Comprehensive Electrostatics and Capacitance Study Guide

Electrostatic Charge, Quantization, and Fundamental Principles

• Electric Charge ($q$):

  • Quantization of Charge: Electric charge is quantized, meaning charge exists in discrete integral multiples of the elementary charge $e$ ($q = n e$). A charge that is a fractional value of the charge on an electron is not possible.
  • Elementary Charge ($e$): The charge of an electron is $e = 1.6 \times 10^{-19}\,\text{C}$.
  • Quantum of Electric Charge in CGS/e.s.u.: The quantum of electric charge in electrostatic units (e.s.u.) is $4.78 \times 10^{-10}\,\text{e.s.u.}$
  • Conversion Between SI and CGS Units: $1\,\text{coulomb} = 3 \times 10^9\,\text{e.s.u.}$
  • Number of Electrons in Charge Quantities:
  • Number of electrons taken out or present in $1\,\text{coulomb}$ of charge:       $n = \frac{1\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.25 \times 10^{18}$
  • Number of electrons present in $8\,\text{coulomb}$ of negative charge:       $n = \frac{8\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 5 \times 10^{19}$
  • Dimensional Formula of Charge: $[Q] = \mathbf{[AT]}$ (where $\text{A}$ represents current in amperes and $\text{T}$ represents time).
  • Current Equivalence Record: $1\,\text{amp}$ is listed in comparative reference as $1.6 \times 10^{-27}\,\text{kg}$.
  • Historical Discovery: The fact of animal electricity was established by Count Alessandro Volta.
  • Conservation: Charge cannot be destroyed or created; there exists a minimum permissible charge on a particle.

• Mass Changes During Charging:

  • When a body is charged, its mass may increase or decrease.
  • A body becomes negatively charged by gaining electrons, which causes its mass to increase.
  • A body becomes positively charged by losing electrons, which causes its mass to decrease.
  • Example: If metallic sphere A is given a positive charge and another identical metallic sphere B of equal initial mass is given an equal negative charge, the mass of sphere B increases while the mass of sphere A decreases.

• Coulomb's Law:

  • Definition: Governs the electrostatic force $F$ acting between two point charges $q_1$ and $q_2$ separated by distance $r$ in vacuum/air:     $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
  • Nature of Force: Coulombian force is both a central force and an electric force.
  • Vector Form of Coulomb's Law:     $\mathbf{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\mathbf{r}|^3} \mathbf{r}$
  • Distance and Charge Scaling Effects:
  • If the distance $r$ between two charges is made half ($r \rightarrow \frac{r}{2}$) and one of the charges is halved ($q_1 \rightarrow \frac{q_1}{2}$), the force becomes:       $F' = \frac{1}{4\pi\epsilon_0} \frac{\left(\frac{q_1}{2}\right) q_2}{\left(\frac{r}{2}\right)^2} = \frac{\frac{q_1 q_2}{2}}{\frac{r^2}{4}} = 2 \left(\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\right) = 2F$       The force becomes double compared to its previous value.
  • If the distance between two charges is made half ($r \rightarrow \frac{r}{2}$) without changing charges, the force increases to $4F$.
  • Effect of Dielectric Medium: Inserting a dielectric material between two positive charges in air reduces the repulsive force ($F_{\text{medium}} = \frac{F_{\text{air}}}{\epsilon_r}$), so the force decreases.

• Four-Charge Square System Equilibrium:

  • For a system with four point charges $q, Q, q, Q$ placed at the four corners of a square, if the net electrostatic force acting on charge $Q$ is zero, the ratio of $Q$ to $q$ is:     $\frac{Q}{q} = 2\sqrt{2}$

Electric Field Intensity, Force, and Field Lines

• Electric Field Intensity (\\mathbf{E}\):

  • Definition: Electric field intensity at a point is defined as the force per unit charge acting on a test charge:     $\mathbf{E} = \frac{\mathbf{F}}{q}$
  • Force on Charge in Field: The force acting on a charge $q$ in an electric field $\mathbf{E}$ is:     $F = qE$
  • Units of Electric Field:
  • $\text{N/C}$ (Newton/Coulomb)
  • $\text{V/m}$ or $\text{volt}\cdot\text{metre}^{-1}$ (Volt/metre)
  • Dimensional Formula:     $[E] = \frac{[F]}{[q]} = \frac{\mathbf{[MLT^{-2}]}}{\mathbf{[AT]}} = \mathbf{[MLT^{-3}A^{-1}]}$
  • Vector Quantity: Electric intensity is a vector physical quantity.

• Electric Field Distributions and Conductors:

  • Surface of a Charged Conductor: The electric field intensity near the outer surface of a charged conductor is:     $E = \frac{\sigma}{\epsilon_0}$     The field is always perpendicular to the surface.
  • Inside a Conductor: If a conductor is placed in an external electric field, the field inside the conductor is zero.
  • Charged Metallic Sphere: On moving from the surface of a charged metallic sphere toward its center, the electric field is zero at all places.
  • Cavity inside Metal (Electrostatic Shielding): The electric field inside a cavity within a metal is always zero.
  • Infinite Thin Sheet: The electric field at a point near an infinite thin sheet of a charged conductor is:     $E = \frac{\sigma}{2\epsilon_0}$
  • Charge Distribution: Charge given to a conductor resides on its outer surface, whereas charge given to a sphere of a bad conductor (insulator) is distributed inside the surface (throughout its volume).

• Electric Field Lines:

  • Electric field lines provide comprehensive information about:
  1. Field strength (density of field lines).
  2. Direction of the electric field at any point.
  3. Nature of charge (originating from positive charge, terminating on negative charge).

Electrostatic Potential and Potential Energy

• Electric Potential (\V\):

  • Relation between Electric Field and Potential: The electric field $\mathbf{E}$ is the negative gradient of electric potential $V$:     $E = -\frac{dV}{dr}$
  • Potential due to Point Charge: The electric potential at distance $r$ from a charge $q$ is:     $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$     Electric potential is inversely proportional to distance $r$ ($V \propto \frac{1}{r}$).
  • Constant Potential Region: In a region of constant electric potential, the electric field is zero ($E = 0$) and there can be no charge inside the region.
  • Equipotential Surfaces: If a uniform electric field exists along the X-axis, the equipotential surfaces are parallel to the YZ-plane.

• Electrostatic Potential Energy (\U\):

  • Two Point Charges: Potential energy of two point charges $q_1$ and $q_2$ separated by distance $r$:     $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$
  • Calculation Example: Two equal positive point charges of $1\,\mu\text{C}$ ($1 \times 10^{-6}\,\text{C}$) kept at a distance of $1\,\text{meter}$ in air have a potential energy of:     $U = \frac{(9 \times 10^9) \times (1 \times 10^{-6}) \times (1 \times 10^{-6})}{1} = 9 \times 10^{-3}\,\text{Joule}$
  • Charges of Opposite Sign: For two point charges of opposite sign ($+q_1$ and $-q_2$), the potential energy $U = \frac{1}{4\pi\epsilon_0} \frac{-q_1 q_2}{r}$ is always negative.

Electric Dipoles: Properties, Potential, Field, and Torque

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

  • Definition: The vector quantity defined as the product of charge magnitude $q$ and separation distance $2\mathbf{l}$:     $\mathbf{p} = q \times 2\mathbf{l}$
  • SI Unit: Coulomb-metre ($\text{C}\cdot\text{m}$).
  • Calculation Example 1: Two equal and opposite charges of $0.2\,\text{C}$ separated by $3.0\,\text{cm}$ ($0.03\,\text{m}$) yield an electric dipole moment of $0.6 \times 10^{-8}\,\text{coulomb-meter}$.
  • Calculation Example 2: Two opposite charges of magnitude $+3.2 \times 10^{-19}\,\text{C}$ and $-3.2 \times 10^{-19}\,\text{C}$ separated by $2.4 \times 10^{-10}\,\text{m}$:     $p = (3.2 \times 10^{-19}\,\text{C}) \times (2.4 \times 10^{-10}\,\text{m}) = 7.68 \times 10^{-29}\,\text{c-m}$
  • Calculation Example 3: Two equal and opposite charges of $5\,\text{coulomb}$ separated by $5.0\,\text{cm}$ ($0.05\,\text{m}$):     $p = 5\,\text{C} \times 0.05\,\text{m} = 0.25\,\text{C}\cdot\text{m} = 25 \times 10^{-2}\,\text{coulomb-metre}$

• Electric Dipole Potential and Field:

  • Broad-Side On (Equatorial) Position: The electric potential due to an electric dipole in broad-side on position is zero.
  • Axial Position at Large Distance: At a large distance $r$ from the center of a small electric dipole, electric potential is proportional to $\frac{1}{r^2}$.

• Dipole in Uniform Electric Field:

  • Resultant Force: The net force on an electric dipole in a uniform electric field is always zero.
  • Torque: The dipole experiences a torque $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$.
  • Work Done in Rotation: Work done in rotating an electric dipole through angle $\theta$ from field direction:     $W = pE (1 - \cos\theta)$
  • Potential Energy: Taking potential energy as zero when perpendicular ($\theta = 90^\circ$) to field $\mathbf{E}$, the potential energy at angle $\theta$ is:     $U = -\mathbf{p} \cdot \mathbf{E} = -pE \cos\theta$

• Dipole-Dipole Electrostatic Energy:

  • The electrostatic energy of a system made of two electric dipoles kept at distance $r$ is proportional to $\frac{1}{r^3}$ ($r^{-3}$).

Gauss's Law, Electric Flux, and Enclosed Charges

• Gauss's Law:

  • Total electric flux $\Phi$ passing through any closed surface is proportional to the total charge enclosed $q_{\text{enclosed}}$:     $\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$
  • Flux from Unit Charge: Total electric flux coming out from a stationary unit positive charge ($q = 1\,\text{C}$) in air:     $\Phi = \frac{1}{\epsilon_0} = \epsilon_0^{-1}$

• Flux Independence of Closed Surface Shape:

  • Total electric flux coming out from a whole surface enclosing charge $+q$ depends only on enclosed charge and is identical regardless of the surface geometry ($S_1, S_2, S_3, S_4$).      

• Enclosed Dipoles:

  • If $n$ electric dipoles are situated inside a closed surface, the net charge enclosed is zero ($q_{\text{net}} = 0$). Thus, total electric flux coming out of the closed surface is zero.

• Units of Electric Flux:

  • SI Unit: $\text{volt-metre}$ (or $\text{V}\cdot\text{m}$).

Capacitance, Dielectrics, and Capacitor Networks

• Capacitance (\C\):

  • Isolated Conductor Ratio: For an isolated conductor, the ratio of total charge to potential is constant:     $C = \frac{\text{Total charge}}{\text{Potential}} = \frac{Q}{V}$
  • Units:
  • SI Unit: Farad ($\text{F}$) or $\text{coulomb/volt}$.
  • Picofarad ($\text{pF}$): Unit of electric capacity ($1\,\text{pF} = 10^{-12}\,\text{F}$).
  • Dependencies: Capacity depends on shape of plates, size of plates, distance between plates, and dielectric medium. Capacity does not depend on charges on plates.
  • Capacity of Spherical Conductor:     $C = 4\pi\epsilon_0 R$
  • Calculation Example: For a spherical conductor of capacity $1.0\,\mu\text{F}$ ($10^{-6}\,\text{F}$):       $R = \frac{C}{4\pi\epsilon_0} = (10^{-6}\,\text{F}) \times (9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2) = 9000\,\text{m} = 9\,\text{km}$
  • For a metallic sphere of radius $R$ charged to potential $V$, charge $q = C V = (4\pi\epsilon_0 R) V$, making $q$ proportional to both $V$ and $R$.

• Capacitor Networks and Combinations:

  • Parallel Combination:
  • Potential difference remains identical across each condenser.
  • Equivalent capacity of two capacitors each of capacity $C$ in parallel: $C_{\text{eq}} = C + C = 2C$.
  • Charge on capacitor $1\,\mu\text{F}$ under potential difference $1\,\text{volt}$: $Q = CV = 1\,\mu\text{F} \times 1\,\text{V} = 1\,\mu\text{C}$.
  • Sum of charges on collecting plate ($+Q$) and condensing plate ($-Q$) of a charged capacitor is zero.
  • Series Combination:
  • Charge remains identical across series condensers while potential divides inversely with capacity.
  • Series Calculation Example: Two capacitors $C_1 = 2\,\mu\text{F}$ and $C_2 = 4\,\mu\text{F}$ connected in series across $1200\,\text{V}$:       $C_{\text{eq}} = \frac{2 \times 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}\,\mu\text{F}$$Q = C_{\text{eq}} V = \left(\frac{4}{3}\,\mu\text{F}\right) \times 1200\,\text{V} = 1600\,\mu\text{C}$       Potential difference across $2\,\mu\text{F}$ capacitor:       $V_1 = \frac{Q}{C_1} = \frac{1600\,\mu\text{C}}{2\,\mu\text{F}} = 800\,\text{V}$
  • Minimum Capacitors Design Problem: Minimum number of $2\,\mu\text{F}$ capacitors needed to obtain $5\,\mu\text{F}$ is 4 (configured as two parallel capacitors of $2\,\mu\text{F}$ giving $4\,\mu\text{F}$, plus two $2\,\mu\text{F}$ capacitors in series giving $1\,\mu\text{F}$, connected in parallel: $4\,\mu\text{F} + 1\,\mu\text{F} = 5\,\mu\text{F}$).
  • Composite High-Voltage Array Problem: Minimum number of capacitors each rated $8\,\mu\text{F}$ and $250\,\text{V}$ required to construct a composite capacitor of $16\,\mu\text{F}$ rated $1000\,\text{V}$:
  • Capacitors in series per row $= \frac{1000\,\text{V}}{250\,\text{V}} = 4$
  • Capacitance per row $= \frac{8\,\mu\text{F}}{4} = 2\,\mu\text{F}$
  • Number of parallel rows $= \frac{16\,\mu\text{F}}{2\,\mu\text{F}} = 8$
  • Total capacitors $= 4 \times 8 = 32$
  • Resistor-Capacitor Circuit Problem: In a DC circuit with a $12\,\text{V}$ source across series resistors $9\,\Omega$ and $3\,\Omega$, with a $6\,\mu\text{F}$ capacitor connected across the $3\,\Omega$ resistor:
  • Circuit current $I = \frac{12\,\text{V}}{9\,\Omega + 3\,\Omega} = 1\,\text{A}$
  • Potential drop across $3\,\Omega$ resistor $V = 1\,\text{A} \times 3\,\Omega = 3\,\text{V}$
  • Charge on $6\,\mu\text{F}$ capacitor $Q = CV = 6\,\mu\text{F} \times 3\,\text{V} = 18\,\mu\text{C}$

• Energy Stored and Dissipated in Capacitors:

  • Stored energy $U = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2C}$.
  • Calculation Example 1: A capacitor of $100\,\mu\text{F}$ charged to $100\,\text{volt}$ stores energy:     $U = \frac{1}{2} (100 \times 10^{-6}\,\text{F}) (100\,\text{V})^2 = 0.5\,\text{joule}$
  • Calculation Example 2: Energy of a charged metallic sphere of radius $18\,\text{cm}$ ($0.18\,\text{m}$) given a charge of $5 \times 10^{-6}\,\text{C}$:     $U = \frac{q^2}{8\pi\epsilon_0 R} = \frac{(5 \times 10^{-6})^2 \times (9 \times 10^9)}{2 \times 0.18} = 0.625\,\text{J} \approx 0.6\,\text{Joule}$
  • Energy Ratio: Energy stored in a capacitor and energy dissipated during charging bear a ratio of 1 : 1.

Dielectric Constants, Permittivity, Electromagnetic Energy, and Applications

• Permittivity of Free Space (\\epsilon_0\):

  • Value: $\epsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}$ (or $8.85 \times 10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$).
  • SI Units: $\text{F}\cdot\text{m}^{-1}$, $\text{Coulomb}^2 / \text{Newton-metre}^2$, or $\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.
  • Dimensional Representation:     $[\epsilon_0] = \mathbf{[M^{-1}L^{-3}T^4A^2]}$

• Dielectric Constant (Relative Permittivity \\epsilon_r\):

  • Value of $\epsilon_r$ in air is 1.
  • Dielectric constant for a metal (perfect conductor) is infinity ($\infty$).

• Electromagnetic Field Energy Densities:

  • Dimensional representation of electrostatic energy density $\frac{1}{2}\epsilon_0 E^2$ is identical to magnetic energy density $\frac{B^2}{2\mu_0}$.

• Charge Densities:

  • Linear Charge Density (\\lambda\): $\lambda = \frac{q}{L}$, unit is Coulomb/metre ($\text{C/m}$).
  • Surface Charge Density (\\sigma\): $\sigma = \frac{q}{A}$, unit is Coulomb/metre$^2$ ($\text{C/m}^2$).

• Van de Graaff Generator:

  • An electrostatic machine designed to build up high voltages of a few million volts (produces high voltage at low current).

• Practical Phenomenon - Bird on High-Power Line:

  • A parrot sitting on a bare high-power line will not be affected practically.
  • Explanation: Both feet of the bird rest on the same conducting line at the exact same potential, resulting in zero potential difference across its body and no current flow.