Comprehensive Study Notes on Electrostatic Potential and Capacitance

INTRODUCTION AND CONSERVATIVE FORCES

  • The concept of potential energy, introduced in Classes XI (Chapters 5 and 7), applies to forces where work done against the force is stored as potential energy. Examples include spring force and gravitational force.

  • When an external force is removed from a body, it moves, gaining kinetic energy (KK) and losing an equal amount of potential energy (UU). The sum of kinetic and potential energies is conserved.

  • Forces that exhibit this property are known as conservative forces.

  • The Coulomb force between two stationary charges is a conservative force. This is expected because, like the gravitational force, it has an inverse-square dependence on distance (1/r21/r^2) and differs primarily in the proportionality constants (masses are replaced by charges).

  • Just as a mass has potential energy in a gravitational field, a charge has electrostatic potential energy in an electrostatic field.

ELECTROSTATIC POTENTIAL ENERGY

  • Consider an electrostatic field E\mathbf{E} produced by a charge QQ at the origin. To move a test charge qq from point RR to point PP against the repulsive force (assuming both Q, q > 0), an external force Fext\mathbf{F}_{ext} must be applied.

  • Assumptions for the test charge:

    • The test charge qq is sufficiently small so it does not disturb the original configuration of charge QQ.

    • The external force Fext\mathbf{F}_{ext} is exactly equal and opposite to the electric force FE\mathbf{F}_E (i.e., Fext=FE\mathbf{F}_{ext} = -\mathbf{F}_E).

    • The charge is moved with infinitesimally slow constant speed, meaning there is no net force or acceleration.

  • Work and Energy Relationship:

    • The work done by the external force is stored as potential energy (UU).

    • Work done by external forces moving a charge qq from RR to PP: WRP=RPFextdr=RPFEdrW_{RP} = \int_R^P \mathbf{F}_{ext} \cdot d\mathbf{r} = -\int_R^P \mathbf{F}_E \cdot d\mathbf{r}.

    • This work increases the potential energy by an amount equal to the potential energy difference between points RR and PP: ΔU=UPUR=WRP\Delta U = U_P - U_R = W_{RP}.

  • Key Characteristics of Electrostatic Potential Energy:

    • The work done depends only on the initial and final positions (RR and PP) and is independent of the path taken. This is the fundamental characteristic of a conservative force.

    • The actual value of potential energy is not physically significant; only the difference ΔU\Delta U matters. An arbitrary constant α\alpha can be added to the potential energy at every point without changing the difference ((UP+α)(UR+α)=UPUR(U_P + \alpha) - (U_R + \alpha) = U_P - U_R).

    • By convention, potential energy is chosen to be zero at infinity (U=0U_\infty = 0).

    • The potential energy of charge qq at point PP is the work done by an external force in bringing the charge qq from infinity to that point: WP=UPU=UPW_{\infty P} = U_P - U_\infty = U_P.

ELECTROSTATIC POTENTIAL (V)

  • Electrostatic potential VV is the work done per unit test charge by an external force to bring a unit positive charge from infinity to a specific point.

  • It is a characteristic of the electric field associated with a charge configuration and is independent of the test charge qq.

  • Mathematical Definition:

    • Potential difference between points PP and RR: VPVR=UPURq=WRPqV_P - V_R = \frac{U_P - U_R}{q} = \frac{W_{RP}}{q}.

    • Potential at point PP (with V=0V_\infty = 0): VP=WPqV_P = \frac{W_{\infty P}}{q}.

  • Important Considerations:

    • Only the potential difference is physically significant.

    • To obtain the potential, it is ideal to use an infinitesimal test charge δq\delta q, calculate the work δW\delta W, and find the ratio δW/δq\delta W / \delta q.

  • Historical Note: The unit of potential is the Volt (V), named after Count Alessandro Volta (1745–1827). Volta was an Italian physicist who established that "animal electricity" (observed by Luigi Galvani) was actually generated by moisture between dissimilar metals, leading to the invention of the voltaic pile (battery).

POTENTIAL DUE TO A POINT CHARGE

  • Consider a charge QQ at the origin. We calculate the work done to bring a unit positive test charge from infinity to point PP at distance rr.

  • At an intermediate point PP' at distance rr', the force on a unit positive charge is: F=Q4πϵ0r2r^\mathbf{F} = \frac{Q}{4\pi\epsilon_0 r'^2} \mathbf{\hat{r}'}.

  • The work done against this force for a small displacement Δr\Delta r' is: ΔW=Q4πϵ0r2Δr\Delta W = -\frac{Q}{4\pi\epsilon_0 r'^2} \Delta r' (negative sign because the displacement is opposite to the force direction).

  • Total work WW (Potential VV) is the integral from infinity to rr:

    • W=rQ4πϵ0r2dr=[Q4πϵ0r]r=Q4πϵ0rW = \int_\infty^r -\frac{Q}{4\pi\epsilon_0 r'^2} dr' = \left[ \frac{Q}{4\pi\epsilon_0 r'} \right]_\infty^r = \frac{Q}{4\pi\epsilon_0 r}.

  • General Expression: V(r)=14πϵ0QrV(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}.

  • Behavior:

    • If Q > 0, then V > 0; if Q < 0, then V < 0.

    • The potential varies inversely with distance (V1/rV \propto 1/r), whereas the electric field varies with the inverse square (E1/r2E \propto 1/r^2).

POTENTIAL DUE TO AN ELECTRIC DIPOLE

  • An electric dipole consists of charges qq and q-q separated by a distance 2a2a. The dipole moment is p\mathbf{p} where p=q×2a|\mathbf{p}| = q \times 2a, pointing from q-q to qq.

  • The potential at point PP is the sum of potentials from both charges (Superposition Principle):

    • V=14πϵ0(qr1qr2)V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right), where r1r_1 is distance from qq and r2r_2 from q-q.

  • By geometry:

    • r12=r2+a22arcos(θ)r_1^2 = r^2 + a^2 - 2ar \cos(\theta)

    • r22=r2+a2+2arcos(θ)r_2^2 = r^2 + a^2 + 2ar \cos(\theta)

  • For large distances (r >> a), using binomial expansion:

    • 1r11r(1+arcos(θ))\frac{1}{r_1} \approx \frac{1}{r} \left( 1 + \frac{a}{r} \cos(\theta) \right)

    • 1r21r(1arcos(θ))\frac{1}{r_2} \approx \frac{1}{r} \left( 1 - \frac{a}{r} \cos(\theta) \right)

  • Substituting these into the potential formula: V=q4πϵ02acos(θ)r2=14πϵ0pcos(θ)r2V = \frac{q}{4\pi\epsilon_0} \frac{2a \cos(\theta)}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{p \cos(\theta)}{r^2}.

  • Vector Form: V=14πϵ0pr^r2V = \frac{1}{4\pi\epsilon_0} \frac{\mathbf{p} \cdot \mathbf{\hat{r}}}{r^2}.

  • Contrasting Features:

    • Dipole potential depends on both distance (rr) and direction (θ\theta).

    • It falls off as 1/r21/r^2, unlike a point charge potential (1/r1/r).

    • Potential on the dipole axis (θ=0,π\theta = 0, \pi) is ±14πϵ0pr2\pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}.

    • Potential on the equatorial plane (θ=π/2\theta = \pi/2) is zero.

POTENTIAL DUE TO A SYSTEM OF CHARGES

  • For charges q1,q2,,qnq_1, q_2, \dots, q_n at distances r1P,r2P,,rnPr_{1P}, r_{2P}, \dots, r_{nP} from point PP, the total potential is: V=V1+V2++Vn=14πϵ0i=1nqiriPV = V_1 + V_2 + \dots + V_n = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^n \frac{q_i}{r_{iP}}.

  • Continuous Charge Distributions: For a distribution with density ρ(r)\rho(\mathbf{r}), the potential is found by integrating over all volume elements dvdv: V=14πϵ0ρ(r)dvrV = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\mathbf{r}) dv}{r}.

  • Uniformly Charged Spherical Shell (Radius RR, Charge qq):

    • Outside the shell (rRr \geq R): Potential is the same as if the charge were at the center: V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}.

    • Inside the shell (r < R): The electric field is zero, so no work is done moving a charge inside. The potential is constant and equal to the value at the surface: V=14πϵ0qRV = \frac{1}{4\pi\epsilon_0} \frac{q}{R}.

EQUIPOTENTIAL SURFACES

  • An equipotential surface is a surface where the potential has a constant value at all points.

  • Properties:

    • For a point charge, equipotential surfaces are concentric spheres centered on the charge.

    • For a uniform electric field, equipotential surfaces are planes perpendicular to the field lines.

    • The electric field is always normal to the equipotential surface. If it weren't, there would be a tangential component of field that requires work to move a charge, contradicting the definition of the surface.

    • Potential difference between any two points on an equipotential surface is zero (ΔV=0\Delta V = 0), hence work done moving a charge on the surface is zero (W=0W = 0).

RELATION BETWEEN FIELD AND POTENTIAL

  • Consider two surfaces AA and BB with potentials VV and V+δVV + \delta V. Let δl\delta l be the perpendicular distance between them.

  • Work done moving a unit charge against the field is Eδl=V(V+δV)=δV|\mathbf{E}| \delta l = V - (V + \delta V) = -\delta V.

  • Relationships:

    • E=δVδl|\mathbf{E}| = -\frac{\delta V}{\delta l}

    • The electric field is in the direction in which the potential decreases steepest.

    • Its magnitude equals the change in potential per unit displacement normal to the equipotential surface.

POTENTIAL ENERGY OF A SYSTEM OF CHARGES (NO EXTERNAL FIELD)

  • Two-Charge System (q1,q2q_1, q_2):

    • Work to bring q1q_1 from infinity to r1=0\mathbf{r}_1 = 0.

    • Work to bring q2q_2 from infinity to r2\mathbf{r}_2 in the field of q1=q2V1(r2)=14πϵ0q1q2r12q_1 = q_2 V_1(\mathbf{r}_2) = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}.

    • Total potential energy: U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}.

  • Three-Charge System (q1,q2,q3q_1, q_2, q_3):

    • Work includes interactions between all pairs: U=14πϵ0(q1q2r12+q1q3r13+q2q3r23)U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right).

  • The total energy is path-independent and characterizes the configuration's state.

POTENTIAL ENERGY IN AN EXTERNAL FIELD

  • Single Charge: In an external potential V(r)V(\mathbf{r}), the potential energy of charge qq is U=qV(r)U = qV(\mathbf{r}).

  • Electron Volt (eV): Energy gained by an electron accelerated by 1 Volt difference (1.6×1019J1.6 \times 10^{-19}\,J).

    • 1keV=103eV1\,keV = 10^3\,eV, 1MeV=106eV1\,MeV = 10^6\,eV, 1GeV=109eV1\,GeV = 10^9\,eV, 1TeV=1012eV1\,TeV = 10^{12}\,eV.

  • System of Two Charges in External Field:

    • Total energy U=q1V(r1)+q2V(r2)+14πϵ0q1q2r12U = q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}.

  • Dipole in Uniform External Field:

    • Experience torque τ=p×E\mathbf{\tau} = \mathbf{p} \times \mathbf{E}.

    • Potential energy U(θ)=pE=pEcos(θ)U(\theta) = -\mathbf{p} \cdot \mathbf{E} = -pE \cos(\theta).

    • Choosing θ=π/2\theta = \pi/2 as zero reference for potential energy simplifies the work calculations.

ELECTROSTATICS OF CONDUCTORS

  1. Inside a conductor, the electrostatic field is zero: Free electrons redistribute to cancel any internal field.

  2. Field at the surface is normal: Tangential components would cause charges to move, violating the static state.

  3. Interior has no excess charge: By Gauss's Law, if E=0E=0 inside, the enclosed charge must be zero. Excess charge resides on the surface.

  4. Potential is constant throughout: Since E=0E=0 inside, no work is done moving a charge; hence VV is the same everywhere, including the surface.

  5. Surface Electric Field: E=σϵ0n^\mathbf{E} = \frac{\sigma}{\epsilon_0} \mathbf{\hat{n}}, where σ\sigma is surface charge density.

  6. Electrostatic Shielding: Inside a cavity of a conductor, the electric field is always zero, regardless of outside charges or the conductor's charge. This protects sensitive instruments.

DIELECTRICS AND POLARISATION

  • Dielectrics are non-conducting substances with no free charge carriers. An external field induces a dipole moment by stretching or reorienting molecules.

  • Non-polar molecules: Centers of positive and negative charges coincide (e.g., H2,O2H_2, O_2). External fields induce dipoles.

  • Polar molecules: Permanent dipole moments exist due to charge separation (e.g., H2O,HClH_2O, HCl). Thermal energy causes random orientation; external fields align them.

  • Polarisation (P\mathbf{P}): Dipole moment per unit volume. For linear isotropic dielectrics: P=ϵ0χeE\mathbf{P} = \epsilon_0 \chi_e \mathbf{E}, where χe\chi_e is electric susceptibility.

  • Inside a dielectric, induced surface charges (bound charges) produce an internal field that opposes and reduces (but does not cancel) the external field.

CAPACITORS AND CAPACITANCE

  • A capacitor is a system of two conductors separated by an insulator.

  • Capacitance (CC): The ratio of charge QQ on one plate to the potential difference VV between them: C=Q/VC = Q / V.

  • Unit: Farad (F) (1F=1C/V1\,F = 1\,C/V). Common sub-multiples: μF(106),nF(109),pF(1012)\mu F (10^{-6}), nF (10^{-9}), pF (10^{-12}).

  • Dielectric Strength: The maximum electric field a dielectric can withstand before breakdown (3×106V/m\approx 3 \times 10^6\,V/m for air).

  • Parallel Plate Capacitor:

    • Two plates of area AA separated by distance dd.

    • Electric field E=σ/ϵ0=Q/(Aϵ0)E = \sigma / \epsilon_0 = Q / (A \epsilon_0).

    • Potential V=Ed=Qd/(Aϵ0)V = Ed = Q d / (A \epsilon_0).

    • Capacitance in vacuum: C0=ϵ0AdC_0 = \frac{\epsilon_0 A}{d}.

  • Effect of Dielectric: Inserting a dielectric of constant KK reduces field to E=E0/KE = E_0 / K and potential to V=V0/KV = V_0 / K, thus increasing capacitance: C=KC0C = K C_0.

  • Dielectric Constant (KK): Ratio of permittivity of substance to permittivity of vacuum (K=ϵ/ϵ0K = \epsilon / \epsilon_0).

COMBINATIONS AND ENERGY

  • Capacitors in Series:

    • Charge QQ is the same.

    • Potential sums: V=V1+V2+V = V_1 + V_2 + \dots

    • Resultant capacitance: 1C=1C1+1C2+\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dots

  • Capacitors in Parallel:

    • Potential VV is the same.

    • Charge sums: Q=Q1+Q2+Q = Q_1 + Q_2 + \dots

    • Resultant capacitance: C=C1+C2+C = C_1 + C_2 + \dots

  • Energy Stored (UU):

    • U=12QV=12CV2=Q22CU = \frac{1}{2} QV = \frac{1}{2} CV^2 = \frac{Q^2}{2C}.

  • Energy Density (uu): Energy per unit volume in a field: u=12ϵ0E2u = \frac{1}{2} \epsilon_0 E^2.

QUESTIONS AND DISCUSSION

  • Example 2.1: Potential at 9 cm from 4×107C4 \times 10^{-7}\,C is 4×104V4 \times 10^4\,V. Work required to bring 2×109C2 \times 10^{-9}\,C from infinity is 8×105J8 \times 10^{-5}\,J. The path taken does not affect the answer due to the conservative nature of the field.

  • Example 2.4: Work to arrange four charges (+q,q,+q,q+q, -q, +q, -q) at square corners (side dd) is W=q24πϵ0d(42)W = \frac{-q^2}{4\pi\epsilon_0 d} (4 - \sqrt{2}). Extra work to bring q0q_0 to center is zero because center potential is zero.

  • Example 2.7 Practical Insights:

    • Combs attract paper via polarisation; wet hair reduces friction/charge.

    • Aircraft tires are conductive to discharge static electricity accumulated during flight/landing.

    • Birds on wires are safe because there is no potential difference between their feet; grounding a human creates a high potential difference, causing fatal shock.

  • Points to Ponder: Potential at a charge's own location is infinite and undefined. Electrostatic shielding works from outside in, but placing a charge inside a cavity does not shield the exterior from that charge.