L
LLLOS.ai
Learn
L

Chapter 1 — Electrostatics

Class 12 · Physics

Overview

This unit studies electric charges at rest and the forces, fields and potentials they create. Starting from Coulomb's law that quantifies the force between two point charges, we build the ideas of superposition, electric field and field lines to visualise how charges influence the space around them. We learn to calculate fields and potentials for various charge distributions: point charges, lines, rings and spherical shells. Electric flux and Gauss's law give a powerful method to find fields for symmetric charge arrangements. Electric potential and potential energy explain work done in moving charges and lead to equipotential surfaces. The electric dipole is introduced as a basic neutral system with a characteristic field and potential. Properties of conductors and insulators in electrostatic equilibrium are discussed. Finally, capacitors, their combinations, dielectrics and the energy stored in capacitors are studied because they are fundamental to circuits and many devices. This unit matters because electrostatics underpins electricity, electronics, molecular interactions and forces in everyday life. It develops vector techniques, integral thinking and physical reasoning required for higher studies in electromagnetism and practical applications in sensors, capacitors, insulation, and design of electronic components.

Learning Objectives

  • Explain and apply Coulomb's law to calculate electrostatic forces between point charges.
  • Use the principle of superposition to find net forces and fields from multiple charges.
  • Define electric field and calculate field distributions for simple charge arrangements.
  • State and apply Gauss's law to evaluate electric fields for high-symmetry charge distributions.
  • Define electric potential and relate potential to electric field; compute potentials for point and extended charges.
  • Describe equipotential surfaces and use them to reason about work done in moving charges.
  • Explain the electric dipole, derive its field and potential on axial and equatorial lines, and compute torque in an external field.
  • Distinguish conductors and insulators in electrostatic equilibrium and explain behaviour of charges on conductors.
  • Describe capacitors, derive expressions for capacitance of simple geometries, and compute energy stored in capacitors.
  • Understand the effect of a dielectric on capacitance and store energy, and apply formulas to capacitor networks.

Topics in this chapter

18 topics · tap a topic title to jump straight to it.

1

Nature of electric charge and conservation

What is electric charge? Electric charge is an intrinsic property of certain particles of matter that produces electric forces. There are two types of charge known as positive and negative. Electrons carry negative charge and protons carry positive charge. Charge comes in discrete units; experimentally the smallest observed free charge is the elementary charge, commonly denoted e. Any macroscopic charge is an integer multiple of e. Charge is a scalar quantity with sign and has the SI unit coulomb (C).

Quantisation and transfer Quantisation means that when we move charges around, we move bundles of elementary charges. In everyday processes such as rubbing a plastic rod with fur, electrons transfer from one material to another producing net negative charge on one object and equal net positive charge on the other. Although objects appear to gain or lose charge, the total charge of the closed system remains constant.

Conservation law Conservation of charge is a fundamental experimental fact: in any physical process the algebraic sum of charges remains unchanged. When charges are separated or recombined, positive and negative transfers always balance so that no net charge is spontaneously created or destroyed in ordinary processes. This is important in electrical circuits and chemistry where transfer of electrons changes chemical identity but not total charge.

Conductors and insulators Materials differ in how easily their charges move. In conductors mobile charges (usually electrons) move freely through the material; in insulators charges are bound to atoms and move negligibly. Semiconductors lie between these extremes. The mobility of charges affects how materials behave when charged: in conductors charges quickly move to surfaces and reach electrostatic equilibrium, while in insulators charges stay where placed and create localised static fields.

Methods of charging There are three standard methods to charge bodies: friction (triboelectric effect), conduction (contact) and induction. Friction transfers electrons due to differing affinities. Conduction allows charge to flow when two conductors touch until potentials equalise. Induction rearranges charges in a conductor by bringing a charged object near and possibly grounding the conductor to leave net charge behind.

Practical consequences Understanding charge and its conservation explains static cling, sparks, lightning and the working of devices such as photocopiers and electrostatic precipitators. Safety practices like earthing remove unwanted charge. The concept of conserved charge also underlies circuit analysis where current is flow of charge, and the continuity equation relates current to changing charge distributions.

📌 Examples
  • Two identical plastic rods rubbed with fur gain equal and opposite charges; measure charge by electroscope to show conservation.
  • A neutral metal sphere touched by a negatively charged rod becomes negatively charged; the final charge equals charge transferred.
  • A charged comb brought near small bits of paper induces attraction; grounding the comb removes induced charge.
  • Transfer of electrons in a simple circuit does not change total charge of the system but moves charge carriers along the conductor.
🧮 Formulas
  1. Charge quantization: q = ne, where n is integer and e = 1.602×10^-19 C
📊 Visual ideas
Diagram showing two objects rubbed producing +Q and -Q; an electroscope indicating charge.
Schematic of induction: charged rod near neutral sphere, showing separation of charges and connection to ground.
🔬2

Coulomb's law

Empirical law and meaning Coulomb's law quantifies the electrostatic force between two stationary point charges. It states that the magnitude of the force between two point charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Direction is along the straight line joining the charges, repulsive for like signs and attractive for unlike signs.

Mathematical expression For two point charges q1 and q2 separated by a distance r in free space, the magnitude of the force is F = k |q1 q2| / r^2, where k = 1/(4πε0). The constant ε0 is the permittivity of free space. In vector form, treating r̂ as unit vector from q1 to q2, the force on q2 due to q1 is F_21 = (1/4πε0) (q1 q2 / r^2) r̂. Here the algebraic sign of q1 q2 determines whether the vector points toward or away from q1.

Role of medium Coulomb's law in a material medium modifies the constant: replace ε0 by ε, the permittivity of the medium, so forces are reduced by the medium's permittivity. This accounts for screening effects and explains why forces in materials differ from those in vacuum.

Assumptions and domain The law applies exactly to point charges and to spherically symmetric charge distributions treated as point charges when observed from outside. It is a static law; when charges move at significant speeds, magnetic effects and retarded potentials from full electromagnetism become relevant. Coulomb's law is experimentally accurate at macroscopic distances and underlies the inverse-square behaviour of the electric field for monopole sources.

Vector treatment in problems When charges are not collinear, resolve forces into components. Choose coordinate axes and compute vector contributions from each charge, then add component-wise. The sign of a charge is important: a negative sign reverses the force direction for a given vector r̂. Use units consistently and convert distances to metres and charges to coulombs.

Comparisons and examples Coulomb forces are typically much stronger than gravitational forces between elementary particles: for two electrons the electrostatic repulsion far exceeds gravitational attraction. This large strength explains why electromagnetic forces dominate chemistry and material properties at atomic scales.

📌 Examples
  • Calculate magnitude and direction of force between +2 μC and -3 μC separated by 5 cm.
  • Two electrons separated by distance r: compute repulsive force and compare to gravitational attraction between them.
  • Find force on a small charge placed near a uniformly charged conducting sphere (treated as point charge if outside).
🧮 Formulas
  1. F = (1/4πε0) * (|q1 q2|/r^2)
  2. Vector form: F_12 = (1/4πε0) * (q1 q2 / r^2) r̂
  3. k = 1/4πε0 ≈ 8.988×10^9 N·m^2/C^2
📊 Visual ideas
Sketch of two point charges with the force vector along the line joining them, showing attraction and repulsion cases.
🔬3

Principle of superposition

Conceptual statement The principle of superposition is fundamental: the net electrostatic force or the net electric field produced by a collection of charges is the vector sum of the forces or fields produced by each charge individually. This works because electrostatic interactions in classical physics are linear — each source contributes independently to the total.

Applying to forces To find the total force on a test charge q0 due to several source charges q1, q2, ... , qn, compute each Coulomb force F_i exerted by qi on q0 using magnitude and direction. Then add all vectors: F_net = Σ_i F_i. When charges are not aligned, break each F_i into components along chosen axes and sum components to get F_net. The order of addition does not matter because vector addition is associative and commutative.

Applying to fields Since field E due to a charge distribution is defined as force per unit test charge, E obeys superposition: E_net = Σ_i E_i. This is often more convenient: compute scalar potentials or vector fields from each source, then sum. For continuous distributions replace sums by integrals: E = ∫ dE where each dE is field contribution from an infinitesimal element dq.

Continuous distributions For a line, surface or volume charge distribution with element dq, the infinitesimal field at a field point is dE = (1/4πε0) (dq r̂ / r^2). To find total E integrate dE vectorially over the distribution. Pay attention to geometry: sometimes symmetry helps cancelling components so integrals simplify. Choose coordinates so that angle relations and distances are easy to express.

Use of symmetry Symmetry simplifies superposition problems greatly. For example, identical charges arranged symmetrically on a circle produce zero net field at centre. In many textbook problems symmetry reduces vector sum to a single non-zero component. Recognising symmetry before computing saves time and reduces algebra.

Limitations Superposition holds for classical electrostatics and in linear media. In non-linear materials, or when quantum effects dominate, simple superposition may not hold in same form. For ICSE/ISC electrostatics problems superposition is always valid and widely used to combine fields and potentials.

📌 Examples
  • Find electric field at centre of a square with equal positive charges at corners using vector addition and symmetry.
  • Compute net force on a charge placed midway between two equal unlike charges separated by a distance using superposition.
  • Determine field on axis of a three-charge linear arrangement by summing contributions from each charge.
🧮 Formulas
  1. Forces: F_net = Σ_i F_i
  2. Fields: E_net = Σ_i E_i
  3. Continuous: E = ∫ (1/4πε0) (dq r̂ / r^2)
📊 Visual ideas
Diagram of four charges at square corners and vector arrows indicating field contributions adding to resultant at centre.
4

Electric field and field lines

Field definition and units Electric field E at a point is defined as the force experienced per unit positive test charge placed at that point: E = F/q0. It is a vector quantity measured in newton per coulomb (N/C) or volt per metre (V/m). The field concept replaces the idea of direct action at a distance by assigning a physical vector to points in space that tells how charges would be pushed or pulled.

Field of a point charge A point charge Q at the origin produces a radial field E(r) = (1/4πε0)(Q/r^2) r̂. The direction is outward for Q positive and inward for Q negative. This inverse-square dependence means field strength decreases rapidly with distance. For a charge in a medium replace ε0 by ε.

Superposition and calculating field For multiple charges, compute field due to each point charge at the field point and add vectors. For continuous distributions dE = (1/4πε0)(dq r̂/r^2) and E = ∫ dE. Often one computes components: choose axes, express r and angles in integration variables and perform vector integration. When symmetry exists, choose coordinates aligned with symmetry to simplify integrals.

Field lines visualisation Field lines are a qualitative way to draw the direction and relative strength of field. Rules: lines start on positive charges and end on negative charges; their density indicates strength; lines never cross; at any point the tangent to a line gives the field direction. For an isolated positive charge, lines are radial; for a dipole, lines leave the positive charge and curve to enter the negative charge, forming closed patterns at infinity.

Relation with equipotentials Equipotential surfaces are everywhere perpendicular to electric field lines. Moving a charge along an equipotential requires no work; moving across equipotentials requires work. For instance, concentric spherical equipotentials accompany a point charge and planes accompany a uniform field. Plotting both together helps visualise gradients and energy changes.

Field near conductors Inside a conductor in electrostatic equilibrium E = 0. Any excess charge resides on the surface and the field just outside is perpendicular to the surface with magnitude E = σ/ε0 (where σ is local surface charge density). Sharp points concentrate surface charge, producing large local fields and supporting corona or spark formation.

Problem solving tips Always check symmetry first. Use vector components and unit vectors carefully. For numerical work convert units properly and choose convenient origins, noting that field due to a spherically symmetric distribution outside behaves like a point charge at the centre.

📌 Examples
  • Compute E at a point on the axis of a ring of charge using integration and sketch corresponding field lines.
  • Sketch field lines for: (i) single positive charge, (ii) dipole, (iii) two equal like charges separated by distance.
  • Find E at midpoint of two equal like charges and explain direction by superposition.
🧮 Formulas
  1. E = F/q0
  2. Point charge: E = (1/4πε0) (Q/r^2) r̂
  3. Continuous: E = ∫ (1/4πε0) (dq r̂ / r^2)
📊 Visual ideas
Field line diagram for an electric dipole showing lines from positive to negative and symmetry about the axis.
Radial field lines from an isolated point charge, with density showing 1/r^2 falloff.
5

Electric flux

Basic idea Electric flux measures how much electric field passes through a surface. For a small area element dA with unit normal n̂, the flux through dA is dΦ = E · dA = E cosθ dA, where θ is the angle between E and the normal. Flux is additive over a surface; total flux through surface S is Φ = ∮_S E · dA. Flux is a scalar quantity whose sign depends on whether field lines leave (positive) or enter (negative) the surface when outward normal is used.

Interpretation Physically, flux counts the net number of field lines crossing a surface in a qualitative picture where line density represents field strength. If the field is uniform and the surface flat, flux simplifies to Φ = E A cosθ. For curved surfaces or non-uniform fields the integral form must be used. Flux depends on orientation: rotating a surface relative to a field changes flux according to cosθ.

Closed vs open surfaces For an open surface the flux gives the net field crossing that patch; for a closed surface (a surface that encloses a volume) flux measures the net outflow of field from the volume. Gauss's law relates closed-surface flux directly to the net charge enclosed, making flux a bridge between field geometry and charge distribution.

Units and usefulness SI unit of flux is N·m^2/C (same as volt·metre). Flux integrals are central to solving many electrostatic problems because they allow converting difficult pointwise field determinations into surface integrals which sometimes simplify if symmetry is present. For example, for radial fields the flux through a sphere becomes trivial because E is constant on the sphere.

Computation strategies For symmetric configurations choose surfaces where E is constant or perpendicular so E can be pulled out of integral. For a sphere concentric with a point charge flux is simply E times area 4πr^2. For a cylinder around an infinite line charge choose the curved surface where E is constant and the flat ends where E is parallel to surface so they contribute zero. For plane sheets use pillbox Gaussian surfaces to capture flux through faces.

Qualitative examples A point charge located outside a closed surface produces zero net flux through that surface because as many field lines enter as leave. If the charge is inside, net flux equals Q/ε0, independent of the shape of the enclosing surface. This insensitivity to shape underlies Gauss's law and makes flux a powerful conceptual tool.

📌 Examples
  • Calculate flux through a square of side a placed at angle θ in uniform field E: Φ = E a^2 cosθ.
  • For a point charge Q at centre of a sphere radius R compute flux through sphere: Φ = Q/ε0 using E = Q/(4πε0 R^2) and surface area 4πR^2.
  • Compute flux through a cylindrical surface in a uniform field parallel to axis; show flux through curved surface is zero if field parallel to axis.
🧮 Formulas
  1. dΦ = E · dA
  2. Φ = ∮_S E · dA
  3. Uniform flat surface: Φ = E A cosθ
📊 Visual ideas
Diagram of field lines crossing a flat surface at angle θ showing area vector and cosθ component.
Spherical surface with radial field lines and area elements to illustrate constant E over sphere.
🔬6

Gauss's law and its proof sketch

Statement of Gauss's law Gauss's law states that the net electric flux through any closed surface equals the net charge enclosed divided by the permittivity of free space: ∮_S E · dA = Q_enclosed / ε0. This is a central relation in electrostatics and is one of Maxwell's equations in differential and integral form.

Reasoning for a point charge Consider a point charge Q at the centre of a sphere of radius r. The field at the sphere is radial and has magnitude E = (1/4πε0)(Q/r^2). The flux through the sphere is E times area 4πr^2, giving Φ = Q/ε0. This exact result motivates the general rule that if a charge is enclosed it contributes to flux in a way independent of the surface shape.

Solid angle argument for arbitrary shapes For a point charge located at some point inside an arbitrary closed surface, imagine dividing the surface into small patches. Each patch sees the charge under some solid angle. The contribution of the charge to flux through each patch equals (Q/4πε0) times that patch's solid angle divided by r^2 times the cosine factor; summing over the entire closed surface recovers the full solid angle 4π and gives total flux Q/ε0. If the charge is outside, contributions with positive and negative signs cancel and net flux is zero. Superposition extends this to many charges.

Utility and symmetry Gauss's law is most useful when the chosen closed surface exploits symmetry so that E has constant magnitude on parts of the surface and is either parallel or perpendicular to surface normals. Typical symmetries are spherical, cylindrical and planar. In such cases surface integrals simplify to algebraic equations and one can solve for E directly without performing complex integrals.

Limitations and caution Gauss's law always holds, but it only yields an easy solution for E when symmetry is present. For asymmetric charge distributions it remains true but evaluating ∮ E·dA requires knowing E at each point which may be as difficult as the original problem. Also idealizations like infinite line or plane are useful approximations but must be used with understanding of limits.

Physical meaning Gauss's law links field lines and source charges: field lines begin and end on charges, and the net number leaving a closed surface counts the net positive charge inside in units set by ε0. This gives a clear geometric interpretation of enclosed charge in terms of field behaviour at the bounding surface.

📌 Examples
  • Use Gauss's law to find E outside and inside a uniformly charged spherical shell.
  • Apply Gauss's law to find E at distance r from an infinite line charge of linear charge density λ.
  • Use pillbox Gaussian surface to find E near an infinite charged conducting plane or infinite uniformly charged plane sheet.
🧮 Formulas
  1. Gauss's law: ∮_S E · dA = Q_enclosed/ε0
📊 Visual ideas
Gaussian sphere with a point charge at centre showing radial E and constant magnitude over surface.
Cylindrical Gaussian surface around infinite line charge showing field lines and flux through curved surface.
🔬7

Applications of Gauss's law: spherical and cylindrical symmetry

Spherical symmetry: shells and solid spheres For a thin spherical shell carrying total charge Q, choose a spherical Gaussian surface concentric with the shell. If the field point is outside the shell (r > R) the enclosed charge is Q and by Gauss's law E(4π r^2) = Q/ε0, so E = (1/4πε0)(Q/r^2), identical to a point charge at the centre. If the field point is inside the shell (r < R) the enclosed charge is zero and E = 0. For a uniformly charged solid sphere of radius R and total charge Q, use a Gaussian sphere of radius r < R. The enclosed charge scales as Q(r^3/R^3), so E(4π r^2) = (Q r^3)/(R^3 ε0), giving E = (1/4πε0)(Q r / R^3). Outside the solid sphere field reduces to the point-charge form.

Physical interpretation The result that a spherical shell produces zero field inside explains shielding: a hollow spherical conductor shields its interior from outside static fields. For a solid sphere the field rising linearly with r inside shows that near the centre the field is weak and increases outward until the surface.

Cylindrical symmetry: infinite line and cylinder For an infinitely long line of charge with linear density λ, use a coaxial cylindrical Gaussian surface of radius r and length L. The curved surface contributes flux E(2π r L) while flat ends contribute zero if field is purely radial. Enclosed charge = λ L, so E = λ/(2π ε0 r) directed radially outward. For a uniformly charged solid cylinder of radius R with volume charge density ρ, inside (r < R) the enclosed charge per length scales as ρ π r^2, leading to E = ρ r /(2 ε0) and outside the field falls as 1/r as in the line case.

Planar symmetry: infinite plane For an infinite plane sheet with surface charge density σ, a pillbox Gaussian surface having faces parallel to the plane captures flux through both faces. If the plane is isolated and symmetric both sides see field magnitude E = σ/(2ε0); for a conducting plane with all charge on one side the field immediately outside is σ/ε0. These results are very useful approximations for plate capacitors when plate separation is small compared to plate dimensions.

Practical cautions The quoted formulas assume ideal infinite or perfectly symmetric distributions. Real finite objects approximate these results near their central regions, but edge effects occur near boundaries. Always check domain of validity when using Gauss's law to ensure chosen Gaussian surface respects symmetry so E can be taken constant or zero on parts of the surface.

📌 Examples
  • Find E inside and outside a uniformly charged solid sphere using Gauss's law and compare with point charge result outside.
  • Calculate E at distance r from an infinite line with λ = 5×10^-9 C/m at r = 2 cm.
  • Use Gauss's law to show field near an infinite conducting plane with surface charge density σ is E = σ/ε0.
🧮 Formulas
  1. Solid sphere (r<R): E = (1/4πε0) (Q r / R^3) r̂
  2. Solid sphere (r>R): E = (1/4πε0) (Q / r^2) r̂
  3. Infinite line: E = λ/(2π ε0 r)
  4. Infinite plane sheet: E = σ/(2ε0) (single sheet)
📊 Visual ideas
Cross-section of uniformly charged solid sphere showing Gaussian sphere of radius r<R and field vector proportional to r.
Cylindrical Gaussian surface around infinite line charge showing radial E and flux through curved surface.
8

Electric potential: definition and relation to field

Definition of potential Electric potential V at a point is the work done per unit positive test charge in bringing it from infinity to that point slowly, without acceleration, with the reference V(∞)=0. It is a scalar quantity measured in volts (V) where 1 V = 1 J/C. Potential simplifies energy calculations because it is scalar and potentials from multiple sources add algebraically.

Relation to field The electric field is the negative gradient of the potential: E = -∇V. In one dimension this reduces to E_x = -dV/dx. This relation shows that where potential decreases rapidly with distance, the corresponding field is large, and that field lines point in the direction of greatest decrease of potential.

Potential of a point charge For a point charge Q located at origin and V(∞)=0, potential at distance r is V(r) = (1/4πε0)(Q/r). Unlike field which falls as 1/r^2, potential falls as 1/r. The potential is positive for Q>0 and negative for Q<0. For a system of point charges potentials add: total V = Σ (1/4πε0)(q_i/r_i).

Equipotential surfaces Equipotential surfaces are loci where V has the same value. They are always perpendicular to electric field lines. For a point charge equipotentials are concentric spheres; for a uniform field they are planes. Moving a charge along an equipotential requires no work because ΔV = 0 along the path.

Work and potential energy Work done by the external agent in moving a charge q from A to B is q[V(B) - V(A)]. Potential energy of a charge q in potential V is U = qV. Conservation of mechanical plus electrostatic potential energy governs motion of charges in electrostatic fields and is useful for problems involving forces and speeds.

Practical use Potential is often easier to compute than field, especially for continuous distributions, because it is scalar. Once V is found one can compute E by differentiation. Potential also clarifies boundary conditions: conductors are equipotentials and knowing surface potentials often suffices to determine fields in regions between conductors.

📌 Examples
  • Compute potential at point on axis of a uniformly charged ring and then find axial electric field by differentiating V.
  • Find potential at distance r from point charge Q: V = (1/4πε0)(Q/r).
  • Calculate work required to bring a charge q from infinity to distance r from Q: W = qV = (1/4πε0)(Q q / r).
🧮 Formulas
  1. Potential difference: V(B) - V(A) = -∫_A^B E · dl
  2. Point charge: V = (1/4πε0) (Q/r) (with V(∞)=0)
  3. Potential energy: U = qV
📊 Visual ideas
Equipotential spherical surfaces around a point charge with radial E lines perpendicular to spheres.
Potential versus r graph for point charge, showing V∝1/r.
🔬9

Potential due to continuous charge distributions

General method For a continuous distribution of charge, the potential at a point is found by summing contributions from all infinitesimal elements of charge. Divide the source into small elements of charge dq. Each element at distance r from the field point contributes dV = (1/4πε0)(dq/r). Integrate over the entire distribution to obtain V = ∫ (1/4πε0)(dq/r). Because potential is scalar, this integral is often simpler than computing the vector field E directly.

Line charge example Consider a uniformly charged rod of length L with linear charge density λ. To find potential at a point on the axis or perpendicular bisector, express dq = λ dx and distance r in terms of integration variable x. Integrate dq/r over x from one end to the other. For an infinite line the potential diverges if referenced to infinity; in such cases we compute potential differences instead of absolute potential.

Ring and disk For a ring of radius a carrying total charge Q, every element of charge is at the same distance √(a^2 + x^2) from a point on the axis at distance x from centre. Thus V_axis = (1/4πε0)(Q/√(a^2 + x^2)). For a uniformly charged disk, treat it as a set of concentric rings of radius r, with charge dq = σ 2π r dr, and integrate dq/√(r^2 + x^2) from r=0 to disk radius to obtain V on axis. Because each ring contributes equally in angular direction, integration reduces to a one-dimensional integral over r.

Surface and volume charges For a uniformly charged spherical shell or solid sphere, integrate over surface or volume to get potential. For a uniformly charged solid sphere one finds inside potential V(r<R) = (1/4πε0)(Q/2R)(3 - r^2/R^2) while outside V(r>R) = (1/4πε0)(Q/r). The potential is continuous at r = R though field may change abruptly.

Choosing reference and limits Always state reference point for potential. For finite distributions the standard reference V(∞)=0 works. For infinite distributions, use potential differences or choose a finite reference point. Exploit symmetry to reduce dimensionality of integrals: rings for disks, shells for spheres, and cylindrical shells for rods.

Using V to get E Once V is obtained, compute field by E = -∇V. Often differentiating a scalar result is algebraically simpler than integrating vector contributions directly. This two-step method—compute V then E—is a standard strategy for many ICSE/ISC problems involving continuous charge distributions.

📌 Examples
  • Derive potential on axis of a uniformly charged ring and then find axial E = -dV/dx.
  • Compute potential at center of uniformly charged ring or disk using appropriate integration.
  • Find potential inside and outside uniformly charged solid sphere and verify continuity at r = R.
🧮 Formulas
  1. Continuous charge: V = ∫ (1/4πε0) (dq/r)
  2. Ring (axis): V = (1/4πε0) (Q/√(a^2 + x^2))
📊 Visual ideas
Diagram of ring and point on axis with distances labelled to be used in integral.
Plot of V versus x along axis of ring showing maximum at centre decreasing with x.
🔬10

Equipotential surfaces and conductors in electrostatic equilibrium

Equipotential surfaces Equipotential surfaces are sets of points where the electric potential has the same value. They are always orthogonal to field lines. If you move a charge along an equipotential no work is done because potential difference is zero. For simple sources equipotentials have characteristic shapes: spheres around a point charge, planes for a uniform field, and more complex closed surfaces for multipole arrangements.

Conductors at electrostatic equilibrium Inside a conductor at electrostatic equilibrium the electric field is zero. Free charges in the conductor redistribute until internal electric fields cancel. Consequently the interior volume of a conductor is an equipotential; every point inside or on the conductor has the same potential. Any excess charge resides entirely on the outer surface of the conductor.

Surface properties On the surface of a conductor the electric field is perpendicular to the surface (no tangential component) and its magnitude just outside is E = σ/ε0, where σ is local surface charge density. Where the surface curvature is high (sharp points) σ is larger and local fields are stronger. This explains concentration of field near points and is exploited in devices such as lightning rods and electrostatic precipitators.

Cavities and shielding A hollow conductor shields its interior from external static fields: if there are no charges inside a cavity the field inside is zero. If a charge is placed inside a cavity, induced charges appear on the cavity surface and on the outer surface so that outside field depends only on total enclosed charge. This shielding principle is used in Faraday cages to protect sensitive electronics from external static fields.

Practical use and measurements Equipotential mapping in labs with conducting paper and voltmeter helps visualise field geometry and locate potential gradients. In engineering conductors form shields, grounding paths, and electrode surfaces that are designed to maintain desired potentials. Recognising conductors as equipotentials simplifies many boundary-value problems because boundary potentials are known and can be used with Laplace or Poisson equations.

📌 Examples
  • Show that electric field inside a hollow conducting sphere with external uniform field is zero and that induced charges appear on outer surface.
  • Explain why charge accumulates at sharp points of a conductor using equipotential and field concentration arguments.
  • Using Gauss's law, show E just outside a charged conductor is σ/ε0 normal to surface.
🧮 Formulas
  1. Surface field of conductor: E_outside = σ/ε0 (normal component just outside)
  2. Inside conductor: E_inside = 0
📊 Visual ideas
Cross-section of conductor with external field showing induced surface charges and zero field inside.
Equipotential lines around a point charge and conductor showing perpendicular intersection with surface.
11

Electric dipole: field, potential and torque

Definition and dipole moment An electric dipole consists of two equal and opposite charges +q and -q separated by small distance 2a. The dipole moment p is a vector defined as p = q(2a) pointing from the negative charge toward the positive charge. Dipoles are fundamental in describing neutral systems with separated charges such as polar molecules and arrangements of charges in dielectrics.

Exact potential and approximate far-field The exact potential at a field point is the algebraic sum of potentials due to each charge: V = (1/4πε0)(q/r_+ - q/r_-). For points far away compared to separation (r ≫ a), a multipole expansion gives the leading term as V ≈ (1/4πε0)(p · r̂ / r^2). This shows the dipole potential falls as 1/r^2, and the corresponding field falls as 1/r^3, which is faster than monopole fields.

Axial and equatorial fields On the axial line (along dipole axis) at distance r from centre with r ≫ a, field magnitude is E_axial ≈ (1/4πε0)(2p/r^3) directed along axis away from positive end. On the equatorial line (perpendicular bisector), field is E_equatorial ≈ (1/4πε0)(p/r^3) directed opposite to p. These relations show both magnitude and directional differences between axial and equatorial points.

Torque and potential energy A dipole in a uniform external electric field E experiences a torque τ = p × E that tends to align p with E. The potential energy of dipole in field is U = -p · E, minimum when p and E are parallel. These expressions explain alignment of polar molecules in external fields and are key to understanding dielectric polarisation at macroscopic level.

Applications and remarks Dipoles explain many physical phenomena: behaviour of molecules in fields, force between neutral atoms at long range (in a simplified picture), and radiation patterns in antennas (time-varying dipoles). Remember dipole approximations are valid when observation distance is large compared to separation; near-field expressions require using exact formulas summing contributions from each charge.

📌 Examples
  • Compute electric field on axial point at distance r from centre of dipole with charges ±q separated by 2a, then take r≫a limit to get E ≈ (1/4πε0)(2p/r^3).
  • Find torque on dipole p in uniform field E making angle θ: τ = pE sinθ and potential energy U = -pE cosθ.
  • Calculate potential at a point on equatorial line exactly by summing contributions from +q and -q and show odd symmetry.
🧮 Formulas
  1. Dipole moment: p = q(2a) (vector from -q to +q)
  2. Axial field (far): E_axial ≈ (1/4πε0) (2p / r^3)
  3. Equatorial field (far): E_equatorial ≈ (1/4πε0) (p / r^3) (direction opposite p)
  4. Potential (far): V ≈ (1/4πε0) (p·r̂ / r^2)
  5. Torque: τ = p × E
  6. Potential energy: U = -p · E
📊 Visual ideas
Dipole diagram with +q and -q separated by 2a, showing dipole moment vector p from - to + and field lines looping from + to -.
Plots of axial and equatorial field magnitudes versus distance r showing 1/r^3 dependence in far field.
12

Electric potential energy and interaction energy

Pair potential energy For two point charges q1 and q2 separated by distance r, the electrostatic potential energy is U = (1/4πε0)(q1 q2 / r) when potential at infinity is taken as zero. The sign of U indicates whether work is required to assemble the charges: like charges give positive U (work required to bring them closer), opposite charges give negative U (energy released when they come together).

Many-charge systems For a system of many charges the total potential energy is the sum over pairs, but to avoid double counting one convenient formula is U = (1/2) Σ_i q_i V_i where V_i is the potential at the location of q_i due to all other charges. This expression follows because each pair energy appears twice in Σ_i q_i V_i and the factor 1/2 corrects that.

Continuous distributions and self-energy For continuous charge distributions with density ρ, energy can be expressed as U = (1/2) ∫ ρ V dτ or equivalently as an integral of energy density in the field: U = (1/2) ε0 ∫ E^2 dτ over all space. For a uniformly charged solid sphere of total charge Q and radius R, assembling the charge yields self-energy U = (3/5)(1/4πε0)(Q^2/R).

Relation to capacitors Energy stored in a capacitor can be obtained by integrating the work done to move small charges dq onto a plate: dW = V dq and integrating from 0 to Q yields U = (1/2) QV. Using Q = CV and V = Q/C gives equivalent forms U = (1/2) C V^2 = (1/2) Q^2 / C. Energy density view gives u = (1/2) ε E^2 in the region of electric field.

Conservative force and work Electrostatic forces are conservative: work done in moving a charge between two points depends only on end points and equals change in potential energy. This property simplifies many mechanical-energy problems where electrical potential energy converts to kinetic energy or vice versa.

Applications and problem tips Use pairwise sums for discrete charges and integrals for continuous ones. Watch signs: bringing like charges together increases potential energy; bringing opposite charges together decreases it. Use energy methods to find equilibrium configurations, minimum-energy arrangements, and forces by differentiating energy with respect to position parameters where appropriate.

📌 Examples
  • Compute potential energy of three point charges at vertices of an equilateral triangle each carrying charge q and separated by a distance a.
  • Derive self-energy of uniformly charged solid sphere and show U = (3/5)(1/4πε0)(Q^2/R).
  • Show energy stored in capacitor by integrating work done during charging: U = (1/2) C V^2.
🧮 Formulas
  1. Two charges: U = (1/4πε0) (q1 q2 / r)
  2. System: U = (1/2) Σ_i q_i V_i
  3. Energy in capacitor: U = (1/2) C V^2 = (1/2) Q^2/C = (1/2) QV
📊 Visual ideas
Graph of potential energy U versus separation r for like and unlike charges showing U∝1/r.
Charging graph showing incremental energy dW = V dq accumulation to total U = (1/2) QV.
🔬13

Capacitance and parallel plate capacitor

Definition and physical meaning Capacitance C is a geometrical property of a conductor pair that measures how much charge Q can be stored for a given potential difference V: C = Q/V. It depends only on the shapes, sizes and separation of conductors and on the medium between them. Capacitance has SI unit farad (F), where 1 F = 1 C/V.

Parallel plate capacitor model The simplest common model is the parallel plate capacitor consisting of two large conducting plates of area A separated by distance d, with d much smaller than plate linear dimensions so edge effects are negligible. For uniform surface charge ±σ on the plates the field between plates is E = σ/ε0 and outside region negligible. The potential difference between plates is V = E d = (σ d)/ε0. Since Q = σ A we get C = Q/V = ε0 A / d for vacuum. If a dielectric with permittivity ε fills the gap, E is reduced by dielectric polarisation and C = ε A / d = κ ε0 A / d where κ is the dielectric constant.

Energy storage and density A charged capacitor stores electrostatic energy. The work done in charging from 0 to Q is U = ∫_0^Q V dq = (1/2) QV. Using Q = CV, energy can be written U = (1/2) C V^2 or U = (1/2) Q^2 / C. Energy can also be viewed as stored in the electric field: energy density u = (1/2) ε E^2 and total energy equals integral of u over volume between the plates. For a parallel plate capacitor with volume V_space = A d, total energy U = (1/2) ε E^2 A d, consistent with capacitor formulas.

Dielectric effects on capacitance Inserting a dielectric increases capacitance by factor κ, because bound charges in dielectric reduce effective field for given free charge so a larger amount of free charge is stored at the same potential. Dielectrics also introduce breakdown limits: each material has maximum field before it becomes conductive, so practical capacitor design balances κ, thickness and breakdown strength.

Limitations of ideal formula The formula C = ε0 A / d assumes uniform field and negligible fringing. For finite plates or larger separations edge effects make field non-uniform and lower effective capacitance slightly. In precision work or microelectronic design fringe fields and plate geometry are included using numerical methods or corrections.

Practical use Capacitors are used for energy storage, filtering, timing circuits and coupling/decoupling signals. Design choices use area, separation and dielectric selection to achieve required capacitance and voltage rating. For high capacitance select large area, small separation and high-κ dielectric but ensure dielectric thickness prevents breakdown at operating voltage.

📌 Examples
  • Calculate capacitance of parallel plate capacitor with A = 0.02 m^2 and d = 1 mm in vacuum: C = ε0 A / d.
  • Find energy stored when such capacitor charged to V = 100 V: U = (1/2) C V^2.
  • Compute field E and surface charge density σ given Q and plate area A: σ = Q/A, E = σ/ε0.
🧮 Formulas
  1. Capacitance: C = Q/V
  2. Parallel plate: C = ε0 A / d (vacuum) or C = ε A / d (with dielectric)
  3. Energy: U = (1/2) C V^2 = (1/2) QV
  4. Energy density: u = (1/2) ε E^2
📊 Visual ideas
Schematic of parallel plate capacitor with area A, separation d, showing uniform E between plates and fringe fields at edges.
Plot of energy density u versus position across dielectric-filled gap showing uniformity if dielectric uniform.
🔬14

Combination of capacitors

Parallel combination When capacitors are connected in parallel their plates at corresponding potentials are joined so the potential difference across each capacitor is the same. Charges on individual capacitors add: Q_total = Σ Q_i = Σ C_i V. Therefore equivalent capacitance C_eq is sum of capacitances: C_eq = Σ C_i. Parallel connection increases effective plate area and is used when larger capacitance at same voltage is desired.

Series combination In series the same charge Q flows through each capacitor but potentials across them add: V_total = Σ V_i. For series capacitors equivalent capacitance obeys 1/C_eq = Σ (1/C_i). For two capacitors C1 and C2 in series, C_eq = (C1 C2)/(C1 + C2). Series connection decreases overall capacitance and is used to obtain higher voltage rating by sharing voltage across components.

Energy in combinations For capacitors connected to a battery, total energy stored is U = (1/2) C_eq V^2 where V is battery voltage. For capacitors charged separately and then connected together without a battery, charge redistributes until potentials equal; during redistribution some energy is dissipated as heat in connecting wires, so final stored energy is less than the initial total. Calculating energy before and after connections helps determine energy lost.

Reduction techniques Complex networks of capacitors can be simplified stepwise by replacing series or parallel groups with their equivalent capacitances. After reducing the network to C_eq between two terminals, charging behavior and energies follow simple formulas. Be careful to identify which capacitors share voltage and which share charge before applying formulas.

Special cases and design notes Two identical capacitors C in parallel give 2C, in series give C/2. When capacitors have dielectrics or different voltage ratings, ensure safe voltage distribution in series by using matched capacitors or additional balancing resistors. For AC circuits, reactance and frequency dependence matter but static capacitance rules for series/parallel combinations remain algebraic.

📌 Examples
  • Find equivalent capacitance of C1 = 4 μF and C2 = 6 μF in series and in parallel.
  • Two capacitors charged to different voltages are then connected in parallel: compute final voltage and energy loss.
  • Given three capacitors in complex network, reduce stepwise to find total C and charge distribution when connected to battery.
🧮 Formulas
  1. Series: 1/C_eq = Σ (1/C_i)
  2. Parallel: C_eq = Σ C_i
  3. Energy: U = (1/2) C V^2
📊 Visual ideas
Circuit diagrams showing two capacitors in series and in parallel with charges and voltages indicated.
Energy versus time sketch for redistribution when charged capacitors are connected without battery showing dissipation.
15

Dielectrics and polarisation

What is a dielectric? A dielectric is an insulating material placed between the plates of a capacitor or otherwise subjected to an electric field. Dielectrics do not conduct free charge but their microscopic charges shift slightly in response to an external field, creating induced dipoles. This process is called polarisation and it reduces the effective electric field inside the material for a given free charge configuration.

Polarisation vector and bound charges Macroscopically polarisation is described by vector P, the dipole moment per unit volume. Polarisation produces bound surface charge density σ_b = P · n̂ on the outer surfaces and bound volume charge density ρ_b = -∇·P inside the material. These bound charges alter the total field and potential produced by free charges and must be accounted for when solving electrostatic problems with dielectrics.

Linear dielectrics and permittivity For many materials in moderate fields, polarization is proportional to the applied field: P = χ_e ε0 E where χ_e is electric susceptibility. The electric displacement vector D is defined as D = ε0 E + P; for linear, homogeneous, isotropic dielectrics D = ε E where ε = ε0 (1 + χ_e) = κ ε0 and κ is the dielectric constant. The presence of dielectric increases capacitance by factor κ: C = κ C_0 for a capacitor whose vacuum capacitance is C_0.

Energy and breakdown Dielectrics change stored energy distribution: energy density becomes u = (1/2) E · D = (1/2) ε E^2 for linear dielectrics. While dielectrics increase capacitance and energy storage at given voltage, each material has a dielectric strength or breakdown field beyond which it becomes conductive and fails. Selection of dielectric balances high κ with high breakdown strength and low loss for AC applications.

Microscopic mechanisms Polarisation mechanisms include electronic displacement (shifting electron clouds relative to nuclei), ionic displacement (relative displacement of ions in lattice) and orientation of permanent molecular dipoles. These mechanisms respond at different frequency ranges and temperatures, which affects dielectric behaviour in AC fields and at high frequencies where orientation may lag causing dielectric loss.

Applications and measurement Dielectrics are central to capacitor design, insulating layers in cables and microelectronic devices. Dielectric constant κ can be measured by inserting sample between capacitor plates and measuring change in capacitance. In design, one chooses material and thickness to achieve desired C while ensuring safe operation at intended voltages and frequencies.

📌 Examples
  • A parallel plate capacitor with vacuum capacitance C0 = ε0 A/d filled with dielectric of κ = 4 gives new capacitance C = 4 C0.
  • Compute bound surface charge density on dielectric surface given P and normal vector using σ_b = P·n̂.
  • Given χ_e for a material, find relative permittivity κ = 1 + χ_e and resulting capacitance increase.
🧮 Formulas
  1. Polarisation: P = χ_e ε0 E (for linear dielectrics)
  2. Displacement: D = ε0 E + P = ε E
  3. Dielectric constant: κ = ε / ε0
  4. Bound surface charge: σ_b = P · n̂
  5. Capacitance with dielectric: C = κ C_0
📊 Visual ideas
Schematic of capacitor with dielectric showing displaced charges inside material and induced bound charges at surfaces.
Plot of capacitance versus dielectric constant κ showing linear increase.
💪16

Forces between charges and on conductors

Basic force expressions A charge q in an electric field E experiences force F = qE. For distributed charges with volume density ρ, force density f = ρ E and total force on a body is F = ∫ ρ E dτ. For surface charges with density σ, integrate surface force densities. These integrals give net mechanical forces due to electrostatic fields on charged bodies.

Electrostatic pressure on conductor surfaces A charged conductor surface experiences electrostatic pressure p equal to the field energy density just outside the surface. Since E_out = σ/ε0, the pressure is p = (1/2) ε0 E_out^2 = σ^2/(2ε0). This pressure acts normal to the surface and tends to push surface elements outward. In practice it explains stresses on charged droplets, charged balloons and high-voltage equipment surfaces where strong fields can deform structures.

Force between capacitor plates Oppositely charged parallel plates attract. One can derive magnitude of force by energy methods or pressure. For fixed charge Q on plates the attractive force per unit area is p = σ^2/(2ε0) and total force F = p A = Q^2/(2ε0 A). Alternatively using U = (1/2) Q^2/C and differentiating with respect to plate separation yields same result. These forces underpin electrostatic actuators and micro-electromechanical devices.

Force on dipoles and in non-uniform fields A dipole in a uniform field experiences a torque but no net force. In a non-uniform field a dipole experiences net force approximately F = (p · ∇)E, pulling it toward regions of stronger field if aligned suitably. This principle explains dielectrophoresis where neutral but polarizable particles move in non-uniform fields and is used in particle manipulation techniques.

Maxwell stress concept (qualitative) For general electromagnetic configurations one can compute forces using the Maxwell stress tensor: integrate the stress over a closed surface surrounding a body to get net electromagnetic force. While tensor methods exceed routine Class 12 calculations, they formalise how fields store and transfer momentum and are useful in advanced problems and design of devices where fields act on bodies in complex ways.

Practical advice Use energy methods or pressure expression for capacitors and conductors, and integrate local force densities for distributed charges. Remember boundary conditions at conductor surfaces when calculating local E and σ, since accurate local values determine pressures and resulting mechanical stresses.

📌 Examples
  • Calculate attractive force between parallel plates with area A, charge ±Q separated by d using pressure p = σ^2/(2ε0).
  • Find force on small dipole in non-uniform electric field using F = (p·∇)E (approximate expression).
  • Compute electrostatic pressure on a spherical shell with surface charge density σ and discuss mechanical stress.
🧮 Formulas
  1. Force on charge: F = qE
  2. Surface pressure: p = σ^2 /(2ε0)
  3. Force between parallel plates: F = (1/2) (Q^2)/(ε0 A) = (1/2) C V^2 / d (as derived)
📊 Visual ideas
Diagram of parallel plate capacitor showing field lines between plates, area A, separation d and direction of attractive force.
Sketch showing surface pressure vectors on charged spherical shell tending to expand it.
🔬17

Experimental methods and instruments

Electroscope and qualitative detection The electroscope is a simple device to detect and compare amounts of charge. It typically consists of a metal rod connected to thin leaves; when charged, leaves diverge because like charges repel. Bringing a charged object near the top without touching shows induction: leaves change divergence due to redistribution of charges. Grounding and touch experiments with an electroscope illustrate conduction and induction charging methods in the laboratory.

Millikan oil-drop concept (historical and conceptual) Millikan's oil-drop experiment measured the elementary charge by balancing gravitational force and electric force on tiny charged oil drops suspended between capacitor plates. Adjusting voltage so that a drop remained stationary gave qE = mg, allowing charge q to be determined. Repeating for many drops showed q was an integer multiple of a fundamental charge e. The experiment teaches precise force balance and use of potentials to control motion.

Equipotential mapping Conducting-paper experiments map equipotential lines by measuring points of equal potential with a voltmeter and marking them. Field lines are drawn perpendicular to equipotentials. This visual technique helps students understand field geometry for dipoles, two plates, and point charges. Setting up plates and connecting to power supply gives steady fields to map and compare with theoretical expectations.

Measuring capacitance and dielectric constant Capacitance of a capacitor can be measured by charging to a known voltage and measuring charge with sensitive electrometers or by AC impedance methods using bridges. Dielectric constant κ of a material is measured by inserting a slab between plates of known area and separation and noting the increase in capacitance: κ = C_with / C_without. Careful measurement accounts for edge effects and alignment.

Field meters and electrometers Modern instruments include field meters that measure surface or ambient electric fields and electrometers that measure charge and potential with very high input impedance. These devices allow quantitative study of static charges, leakage currents, and small capacitances in laboratory and industrial settings.

Safety and practice When working with electrostatics and high voltages use grounding, insulating supports and avoid flammable vapours. Discharge capacitors safely before handling. In teaching labs design experiments with low stored energy to avoid shocks. Observing phenomena like corona, sparks and discharge visually reinforces theoretical learning about strong local fields and breakdown.

📌 Examples
  • Describe an experiment using conducting paper to map equipotentials of a dipole and sketch corresponding field lines.
  • Explain Millikan-style balance conceptually: equate electrical force qE with weight mg to find q = mg/E when drop floats.
  • Method to measure dielectric constant by measuring capacitance before and after inserting slab between parallel plates.
🧮 Formulas
  1. Charge from balance: q = mg/E (concept used in Millikan-type balance when drop is suspended)
  2. Capacitance change with dielectric: C = κ C_0 (used for measurement of κ)
📊 Visual ideas
Schematic of conducting paper experiment with equipotential lines drawn and probes measuring voltage between points.
Diagram of Millikan oil drop set-up showing capacitor plates, oil drop, and balancing forces.
🔬18

Boundary value problems and uniqueness theorem (overview)

Electrostatics as a boundary-value problem Many electrostatic problems reduce to finding the potential function V(r) in a region given the charge distribution ρ(r) and values of V or its normal derivative on the boundary surfaces. In regions with charges Poisson's equation holds: ∇^2 V = -ρ/ε0; in charge-free regions Laplace's equation ∇^2 V = 0 applies. Solving these equations with appropriate boundary conditions gives the electrostatic potential and hence the field. While full analytic solutions require methods beyond Class 12 scope, understanding the setup and uniqueness principles guides problem solving and use of special methods.

Uniqueness theorem (qualitative) The uniqueness theorem states that the solution to Poisson's or Laplace's equation in a specified volume is unique if the potential is specified on the boundary (Dirichlet condition) or if the normal derivative is specified on the boundary (Neumann condition) together with overall charge constraints. Practically this means if you propose a potential satisfying the governing equation and the boundary conditions, it must be the physical solution. This theorem is powerful because it allows constructing solutions by convenient methods and trusting they are correct.

Method of images (introductory) The method of images uses the uniqueness theorem: replace conductors by imaginary charges (images) arranged so that the boundary condition on the conductor surface is met. For example, a point charge q at distance d above an infinite grounded conducting plane can be modelled by placing an image charge -q at the mirror point below the plane. The potential from real and image charges is zero on the plane, satisfying the grounded condition; uniqueness then guarantees this is the correct physical potential in the region above the plane. The method is limited to geometries where appropriate images can be placed, such as infinite planes and certain spheres.

Use and limitations While the method of images yields elegant solutions for a few classical problems, general boundaries require advanced techniques or numerical methods. Nevertheless, uniqueness gives confidence that clever constructions or symmetry-based trial solutions are legitimate if they meet the equations and boundaries. For Class 12, image-method problems and reasoning about boundary conditions appear occasionally and form useful extensions to Gauss's law and potential concepts.

📌 Examples
  • Explain how an image charge of opposite sign located mirror-wise produces V=0 on infinite grounded conducting plane for a point charge above it.
  • Use image method to compute force on a point charge near grounded conducting plane by treating as attraction to image charge.
  • State why uniqueness theorem assures correctness of solution once boundary conditions are met.
🧮 Formulas
  1. Poisson's equation: ∇^2 V = -ρ/ε0 (conceptual)
  2. Laplace's equation (in charge-free region): ∇^2 V = 0
📊 Visual ideas
Diagram of point charge above conducting plane and its image below plane, showing field lines and induced charges.
Sketch of equipotentials near a grounded plane and a single charge showing V=0 at plane.

Key Concepts

Electric charge
A property of matter that causes it to experience a force in an electric field, quantized in units of e.
Coulomb's law
The force between two point charges is F = (1/4πε0) q1 q2 / r^2 along the line joining them.
Superposition principle
Net electric force or field is the vector sum of contributions from individual charges.
Electric field
Force per unit positive test charge at a point, E = F/q.
Electric flux
Surface integral of E·dA giving measure of field lines passing through a surface.
Gauss's law
Net flux through a closed surface equals enclosed charge divided by ε0: ∮E·dA = Q_enclosed/ε0.
Electric potential
Work done per unit positive charge moving from infinity to a point, V = -∫E·dl.
Equipotential surface
A surface on which the electric potential is the same at every point.
Dipole moment
Vector p = q × separation (from negative to positive) characterising an electric dipole.
Capacitance
The ratio of charge stored to potential difference: C = Q/V.
Dielectric constant
Relative permittivity κ = ε/ε0 describing how a dielectric increases capacitance.
Electric potential energy
Energy of a charge configuration due to electrostatic interactions, e.g., U = (1/4πε0) q1 q2 / r for two charges.
Conductors in electrostatic equilibrium
Conductors have no internal electric field and their entire bulk is at constant potential with excess charge on surface.
Surface charge density
Charge per unit area on a surface, denoted σ.
Electric displacement
Vector D = ε0 E + P used to relate free charge to field in presence of dielectrics.
Maxwell stress (qualitative)
Tensor method to compute electromagnetic forces on bodies by integrating stress over surfaces.

Practice Questions

  1. Calculate the force between two point charges +3 μC and -6 μC separated by 5 cm. / दो बिंदु आवेश +3 μC और -6 μC, जो 5 सेमी अलग हैं, के बीच बल की गणना कीजिए।
    Show answer

    Use Coulomb's law: F = (1/4πε0) |q1 q2| / r^2. Here q1 = 3×10^-6 C, q2 = -6×10^-6 C, r = 0.05 m. Magnitude F = (8.988×10^9)(3×10^-6×6×10^-6)/(0.05^2) = (8.988×10^9)(18×10^-12)/(0.0025) ≈ (8.988×10^9)(7.2×10^-9) ≈ 64.7 N. Direction: attractive (since charges opposite) along line joining them. / कुल्हाड़ा का नियम लगाएँ: F = (1/4πε0) |q1 q2| / r^2. q1 = 3×10^-6 C, q2 = -6×10^-6 C, r = 0.05 m. परिमाण F ≈ 64.7 N. दिशा: परस्पर आकर्षित करेंगी।

  2. A point charge q is placed at centre of a hollow conducting spherical shell. What is the field inside the conducting material and what charges appear on inner and outer surfaces? / एक बिंदु आवेश q को खोखले चालक गोलाकार खोल के केंद्र में रखा गया है। चालक के भीतर क्षेत्र क्या होगा और अंदर तथा बाहर की सतहों पर कौन से आवेश प्रकट होंगे?
    Show answer

    Field inside the conducting material is zero (E = 0) because conductor in electrostatic equilibrium has no internal field. The inner surface acquires charge -q to cancel field inside conductor, while the outer surface acquires charge +q so that total charge on conductor equals induced charges and preserves overall charge neutrality if shell initially neutral. / चालक के भीतर क्षेत्र शून्य होता है। आंतरिक सतह पर -q आवेश उत्पन्न होता है और बाहरी सतह पर +q आवेश आता है ताकि कुल आवेश संतुलित रहे।

  3. Use Gauss's law to find electric field at distance r from an infinite line charge with linear charge density λ. / एक अनंत रेखीय आवेश जिसकी रेखीय आवेश घनत्व λ है, से दूरी r पर विद्युत क्षेत्र Gauss के नियम का प्रयोग करके निकालिए।
    Show answer

    Choose a cylindrical Gaussian surface of radius r and length L coaxial with the line. Field is radial and constant on curved surface; flux = E(2π r L). Enclosed charge = λ L. By Gauss: E(2π r L) = λ L/ε0 ⇒ E = λ/(2π ε0 r) directed radially outward for λ>0. / सह-अक्षीय बेलनिक Gaussian सतह लें। ∮E·dA = E(2π r L) = λ L/ε0, अतः E = λ/(2π ε0 r) बाहर की दिशा में।

  4. Find potential at a point on the axis of a ring of radius a carrying charge Q, at distance x from centre. / त्रिज्या a वाले आवेश Q वाले रिंग के अक्ष पर केन्द्र से दूरी x पर बिंदु का विभव निकालिए।
    Show answer

    Every element of ring is at distance √(a^2 + x^2) from the point; potential contributions add scalar: V = (1/4πε0) ∫ dq / r = (1/4πε0) (Q / √(a^2 + x^2)) taking V(∞)=0. / रिंग के प्रत्येक तत्व की दूरी √(a^2 + x^2) है, अतः V = (1/4πε0)(Q/√(a^2 + x^2)).

  5. Two identical capacitors each of capacitance C are connected in series and charged by a battery of voltage V. Find charge on each capacitor and total energy stored. / दो समान धारिता वाले कैपेसिटर, प्रत्येक C, श्रृंखला में जुड़े हैं और V वोल्ट की बैटरी से चार्ज किए गए हैं। प्रत्येक कैपेसिटर पर आवेश और कुल संग्रहीत ऊर्जा निकालिए।
    Show answer

    In series both capacitors have same charge Q. Equivalent capacitance C_eq = C/2. Charge Q = C_eq V = (C/2) V. So each capacitor has charge Q = (C V)/2. Energy stored U = (1/2) C_eq V^2 = (1/2)(C/2)V^2 = (1/4) C V^2. Alternatively sum energies of individual capacitors each (1/2) C (V/2)^2 = 2 × (1/2) C (V^2/4) = (1/4) C V^2. / श्रृंखला में समान आवेश Q रहता है; C_eq = C/2 ⇒ Q = (C/2) V. प्रत्येक पर Q = (C V)/2। कुल ऊर्जा U = (1/4) C V^2।

  6. A dipole of moment p is placed in uniform electric field E making angle θ with p. Find torque and potential energy. / क्षण p वाला एक डाइपोल एक समान इलेक्ट्रिक क्षेत्र E में रखा है, जो p के साथ कोण θ बनाता है। टॉर्क और संभावित ऊर्जा निकालिए।
    Show answer

    Torque magnitude τ = pE sinθ and direction tends to rotate dipole to align p with E (vector τ = p × E). Potential energy U = -p · E = -p E cosθ; minimum when p parallel to E (θ = 0) and maximum when anti-parallel (θ = π). / टॉर्क τ = pE sinθ (दिशा p × E)। संभावित ऊर्जा U = -pE cosθ।

  7. Calculate capacitance of parallel plate capacitor with plate area 0.01 m^2 and separation 2 mm in vacuum. / प्लेट क्षेत्र 0.01 m^2 और पृथक दूरी 2 mm वाले वैक्यूम में परैलल प्लेट कैपेसिटर की धारिता निकालिए।
    Show answer

    Use C = ε0 A / d. ε0 = 8.854×10^-12 F/m, A = 0.01 m^2, d = 2×10^-3 m. So C = (8.854×10^-12)(0.01)/(2×10^-3) = (8.854×10^-14)/(2×10^-3) = 4.427×10^-11 F ≈ 44.3 pF. / C = ε0 A / d ⇒ C ≈ 4.43×10^-11 F ≈ 44.3 pF।

  8. A small test charge q0 experiences force F at a point. Explain how to obtain electric field at that point and the potential difference between two nearby points A and B. / एक छोटे परीक्षण आवेश q0 को किसी बिंदु पर बल F अनुभव होता है। उस बिंदु पर विद्युत क्षेत्र कैसे निकालेंगे और दो निकटवर्ती बिंदु A और B के बीच विभवांतर कैसे पाएँगे?
    Show answer

    Electric field E at that point is E = F / q0 directed along force on positive test charge; use sign accordingly if q0 negative. Potential difference V(B)-V(A) = -∫_A^B E · dl. For small separation approximate ΔV ≈ -E · Δl if E nearly constant between A and B. / E = F/q0। विभवांतर V(B)-V(A) = -∫_A^B E·dl; यदि छोटे निकटतम बिंदु हैं और E लगभग स्थिर है तो ΔV ≈ -E·Δl।

  9. Explain qualitatively why the electric field inside a hollow conductor is zero and what happens when an external charge is brought near the conductor. / गुणात्मक रूप से बताइए कि एक खोखले चालक के अंदर विद्युत क्षेत्र शून्य क्यों है और जब चालक के पास एक बाहरी आवेश लाया जाता है तो क्या होता है।
    Show answer

    In electrostatic equilibrium free charges in conductor move until internal field is zero; if any internal field existed charges would flow. Thus interior region is field-free and conductor is equipotential. When an external charge is brought near, charges in conductor redistribute: opposite sign induced on nearest surface and like sign on far surface, creating internal fields that cancel external field within conductor. If conductor is grounded, excess like charge can flow to ground leaving induced opposite charge on surface. / चालक के अंदर स्वतंत्र आवेश इस तरह व्यवस्थित हो जाते हैं कि आंतरिक क्षेत्र शून्य हो; अन्यथा आवेश बहते रहते। बाहरी आवेश के आने पर सतहों पर प्रेरित आवेश उत्पन्न होते हैं जो अंदर के क्षेत्र को शून्य कर देते हैं।

Related Laws & Principles

Explore all

Foundational laws & principles connected to this chapter — tap to open in the Laws Explorer.

Loading related laws…
Sourced from 0 content files · LLOS Learn · browse all chapters