molar mass of $\text{H}_2\text{O} = 18\,\text{g}$; Moles in $250\,\text{g} = \frac{250}{18} = 13.89\,\text{moles}$.

  • Number of molecules $N = \left(\frac{250}{18}\right) \times 6.02 \times 10^{23} = 8.36 \times 10^{24}\text{ molecules}$.

  • Each molecule has 2 hydrogen protons + 8 oxygen protons = 10 protons and 10 electrons.

  • Total charge magnitude:          \n    q = 8.36 \times 10^{24} \times 10 \times 1.6 \times 10^{-19}\,\text{C} = 1.34 \times 10^7\,\text{C}\n         

• Example 1.3:

  • Problem: (a) Compare magnitude ratios of electric to gravitational force for (i) electron-proton and (ii) two protons. (b) Calculate accelerations of electron and proton at $1\,\text{\AA}$ ($10^{-10}\,\text{m}$) separation ($m_p = 1.67 \times 10^{-27}\,\text{kg}, m_e = 9.11 \times 10^{-31}\,\text{kg}$).

  • Solution:

  • (a) (i) Electron-proton force ratio:          \n    \frac{F_e}{F_G} = \frac{\frac{e^2}{4 \pi \varepsilon_0 r^2}}{\frac{G m_p m_e}{r^2}} = \frac{e^2}{4 \pi \varepsilon_0 G m_p m_e} = 2.4 \times 10^{39}\n         

  • (a) (ii) Two protons force ratio:          \n    \frac{F_e}{F_G} = \frac{e^2}{4 \pi \varepsilon_0 G m_p^2} = 1.3 \times 10^{36}\n         

  • (b) Force at $1\,\text{\AA}$:          \n    |F| = (8.987 \times 10^9) \times \frac{(1.6 \times 10^{-19})^2}{(10^{-10})^2} = 2.3 \times 10^{-8}\,\text{N}\n         

  • Acceleration of electron:          \n    a_e = \frac{2.3 \times 10^{-8}\,\text{N}}{9.11 \times 10^{-31}\,\text{kg}} = 2.5 \times 10^{22}\,\text{m/s}^2\n         

  • Acceleration of proton:          \n    a_p = \frac{2.3 \times 10^{-8}\,\text{N}}{1.67 \times 10^{-27}\,\text{kg}} = 1.4 \times 10^{19}\,\text{m/s}^2\n         

  • Acceleration due to gravity ($9.8\,\text{m/s}^2$) is completely negligible compared to electrostatic acceleration.

• Example 1.4:

  • Problem: Charged spheres A ($q$) and B ($q'$) separated by $10\,\text{cm}$ experience repulsion $F$. Spheres touched by identical uncharged spheres C and D respectively, then C and D removed. Separation halved to $5.0\,\text{cm}$. Calculate new force $F'$.

  • Solution:

  • Initial force $F = \frac{1}{4 \pi \varepsilon_0} \frac{q q'}{r^2}$.

  • After touching uncharged identical spheres C and D, charge splits symmetrically: $q_A = q/2, q_B = q'/2$

  • New distance $r' = r/2 = 5.0\,\text{cm}$.

  • New force:          \n    F' = \frac{1}{4 \pi \varepsilon_0} \frac{(q/2)(q'/2)}{(r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\frac{q q'}{4}}{\frac{r^2}{4}} = \frac{1}{4 \pi \varepsilon_0} \frac{q q'}{r^2} = F\n         

  • The electrostatic force remains unaltered.

• Example 1.5:

  • Problem: Three equal charges $q_1 = q_2 = q_3 = q$ at vertices of an equilateral triangle of side $l$. Find net force on charge $Q$ at centroid O.

  • Solution:

  • Distance from each vertex to centroid $AO = BO = CO = \frac{l}{\sqrt{3}}$.

  • Force magnitudes $F_1 = F_2 = F_3 = \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{(l/\sqrt{3})^2} = \frac{3 q Q}{4 \pi \varepsilon_0 l^2}$.

  • Force vectors $\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3$ are directed along OA, OB, OC at $120^\circ$ to each other.

  • Resultant of $\mathbf{F}_2$ and $\mathbf{F}_3$ equals $\frac{3 q Q}{4 \pi \varepsilon_0 l^2}$ directed along AO, exactly balancing $\mathbf{F}_1$

  • Total net force on $Q$ is strictly zero ($\mathbf{F}_{\text{net}} = 0$).

• Example 1.6:

  • Problem: Charges $q, q, -q$ at vertices A, B, C of an equilateral triangle side $l$. Find force on each charge.

  • Solution:

  • Base magnitude $F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{l^2}$.

  • Force on A ($q$): $\mathbf{F}_1 = F \mathbf{\hat{r}}_1$ parallel to BC.

  • Force on B ($q$): $\mathbf{F}_2 = F \mathbf{\hat{r}}_2$ parallel to AC.

  • Force on C ($-q$): $\mathbf{F}_3 = \sqrt{3} F \mathbf{\hat{n}}$ bisecting $\angle \text{ACB}$.

  • Sum of forces $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$, consistent with Newton's Third Law.

• Example 1.7:

  • Problem: Electron falls $1.5\,\text{cm}$ in uniform field $E = 2.0 \times 10^4\,\text{N/C}$. Field reversed; proton falls same distance. Find fall times.

  • Solution:

  • Electron fall time:          \n    t_e = \sqrt{\frac{2 h m_e}{e E}} = \sqrt{\frac{2 \times 0.015 \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}} = 2.9 \times 10^{-9}\,\text{s}\n         

  • Proton fall time:          \n    t_p = \sqrt{\frac{2 h m_p}{e E}} = \sqrt{\frac{2 \times 0.015 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}} = 1.3 \times 10^{-7}\,\text{s}\n         

  • Heavier particle takes longer time to fall, contrasting free fall under gravity.

• Example 1.8:

  • Problem: Dipole charges $q_1 = +10^{-8}\,\text{C}, q_2 = -10^{-8}\,\text{C}$ separated by $0.1\,\text{m}$. Calculate electric fields at point A (midpoint), point B ($0.05\,\text{m}$ left of $q_1$), point C ($0.1\,\text{m}$ from both charges).

  • Solution:

  • At Point A (midpoint, $r = 0.05\,\text{m}$ from both):          \n    E_{1A} = E_{2A} = \frac{(9 \times 10^9)(10^{-8})}{(0.05)^2} = 3.6 \times 10^4\,\text{N/C} \implies E_A = E_{1A} + E_{2A} = 7.2 \times 10^4\,\text{N/C} \quad (\text{right})\n         

  • At Point B ($r_1 = 0.05\,\text{m}, r_2 = 0.15\,\text{m}$):          \n    E_{1B} = 3.6 \times 10^4\,\text{N/C} \, (\text{left}), \quad E_{2B} = \frac{(9 \times 10^9)(10^{-8})}{(0.15)^2} = 4 \times 10^3\,\text{N/C} \, (\text{right})\n              \n    E_B = E_{1B} - E_{2B} = 3.2 \times 10^4\,\text{N/C} \quad (\text{left})\n         

  • At Point C ($r = 0.1\,\text{m}$):          \n    E_{1C} = E_{2C} = \frac{(9 \times 10^9)(10^{-8})}{(0.10)^2} = 9 \times 10^3\,\text{N/C}\n              \n    E_C = 2 E_{1C} \cos(60^\circ) = 9 \times 10^3\,\text{N/C} \quad (\text{right})\n         

• Example 1.9:

  • Problem: Charges $\pm 10\,\mu\text{C}$ placed $5.0\,\text{mm}$ apart. Find field at (a) point P on axis $15\,\text{cm}$ from center, (b) point Q on equator $15\,\text{cm}$ from center.

  • Solution:

  • Dipole moment $p = (10^{-5}\,\text{C})(5 \times 10^{-3}\,\text{m}) = 5 \times 10^{-8}\,\text{C}\cdot\text{m}$.

  • (a) Axial field:          \n    E_P = \frac{1}{4 \pi \varepsilon_0} \frac{2 p}{r^3} = \frac{(9 \times 10^9)(2 \times 5 \times 10^{-8})}{(0.15)^3} = 2.6 \times 10^5\,\text{N/C} \quad (\text{along } \mathbf{p})\n         

  • (b) Equatorial field:          \n    E_Q = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} = \frac{(9 \times 10^9)(5 \times 10^{-8})}{(0.15)^3} = 1.33 \times 10^5\,\text{N/C} \quad (\text{opposite } \mathbf{p})\n         

• Example 1.10:

  • Problem: Electric field $E_x = \alpha x^{1/2}$, $E_y = E_z = 0$ where $\alpha = 800\,\text{N/C}\cdot\text{m}^{1/2}$. Cube edge $a = 0.1\,\text{m}$ placed with left face at $x = a$, right face at $x = 2a$. Calculate (a) flux through cube, (b) charge inside.

  • Solution:

  • Field at left face $E_L = \alpha a^{1/2}$; Flux $\Phi_L = -E_L a^2 = -\alpha a^{5/2}$.

  • Field at right face $E_R = \alpha (2a)^{1/2}$; Flux $\Phi_R = E_R a^2 = \alpha \sqrt{2} a^{5/2}$.

  • Net flux $\Phi = \Phi_R + \Phi_L = \alpha a^{5/2}(\sqrt{2} - 1)$.          \n    \Phi = 800 (0.1)^{5/2} (1.414 - 1) = 1.05\,\text{N}\cdot\text{m}^2/\text{C}\n         

  • Charge inside $q = \varepsilon_0 \Phi = (8.854 \times 10^{-12})(1.05) = 9.27 \times 10^{-12}\,\text{C}$.

• Example 1.11:

  • Problem: Field $E = +200 \mathbf{\hat{i}}\,\text{N/C}$ for $x > 0$ and $-200 \mathbf{\hat{i}}\,\text{N/C}$ for $x < 0$. Cylinder length $20\,\text{cm}$, radius $5\,\text{cm}$ centered at origin along x-axis. Find (a) flux through flat faces, (b) flux through side, (c) net outward flux, (d) net charge.

  • Solution:

  • (a) Left face ($x = -10\,\text{cm}$): $\mathbf{E} = -200 \mathbf{\hat{i}}$, $\Delta \mathbf{S} = -\Delta S \mathbf{\hat{i}}$.          \n    \Phi_L = (+200) \pi (0.05)^2 = +1.57\,\text{N}\cdot\text{m}^2/\text{C}\n              Right face ($x = +10\,\text{cm}$): $\mathbf{E} = +200 \mathbf{\hat{i}}$, $Delta \mathbf{S} = +\Delta S \mathbf{\hat{i}}.\n    \n         \Phi_R = (+200) \pi (0.05)^2 = +1.57\,\text{N}\cdot\text{m}^2/\text{C}     \n    \n - (b) Side face: \mathbf{E} \perp d\mathbf{S} \implies \Phi_{\text{side}} = 0\n - (c) Net flux \Phi = 1.57 + 1.57 + 0 = 3.14\,\text{N}\cdot\text{m}^2/\text{C}.\n - (d) Net charge q = \varepsilon_0 \Phi = (8.854 \times 10^{-12})(3.14) = 2.78 \times 10^{-11}\,\text{C}.\n- **Example 1.12:**\n - *Problem:* Model atom: point nucleus charge +Ze$surrounded by uniform negative charge sphere radius$R$. Find electric field$E(r)$for (i)$r < R$, (ii)$r > R\n - *Solution:*\n - Charge density \rho = -\frac{Ze}{\frac{4}{3} \pi R^3} = -\frac{3 Ze}{4 \pi R^3}.\n - (i) For r < R$: Enclosed charge$q_{\text{enclosed}} = Ze + \rho \left(\frac{4}{3} \pi r^3\right) = Ze \left(1 - \frac{r^3}{R^3}\right).\n - Applying Gauss's Law: E (4 \pi r^2) = \frac{Ze}{\varepsilon_0} \left(1 - \frac{r^3}{R^3}\right)\n    \n         E(r) = \frac{Ze}{4 \pi \varepsilon_0} \left( \frac{1}{r^2} - \frac{r}{R^3} \right) \quad (r < R)     \n    \n - (ii) For r > R$: Net enclosed charge$q_{\text{enclosed}} = Ze - Ze = 0.\n    \n         E(r) = 0 \quad (r > R)     \n    \n\n# Summary Table of Physical Quantities\n\n| Physical Quantity | Symbol | Dimensions | SI Unit | Mathematical Definition / Remarks |\n| :--- | :--- | :--- | :--- | :--- |\n| Vector Area Element | \Delta \mathbf{S}$|$[\text{L}^2]$|$\text{m}^2$|$\Delta \mathbf{S} = \Delta S \mathbf{\hat{n}} |\n| Electric Field | \mathbf{E}$|$[\text{M L T}^{-3} \text{A}^{-1}]$|$\text{N/C}$or$\text{V/m}$|$\mathbf{E} = \lim_{q \to 0} (\mathbf{F}/q) |\n| Electric Flux | \Phi$|$[\text{M L}^3 \text{T}^{-3} \text{A}^{-1}]$|$\text{N}\cdot\text{m}^2/\text{C}$or$\text{V}\cdot\text{m}$|$\Phi = \int \mathbf{E} \cdot d\mathbf{S} |\n| Dipole Moment | \mathbf{p}$|$[\text{L T A}]$|$\text{C}\cdot\text{m}$|$\mathbf{p} = q (2a) \mathbf{\hat{p}}$(from$-q$to$+q) |\n| Linear Charge Density | \lambda$|$[\text{L}^{-1} \text{T A}]$|$\text{C/m}$|$\lambda = \Delta Q / \Delta l |\n| Surface Charge Density | \sigma$|$[\text{L}^{-2} \text{T A}]$|$\text{C/m}^2$|$\sigma = \Delta Q / \Delta S |\n| Volume Charge Density | \rho$|$[\text{L}^{-3} \text{T A}]$|$\text{C/m}^3$|$\rho = \Delta Q / \Delta V$$ |