CLASS 12 PHYSICS · CHAPTER 1 · STUDY NOTES
Electric Charges and Fields
An in-depth exploration of electrostatics, covering the nature of electric charges, Coulomb's Law, the concept of electric fields, dipoles, and the application of Gauss's Law to symmetric charge distributions.
EXPLORE
Follow charge as it moves between materials
Explain charging, charge detection and why a conductor needs insulation.LOOK · PREDICT · COMPARE
Transfer, detect and retain charge
Both objects begin neutral. Count the three electrons that will move from glass to silk.
- Electrostatics
- The study of electric forces, fields and potentials associated with charges at rest.
Rubbing glass with silk transfers some electrons from glass to silk. Glass becomes positive and silk negative; no net charge is created. “Neutral” means balanced positive and negative charge, not an absence of charged particles.
A gold-leaf electroscope detects charge: charging the knob also charges its connected leaves, which repel. Excess charge can spread over a conductor; on an insulator it remains near where it was deposited.
Let’s look a little closer.
Go deeper: A discharge is a change from the static situation
A crackle or spark is a discharge of accumulated charge. During that event charge moves, so the event itself is not a static configuration. The illustrations compare selected before/after states, not a microscopic rubbing mechanism or measured time course.
A metal object held in a conducting hand can lose charge through the body to Earth. An insulating handle interrupts that path. In metals the mobile carriers are electrons; the conductor picture assumes an isolated body and electrostatic equilibrium. Earthing permits charge exchange with Earth, so the object alone is no longer an isolated system.
Take a moment. What do you think?
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Question
A neutral glass–silk pair is isolated. After rubbing, the silk has gained five electrons. What are the two net charges and their sum?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 1, 2, 3, 4.
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Count charge, then choose the right density
Distinguish signed addition, conservation, quantisation and continuous charge models.LOOK · PREDICT · COMPARE
From whole charges to a continuous density
Like charges repel. The forces are equal and opposite along the joining line.
- Quantisation of free charge
- The net charge of an ordinary body changes in whole multiples q = ne of the elementary charge magnitude e.
| Model quantity | Meaning | Unit |
|---|---|---|
| λ | Charge / length | C/m |
| σ | Charge / area | C/m² |
| ρ | Charge / volume | C/m³ |
Charge is a signed scalar: add charges algebraically. Like charges repel and unlike charges attract. An isolated system conserves total charge even when its charged particles are redistributed.
Use e ≈ 1.60 × 10⁻¹⁹ C. Microscopic charge changes come in whole-electron steps. For a macroscopic distribution, average over many particles: λ = ΔQ/Δℓ in C/m, σ = ΔQ/ΔA in C/m² and ρ = ΔQ/ΔV in C/m³.
Go deeper: A smooth density does not make electrons divisible
A small macroscopic line, area or volume element can still contain many charged particles. Continuous charge density is an averaging model, not the claim that elementary charges are physically smeared into fractions. For uniform density Q = λℓ, σA or ρV; for varying density sum or integrate the local elements.
For a volume element, ΔE = kρΔV r̂/r², where r points from that element to the observation point. Add these fields as vectors. Scalar character and conservation are different properties: one describes how charge adds; the other describes constancy of the isolated total over time.
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Question
A uniformly charged patch has σ = +3 μC/m² and area 0.020 m². Find its charge and explain why you multiply rather than divide.
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 2, 3, 4, 5, 6, 28, 29, 40, 41.
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Watch separation change the force
Calculate a signed interaction while keeping force and position scales separate.LOOK · PREDICT · COMPARE
A little more distance, much less force
In vacuum: q₁ = +2 μC, q₂ = +3 μC, r = 0.2 m. |F| = 1.3500 N. At twice the separation, the force magnitude is one quarter. Positions and force arrows use separate scales.
- Coulomb’s law in vacuum
- For two stationary point charges separated by r, the force magnitude is k|q₁q₂|/r² along their joining line.
In the model q₁ = +2 μC and |q₂| = 3 μC, with k ≈ 9.0 × 10⁹ N m²/C². At r = 0.20 m, each force has magnitude 1.35 N. Doubling r makes the force one quarter as large.
Reversing q₂ changes repulsion to attraction without changing the magnitude. The forces on the two charges are equal and opposite. A charge can therefore experience a different acceleration if its mass is different.
Go deeper: Use a vector for direction and a magnitude for size
Writing r̂₁₂ from charge 1 towards charge 2, the force on 2 is F₂←₁ = kq₁q₂r̂₁₂/r². A negative product reverses this direction. The point-charge approximation needs object dimensions small relative to separation; charges in matter require additional material response.
Coulomb compared separations and charge amounts with a torsion balance. Identical isolated conducting spheres can share an initial charge under a symmetric arrangement. Electrical and gravitational pair forces both have inverse-square distance dependence, but electric forces can attract or repel, whereas the gravitational pair force attracts.
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Question
In this model, move from 0.20 m to 0.40 m and reverse q₂. What changes in magnitude and direction?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 6, 7, 8, 9, 10, 11, 42.
EXPLORE
Separate the source field from the test-charge force
Add source fields as vectors and predict the force on either sign of test charge.LOOK · PREDICT · COMPARE
One location, several source fields
E points away from +. A positive test charge feels force along E. Changing the test charge does not change the source field.
- Electric field
- The force per unit positive test charge at a point, with the source configuration undisturbed; its SI unit is N/C.
For a point source Q, |E| = k|Q|/r²: outward from +Q, inward towards −Q. A small test charge feels F = qE. Reversing the test sign reverses F, while the source field stays the same.
For several fixed sources, add their individual field vectors at the same observation point. The superposition picture uses one shared arrow scale: the violet diagonal is E₁ + E₂, and orange is the force direction on the test charge.
Go deeper: Superposition preserves each contribution
For fixed source positions, introducing another source does not replace the individual Coulomb field. The net field is the vector sum; equal opposite contributions can cancel at a point, without either source disappearing. The test must be small enough not to rearrange an unconstrained source configuration.
Fields describe the surrounding electrical environment even when no test charge is present. Beyond electrostatics, changing electromagnetic disturbances propagate with a finite signal speed and can carry energy. That connection does not mean a static field is a three-stage conveyor from source to test charge.
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Question
Two source fields at P are 300 N/C right and 400 N/C up. What is the resultant magnitude, and which way does a −2 μC test-charge force point?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 11, 12, 13, 14, 15, 16, 17, 18, 19.
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Read a field map rather than a particle track
Interpret tangents, density, null points and uniform-field geometry.LOOK · PREDICT · COMPARE
Follow the local field direction
Lines follow the calculated two-charge field. Arrows run from + towards − or the edge of this view. Lines are a direction map, not particle tracks. They never cross in a charge-free region.
- Electric field line
- A curve whose directed tangent follows the local net electric field wherever that field has a direction.
Arrows go away from positive charge and towards negative charge, or continue beyond the pictured region. The closer the lines in a consistent map, the stronger the field; the number chosen for a drawing is not a measured particle count.
Electrostatic lines do not form closed loops or cross in a region with a well-defined nonzero field. Between equal positive sources, the midpoint has E = 0 and no field direction. A uniform field uses parallel, equally spaced lines.
Go deeper: A two-dimensional page represents a three-dimensional field
For a point charge, equal solid angles carry the same selected number of radial lines. The cross-sectional area grows as r², so relative line density decreases as 1/r². A fixed-size area farther away therefore intersects fewer lines; do not count a page drawing as literal flux.
A tangent gives an instantaneous force direction for a positive test charge, not its complete trajectory. A moving particle’s velocity and inertia also affect its path. The two-charge curves are calculated direction traces; singular charge centres and the zero-field midpoint are excluded from the trace.
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Question
A moving positive particle crosses a drawn field line with sideways velocity. Must it immediately turn and follow the line exactly?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 19, 20, 21, 41.
EXPLORE
Compare a dipole’s field with its response to another field
Distinguish axial/equatorial fields, torque, polarisation and unequal-force effects.LOOK · PREDICT · COMPARE
Zero total charge can still mean a field
On the axis beyond +q, the nearer positive charge gives the larger field: the net field is along p. Far away (r ≫ a), E ≈ 2kp/r³.
- Electric dipole moment
- p = q(2a), directed from −q to +q for a pair separated by 2a; its unit is C m.
A dipole has zero net charge but a nonzero field. Beyond the positive end, the nearer positive charge gives the larger contribution; at an equatorial point, transverse components cancel and the field points opposite p.
In a uniform external field, the pair feels equal opposite forces with zero net force. Their separated lines of action give τ = p × E, with magnitude pE sinθ. A non-uniform field can also give a net force.
Go deeper: Separate exact fields, far-field approximations and molecular moments
On the axis outside the pair, E = 2kpr/(r² − a²)² along p. In the equatorial plane its magnitude is kp/(r² + a²)^(3/2), opposite p. Only for r ≫ a do these become 2kp/r³ and kp/r³; the axial/equatorial factor of two is a far-field comparison, not universal at every r.
A permanent molecular dipole exists without an applied field, as in water; other molecules can acquire an induced moment. A charged comb can polarise neutral paper, and its non-uniform field can attract the nearer induced charge more strongly. Torque vanishes at both 0° and 180° in a uniform field; this does not make the antiparallel orientation stable.
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Question
An original dipole has p = 6 × 10⁻⁹ C m in a uniform field E = 2 × 10⁴ N/C at 90°. What are its net force and torque magnitude?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 23, 24, 25, 26, 27, 28, 42.
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Tilt an area, then close the surface
Compute signed flux and distinguish enclosed charge from every source of the field.LOOK · PREDICT · COMPARE
The surface normal decides the flux
θ is measured between E and the area normal n, not the surface. Φ/(EA) = cos θ = 0.50. The sign tells outward or inward orientation.
- Electric flux
- The surface integral of E·dA; for a flat area in a uniform field, Φ = EA cosθ with θ measured from its chosen normal.
Flux is a signed field–area quantity in N m²/C, not a literal fluid flow or the exact number of drawn lines. At θ = 90° it is zero; reversing the normal reverses its sign. Closed surfaces use outward normals.
Gauss’s law gives ∮E·dA = Q_enclosed/ε₀ for any closed surface. E includes contributions from inside and outside charges, but only net enclosed charge appears on the right. Zero net flux does not require zero field or an empty interior.
Go deeper: The law is general; the shortcut needs symmetry
A centred point charge gives E = kq/r² on a sphere and area 4πr², so their product is q/ε₀. Enlarging that sphere decreases E while increasing area by compensating factors. An outside charge contributes inward and outward flux whose integral cancels; equal line counts alone are not a quantitative proof.
For a varying field or curved surface, add small signed E·ΔA contributions and take the integral limit. Do not replace the integral with E times total area unless symmetry and normal alignment justify it. A Gaussian surface is imagined, not a conducting shell, and must not pass through a discrete point charge’s singular location.
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Question
A closed surface contains +7 nC and −7 nC, with another positive charge outside. What is its net flux, and must E vanish on the surface?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 21, 22, 23, 29, 30, 31, 32.
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Let symmetry choose the Gaussian surface
Derive and compare the wire, sheet and thin-shell fields without confusing their distance laws.LOOK · PREDICT · COMPARE
Different symmetry, different distance law
Field / reference field: 0.667 (reference: r = 1)
A coaxial Gaussian cylinder: radial E gives flux through the curved side only. End caps contribute zero. E(2πrℓ) = λℓ/ε₀.
- Gaussian surface
- An imagined closed surface used to apply Gauss’s law to a specified charge distribution.
| Uniform ideal source | Symmetry shortcut | Magnitude |
|---|---|---|
| Infinite wire | Coaxial cylinder; side contributes | |λ|/(2πε₀r) |
| Infinite sheet | Pillbox; two end faces contribute | |σ|/(2ε₀) |
| Thin shell exterior | Concentric sphere | k|Q|/r² |
| Thin shell interior | Symmetry and zero enclosed charge | 0 |
For a uniform infinite wire, a coaxial cylinder has flux E(2πrℓ) through its side and zero through its caps. With enclosed charge λℓ, |E| = |λ|/(2πε₀r). For a uniform infinite sheet, a pillbox has two contributing faces: 2EA = σA/ε₀.
The sheet gives |E| = |σ|/(2ε₀), independent of distance. A uniformly charged thin spherical shell gives E = 0 inside and |E| = k|Q|/r² outside. Use the graph to compare these distinct distance laws.
Go deeper: Zero enclosed charge alone does not prove zero local field
For the shell interior, spherical symmetry gives equal radial field magnitude over a concentric Gaussian sphere. That symmetry, together with zero enclosed charge, implies E = 0. This argument does not apply to an arbitrary empty Gaussian surface near external charges.
An infinite wire or sheet is an idealisation; finite versions approximate these results only where end or edge effects are negligible. A thin charged surface has different one-sided field limits; NCERT’s stated surface value adopts the outside limit. Do not apply the shell’s zero-interior result to a solid insulating sphere with volume charge: its enclosed charge generally changes with radius.
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Question
At twice the chosen reference distance, what fractions of the reference field remain for the infinite wire, infinite sheet and outside of a thin shell?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 28, 29, 33, 34, 35, 36, 37, 39, 40.