These electric charges and fields class 12 notes compress Chapter 1 of the NCERT Physics textbook into one revision page. Reference: NCERT Class 12 Physics textbook, chapter Electric Charges and Fields.
The page follows a teacher’s order: charging first, then the three properties of charge, Coulomb’s law, the electric field, field lines, flux, the dipole, and finally Gauss’s law with its applications. Every formula below carries its symbols and SI units, and four worked examples use fresh numbers so you practise the method, not memory.
Electric Charges and Fields Class 12 Notes: Chapter Map
The chapter builds in a strict order: charge basics first, then the force law, then the field picture, then flux and Gauss’s law. Every later idea sits on the earlier ones.
| Topic | What it gives you | NCERT pages |
|---|---|---|
| Electrostatics and charging | Why bodies become electrified; two kinds of charge | pp. 1–3 |
| Basic properties of charge | Additivity, conservation, quantisation | pp. 4–6 |
| Coulomb’s law | Force between two point charges | pp. 6–10 |
| Superposition of forces | Net force from many charges | pp. 11–13 |
| Electric field | Field picture and point-charge field | pp. 14–18 |
| Field lines | Mapping the field pictorially | pp. 19–21 |
| Electric flux | Count of field lines through a surface | pp. 21–22 |
| Electric dipole | Field of two opposite charges, torque | pp. 23–27 |
| Continuous charge distributions | Line, surface and volume densities | pp. 28–29 |
| Gauss’s law and applications | Wire, sheet and shell fields | pp. 29–36 |
Every derivation in this chapter rests on Coulomb’s law plus the superposition principle. You can verify any page reference or formula against the official NCERT Class 12 Physics Part I PDF.
Charging by Rubbing: How Bodies Become Electrified
Around 600 BC, Thales of Miletus found that amber rubbed with wool attracts light objects. Our word electricity comes from elektron, the Greek word for amber (NCERT, p. 1).
Experiments show there are exactly two kinds of charge. Like charges repel; unlike charges attract. Benjamin Franklin named them: a glass rod rubbed with silk is positive, and the silk is negative (NCERT, p. 2).
Rubbing transfers electrons, nothing else. A glass rod loses electrons to silk, so the rod becomes positive and the silk negative. No new charge is created (NCERT, p. 3).
The gold-leaf electroscope (Fig. 1.2) detects charge: a charged object touches the metal knob, charge flows to two thin gold leaves, and they diverge. The divergence shows the amount of charge (NCERT, p. 3).
Why do some bodies charge and others not? It comes down to how free the electrons are (NCERT, pp. 3–4):
- Conductors (metals, human body, earth) have electrons comparatively free to move; a charge spreads over the whole surface.
- Insulators (glass, plastic, nylon, wood) hold charge where it was placed.
- A plastic comb charges because plastic is an insulator. A metal spoon does not because the charge leaks through your hand to the earth — both are conductors.
This is the everyday physics behind sweater sparks in dry weather, car-door shocks, and a charged comb picking up paper bits.
The Three Basic Properties of Electric Charge
All charge obeys three quantitative rules. Remember them in order with the device “Quantise, Add, Conserve” (QAC).
- Additivity: total charge of a system is the algebraic sum, with proper signs. Five charges +1, +2, −3, +4, −5 add to −1 in arbitrary units (NCERT, p. 4).
- Conservation: total charge of an isolated system stays constant. Rubbing only transfers electrons. Even when a neutron turns into a proton plus an electron, the total charge is zero before and after (NCERT, pp. 4–5).
- Quantisation: charge always comes in steps of \( e \):
\[ q = ne \]
where \( n \) is an integer, positive or negative, and \( e = 1.6 \times 10^{-19}\ \text{C} \) (NCERT, p. 5). About \( 6 \times 10^{18} \) electrons make −1 C. Smaller units are common in electrostatics: \( 1\ \mu\text{C} = 10^{-6}\ \text{C} \) and \( 1\ \text{mC} = 10^{-3}\ \text{C} \).
Why is quantisation invisible in daily life? A dotted line viewed from a distance looks continuous. At the macroscopic scale, charges of a few \( \mu\text{C} \) contain roughly \( 10^{13} \) times \( e \), so the grainy steps are lost and charge appears continuous (NCERT, pp. 5–6).
One coulomb is a very large unit in electrostatics. NCERT Example 1.1 shows that collecting 1 C would take roughly 200 years if \( 10^9 \) electrons move out every second (NCERT, p. 6). The coulomb itself is defined through the ampere, tying this chapter to the current electricity notes.
Coulomb’s Law: Force Between Two Point Charges
Coulomb measured that the electrostatic force between two point charges varies inversely as the square of the distance and directly as the product of the magnitudes, acting along the line joining them (NCERT, p. 6). A point charge means the linear size of the body is much smaller than the separation.
Scalar form:
\[ F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2} \]
Vector form (NCERT, p. 8):
\[ \mathbf{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{\mathbf{r}}_{21} \]
Symbols: \( q_1, q_2 \) in coulomb, \( r \) in metre, \( F \) in newton. Constants: \( k \approx 9 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2} \) and \( \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\ \text{N}^{-1}\ \text{m}^{-2} \), the permittivity of free space (NCERT, p. 8).
The vector form handles signs automatically: same-sign charges repel, opposite signs attract, and \( \mathbf{F}_{12} = -\mathbf{F}_{21} \), agreeing with Newton’s third law (NCERT, p. 8).
Coulomb used a torsion balance. To change charge he touched a charged sphere to an identical uncharged one, so by symmetry each got \( q/2 \) — a neat use of additivity and conservation (NCERT, p. 7).
The electric force is enormous compared with gravity: between an electron and a proton it is about \( 2.4 \times 10^{39} \) times the gravitational force (NCERT, p. 9).
Rule: use absolute values in the scalar form and decide attraction or repulsion by the signs. Never substitute signed charges into the magnitude formula.
Superposition Principle: Net Force From Many Charges
The force on a charge due to several others is the vector sum of the forces due to each charge taken one at a time. The presence of other charges does not alter any individual pair force (NCERT, p. 11).
This is not an obvious rule. It asserts two things: each pair force is unchanged, and there are no extra three-body or four-body forces (NCERT, p. 41).
For \( n \) charges, the net force on \( q_1 \) is (NCERT, Eq. 1.5, p. 12):
\[ \mathbf{F}_1 = \frac{q_1}{4\pi\varepsilon_0} \sum_{i=2}^{n} \frac{q_i}{r_{1i}^2} \hat{\mathbf{r}}_{1i} \]
The classic pattern is the equilateral triangle. Three equal charges \( q \) at the vertices put a charge \( Q \) at the centroid: by symmetry the three forces are equal and their resultant is zero (NCERT, Example 1.5, p. 12). With charges \( q, q, -q \), the forces do not cancel and each charge feels a net force (NCERT, Example 1.6, p. 13).
Method: draw each pair force, resolve into components, and add vectorially with the parallelogram law. The diagram is the answer’s backbone.
Electric Field: Definition, Direction and the Field Picture
The field concept answers a puzzle: if a test charge is removed, is anything left at that point? The answer is yes — the source charge produces an electric field everywhere around it, and a test charge placed there feels its force (NCERT, p. 14).
Definition (NCERT, Eq. 1.9, p. 14):
\[ \mathbf{E} = \lim_{q \rightarrow 0} \left( \frac{\mathbf{F}}{q} \right) \]
The limit \( q \rightarrow 0 \) keeps the test charge from disturbing the source. The source charge \( Q \) stays fixed; the test charge \( q \) senses the field. The field is independent of \( q \) because \( \mathbf{F} \) is proportional to it (NCERT, pp. 14–15).
Field of a point charge (NCERT, Eq. 1.6, p. 14):
\[ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}} \]
Direction: radially outward for positive \( Q \), radially inward for negative \( Q \). Magnitude depends only on \( r \), so the field is the same on a sphere centred on the charge — this is spherical symmetry (NCERT, p. 15). Unit: \( \text{N/C} \).
Field of a system of charges is a vector sum point by point (NCERT, Eq. 1.10, p. 15).
Physical significance (favourite short answer, NCERT, p. 16): in electrostatics the field is convenient but not necessary. Its real value appears with time-dependent phenomena. A change in one charge reaches another only at speed \( c \), so there is a time delay between cause and effect.
The field picture — a moving charge sends out waves that propagate, reach the other charge and exert force — accounts for that delay. Fields carry energy and have their own dynamics. Faraday introduced the concept.
Electric Field Lines: The Four Rules
Draw arrows pointing along \( \mathbf{E} \), with lengths proportional to the field strength. Then connect arrows that point the same way — the resulting curve is a field line. The tangent at any point gives the direction of \( \mathbf{E} \); an arrow fixes the direction; the closeness of lines shows the strength (NCERT, p. 19).
Four properties always hold (NCERT, p. 21):
- Lines start on positive charges and end on negative charges, or go to infinity for a single charge.
- In a charge-free region, lines are continuous with no breaks.
- Two lines never cross — the field would then have two directions at one point.
- Electrostatic field lines never form closed loops; that follows from the conservative nature of the field, studied in the electrostatic potential and capacitance chapter notes.
Because the number of lines through a given solid angle is constant, the line density falls as \( 1/r^2 \) — matching the inverse-square law (NCERT, p. 20). Faraday invented the picture and called them lines of force, but field line is the better name (NCERT, p. 20).
Memory device: “Start at +, Stop at −, never Cross, never Loop.”
Electric Flux and the Three Charge Densities
Flux \( \Delta\phi \) through a small area element (NCERT, Eq. 1.11, p. 22):
\[ \Delta\phi = \mathbf{E} \cdot \Delta\mathbf{S} = E\,\Delta S \cos\theta \]
where \( \theta \) is the angle between \( \mathbf{E} \) and the area-vector normal. For a closed surface, the area vector points along the outward normal by convention (NCERT, p. 22). Unit: \( \text{N m}^2/\text{C} \).
Fishing net analogy: flux is like the number of arrows passing through a net held in the field. Hold it straight on and maximum arrows pass. Tilt it (change \( \theta \)) and fewer pass. Hold it edge-on (\( \theta = 90^\circ \)) and none pass at all — exactly what \( E\,\Delta S\cos\theta \) says.
For continuous charge, a macroscopically small element still contains a huge number of charges, so the charge looks continuous. The three densities (NCERT, pp. 28–29):
| Density | Definition | Unit |
|---|---|---|
| Linear, \( \lambda \) | \( \Delta Q = \lambda\,\Delta l \) | \( \text{C/m} \) |
| Surface, \( \sigma \) | \( \Delta Q = \sigma\,\Delta S \) | \( \text{C/m}^2 \) |
| Volume, \( \rho \) | \( \Delta Q = \rho\,\Delta V \) | \( \text{C/m}^3 \) |
The field of a continuous distribution is found by summing Coulomb’s-law contributions from every element (NCERT, Eq. 1.27, p. 29).
Electric Dipole: Field, Dipole Moment and Torque
An electric dipole is a pair of equal and opposite charges \( q \) and \( -q \) separated by \( 2a \) (NCERT, p. 23). Its total charge is zero, but its field is not — the two fields never cancel exactly because the charges are separated.
The dipole moment (NCERT, Eq. 1.19, p. 24) is a vector of magnitude \( q \times 2a \), directed from \( -q \) to \( q \):
\[ \mathbf{p} = q \times 2a \,\hat{\mathbf{p}} \]
At large distances (\( r \gg a \)) the dipole field takes simple forms (NCERT, pp. 24–25):
- On the axis: \( \mathbf{E} = \dfrac{2\mathbf{p}}{4\pi\varepsilon_0 r^3} \), along \( \mathbf{p} \).
- On the equatorial plane: \( \mathbf{E} = -\dfrac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \), opposite \( \mathbf{p} \).
Key contrast: a point-charge field falls as \( 1/r^2 \), a dipole field as \( 1/r^3 \). The dipole field decays faster because the fields of \( q \) and \( -q \) nearly cancel at large distance.
| Property | Point charge \( Q \) | Dipole \( \mathbf{p} \) |
|---|---|---|
| Large-distance fall-off | \( 1/r^2 \) | \( 1/r^3 \) |
| Direction | Radial, only in/out | Depends on angle; axis vs equator differ |
| Depends on | \( Q \) and \( r \) | \( \mathbf{p} \), \( r \), and the angle with \( \mathbf{p} \) |
In a uniform field, the net force on a dipole is zero, but the two forces act at different points, producing a torque (NCERT, Eq. 1.22, p. 27):
\[ \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}, \quad \tau = pE\sin\theta \]
The torque rotates the dipole to align \( \mathbf{p} \) with \( \mathbf{E} \); when aligned, the torque is zero. In a non-uniform field a net force appears: parallel to \( \mathbf{E} \) the dipole is pulled toward stronger field, antiparallel toward weaker (NCERT, p. 27).
Real-life application (NCERT, pp. 27–28): a comb run through dry hair picks up paper bits. The comb is charged by friction; the paper is not. The comb polarises the paper — induces a net dipole moment in the field direction — and because the comb’s field is non-uniform, the induced dipole feels a net force toward the comb.
The same polarisation-and-pull mechanism attracts any uncharged light object to a charged one.
Some molecules have a permanent dipole: water is polar. Others, like \( \text{CO}_2 \) and \( \text{CH}_4 \), have zero dipole moment in the absence of a field (NCERT, p. 25).
Gauss’s Law and Its Three Symmetric Applications
Take a sphere of radius \( r \) enclosing a point charge \( q \) at its centre. \( \mathbf{E} \) is radial and constant in magnitude over the sphere, so the flux is \( E \times 4\pi r^2 = \dfrac{q}{4\pi\varepsilon_0 r^2} \times 4\pi r^2 = \dfrac{q}{\varepsilon_0} \) (NCERT, p. 30). This simple case generalises to Gauss’s law (NCERT, Eq. 1.31, p. 30):
\[ \Phi = \frac{q}{\varepsilon_0} \]
The total electric flux through any closed surface equals the enclosed charge divided by \( \varepsilon_0 \). Key rules (NCERT, pp. 30–31):
- The law holds for any closed surface, whatever its shape.
- \( q \) on the right is only the charge inside; the field on the left is due to all charges, inside and outside.
- The chosen surface is the Gaussian surface — do not let it pass through a discrete charge.
- Choose a surface matched to the symmetry; the law is a test of the inverse-square behaviour.
| Configuration | Gaussian surface | Result |
|---|---|---|
| Infinite line, linear density \( \lambda \) | Coaxial cylinder | \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \), radial |
| Infinite plane sheet, density \( \sigma \) | Pillbox (parallelepiped) | \( E = \dfrac{\sigma}{2\varepsilon_0} \), normal to sheet, distance-independent |
| Thin spherical shell, charge \( q \), radius \( R \) | Concentric sphere | \( E = \dfrac{q}{4\pi\varepsilon_0 r^2} \) for \( r \geq R \); \( E = 0 \) for \( r \lt R \) |
Inside a uniformly charged spherical shell the field is zero at every interior point, not just at the centre — the Gaussian sphere inside encloses no charge (NCERT, pp. 35–36). The wire and sheet results are exact for infinite objects and approximate for the middle of long or large ones (NCERT, pp. 34–35).
See all of these fields applied again in the Class 12 Physics notes.
Physics Terms and Definitions: Quick Reference Table
This chapter’s vocabulary reappears in every later chapter of electrostatics. Revise it in one pass.
| Term | Meaning | Example |
|---|---|---|
| Electrostatics | Study of forces, fields and potentials from static charges | Sweater sparks on a dry day |
| Point charge | Charged body whose size is far smaller than the distances to other charges | Metal spheres in a torsion balance |
| Conductor | Material with electrons comparatively free to move | Metals, human body, earth |
| Insulator | Material that resists the passage of charge | Glass, plastic, nylon, wood |
| Quantisation of charge | Charge is an integral multiple of \( e \): \( q = ne \) | 20 electrons give \( -3.2 \times 10^{-18}\ \text{C} \) |
| Additivity of charge | Total charge is the algebraic sum of all charges | +1, +2, −3, +4, −5 give −1 |
| Conservation of charge | Total charge of an isolated system stays constant | Rubbing transfers electrons, creates none |
| Source charge | Charge that produces the field under study | A glass rod in a field calculation |
| Test charge | Small charge that senses the field without disturbing it | \( q \) in \( \mathbf{E} = \mathbf{F}/q \), taken to zero |
| Electric field | Force per unit positive test charge at a point | \( 9 \times 10^9\ \text{N/C} \) at 1 m from 1 C |
| Electric field line | Curve whose tangent gives the field direction | Radial lines leaving a positive charge |
| Electric flux | Count of field lines through a surface: \( \mathbf{E} \cdot \Delta\mathbf{S} \) | \( 1.7 \times 10^5\ \text{N m}^2/\text{C} \) in Example 4 |
| Electric dipole | Pair of equal and opposite charges separated by \( 2a \) | \( \pm 2\ \mu\text{C} \) separated by 4 cm |
| Dipole moment | \( \mathbf{p} = q \times 2a\,\hat{\mathbf{p}} \), from \( -q \) to \( q \) | \( 8 \times 10^{-8}\ \text{C m} \) in Example 3 |
| Gaussian surface | Imaginary closed surface chosen for Gauss’s law | Concentric sphere around a shell |
| Charge densities | \( \lambda, \sigma, \rho \): charge per unit length, area, volume | Wire with \( \lambda = 5\ \mu\text{C/m} \) |
Formula Sheet: Symbols, Units and When to Use Each
Every examinable formula of the chapter in one table. Verify each against the pages cited before the exam.
| Formula | What each symbol means | SI unit | Page |
|---|---|---|---|
| \( q = ne \) | \( q \) total charge, \( n \) integer, \( e \) basic charge | C | p. 5 |
| \( F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2} \) | \( q_1, q_2 \) charges, \( r \) separation | N | p. 8 |
| \( \mathbf{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{21}^2}\hat{\mathbf{r}}_{21} \) | \( \hat{\mathbf{r}}_{21} \) unit vector from 1 to 2 | N | p. 8 |
| \( \mathbf{E} = \mathbf{F}/q \) | \( \mathbf{F} \) force, \( q \) test charge | N/C | p. 14 |
| \( \mathbf{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\hat{\mathbf{r}} \) | \( Q \) source charge, \( r \) distance | N/C | p. 14 |
| \( \Delta\phi = \mathbf{E}\cdot\Delta\mathbf{S} = E\,\Delta S\cos\theta \) | \( \theta \) angle between \( \mathbf{E} \) and the normal | N m²/C | p. 22 |
| \( \mathbf{E}_{\text{axis}} = \dfrac{2\mathbf{p}}{4\pi\varepsilon_0 r^3} \) | \( \mathbf{p} \) dipole moment, \( r \gg a \) | N/C | p. 24 |
| \( \mathbf{E}_{\text{equator}} = -\dfrac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \) | \( r \gg a \), opposite \( \mathbf{p} \) | N/C | p. 24 |
| \( \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E} \), \( \tau = pE\sin\theta \) | \( \theta \) angle between \( \mathbf{p} \) and \( \mathbf{E} \) | N m | p. 27 |
| \( \Phi = q/\varepsilon_0 \) | \( q \) net charge enclosed by closed surface | N m²/C | p. 30 |
| \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \) | \( \lambda \) linear density, \( r \) perpendicular distance | N/C | p. 34 |
| \( E = \dfrac{\sigma}{2\varepsilon_0} \) | \( \sigma \) surface density | N/C | p. 34 |
| \( E = \dfrac{q}{4\pi\varepsilon_0 r^2}\ (r \geq R) \); \( E = 0\ (r \lt R) \) | \( q \) total shell charge, \( R \) shell radius | N/C | pp. 35–36 |
Constants box: \( e = 1.6 \times 10^{-19}\ \text{C} \); \( k \approx 9 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2} \); \( \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\ \text{N}^{-1}\ \text{m}^{-2} \).
Worked Examples With Original Numbers
Worked Example 1: Quantisation of Charge
Method: charge comes in whole steps of \( e \); find \( n = q/e \).
- Step 1: A body carries a net charge of \( -3.2 \times 10^{-18}\ \text{C} \).
- Step 2: Use magnitudes: \( n = \dfrac{3.2 \times 10^{-18}}{1.6 \times 10^{-19}} = 20 \).
Final answer: the charge is made of 20 electrons. The negative sign means 20 excess electrons on the body.
Worked Example 2: Coulomb’s Law Force and Direction
Method: convert units first, use the scalar law with absolute values, then state attraction or repulsion.
- Step 1: Convert: \( 4\ \mu\text{C} = 4 \times 10^{-6}\ \text{C} \), \( 6\ \mu\text{C} = 6 \times 10^{-6}\ \text{C} \), \( 30\ \text{cm} = 0.30\ \text{m} \).
- Step 2: Substitute:
\[ F = \frac{(9 \times 10^9)(4 \times 10^{-6})(6 \times 10^{-6})}{(0.30)^2} = \frac{9 \times 10^9 \times 24 \times 10^{-12}}{0.09} = 2.4\ \text{N} \]
Final answer: the force has a magnitude of 2.4 N and is attractive, because the charges are unlike. Doubling the separation would quarter the force — a quick inverse-square check.
Worked Example 3: Dipole Field on Its Axis
Method: axial far-field formula, valid because \( r \gg a \).
Step 1: A dipole has charges \( \pm 2\ \mu\text{C} \) separated by 4 cm, so \( 2a = 0.04\ \text{m} \).
Dipole moment:
\[ p = (2 \times 10^{-6})(0.04) = 8 \times 10^{-8}\ \text{C m} \]
Step 2: Field on the axis at \( r = 0.20\ \text{m} \) from the centre:
\[ E = \frac{2kp}{r^3} = \frac{2(9 \times 10^9)(8 \times 10^{-8})}{(0.20)^3} = \frac{1.44 \times 10^3}{8 \times 10^{-3}} = 1.8 \times 10^5\ \text{N/C} \]
Final answer: \( 1.8 \times 10^5\ \text{N/C} \) along the dipole moment. The far-field condition holds: 20 cm ≫ 4 cm.
Worked Example 4: Flux Through a Closed Surface by Gauss’s Law
Method: flux depends only on the net enclosed charge, never the surface shape.
Step 1: A closed surface encloses \( +2\ \mu\text{C} \) and \( -0.5\ \mu\text{C} \).
Net charge:
\[ q = (+2\ \mu\text{C}) + (-0.5\ \mu\text{C}) = 1.5\ \mu\text{C} = 1.5 \times 10^{-6}\ \text{C} \]
Step 2: Apply Gauss’s law:
\[ \Phi = \frac{q}{\varepsilon_0} = \frac{1.5 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.7 \times 10^5\ \text{N m}^2/\text{C} \]
Final answer: \( 1.7 \times 10^5\ \text{N m}^2/\text{C} \). The flux is set entirely by the net enclosed charge — a cube, sphere or any other closed surface gives the same value.
Common Mistakes Students Make (With Corrections)
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing \( F \propto 1/r \) | \( F \propto 1/r^2 \) — the inverse-square dependence is Coulomb’s law | Double \( r \); the force must fall to \( 1/4 \) |
| Substituting signed charges into the scalar magnitude formula | Use absolute values in \( |q_1 q_2| \); signs decide attraction or repulsion | Work with magnitudes, then state “attractive/repulsive” separately |
| Claiming a charge of \( 2.4 \times 10^{-19}\ \text{C} \) exists | Charge is quantised: \( q = ne \), \( n \) a whole number | \( 2.4/1.6 = 1.5 \), not an integer → impossible |
| Field inside a shell zero only at the centre | Gauss’s law gives \( E = 0 \) at every interior point | A Gaussian sphere anywhere inside the shell encloses no charge |
| Counting outside charges in the \( q \) of Gauss’s law | \( q \) is only the net charge inside the surface | The field on the surface is due to all charges, but \( q \) counts only enclosed ones |
| Field lines form closed loops | Electrostatic field lines never close — they run from \( + \) to \( – \) | A closed loop would imply a non-conservative field, which electrostatics forbids |
| Unit of flux written as N/C | Flux is \( \text{N m}^2/\text{C} \); N/C is the unit of the field | Dimensions of \( \mathbf{E}\cdot\Delta\mathbf{S} \): \( (\text{N/C})\times\text{m}^2 \) |
| Using \( E = \sigma/(2\varepsilon_0) \) at the edge of a finite sheet | Exact only for an infinite sheet; approximate in the middle of a large sheet | Near the ends the field is not normal to every face of the pillbox |
Exam Notes: Steps That Earn the Mark
Observed marking patterns for this chapter — no year guarantees, just what a careful examiner usually looks for.
- In every force or field question, state the direction of each vector before adding. The direction statement earns the mark even if the arithmetic slips.
- Superposition questions: drawing the vector diagram and resolving along axes is the step that carries the marks.
- Gauss’s law questions: naming the symmetry (spherical, cylindrical, planar) and the Gaussian surface used is a required step. A bare formula without the surface choice often loses a mark.
- Dipole derivations: the examinable step is the far-field approximation \( r \gg a \), which converts the exact expression into \( 2\mathbf{p}/(4\pi\varepsilon_0 r^3) \) on the axis and \( -\mathbf{p}/(4\pi\varepsilon_0 r^3) \) on the equator.
- Numerical hygiene: converting cm → m and \( \mu\text{C} \) → C before substituting is the first credit step.
- Scalar form of Coulomb’s law: magnitude with absolute values plus a separate “attractive/repulsive” statement is safer than signing the force.
Mapping of the chapter’s standard derivations to their pages, with the step that carries the mark:
| Derivation | Textbook pages | Step that carries the mark |
|---|---|---|
| Field of a point charge | pp. 14–15 | Take the test charge \( q \rightarrow 0 \) in \( \mathbf{E} = \mathbf{F}/q \) |
| Dipole field on the axis | pp. 23–24 | Apply \( r \gg a \) to get \( E = 2p/(4\pi\varepsilon_0 r^3) \) |
| Dipole field on the equator | p. 24 | State the axial components cancel, giving \( E = -p/(4\pi\varepsilon_0 r^3) \) |
| Gauss’s law flux through a sphere | pp. 29–30 | Recognise \( E \) is constant, so flux \( = E \times 4\pi r^2 \) |
| Field of an infinite wire | pp. 33–34 | Choose a coaxial cylinder; state flux through the flat ends is zero |
| Field of an infinite sheet | p. 34 | Choose a pillbox; state flux is only through the two faces, \( 2EA = \sigma A/\varepsilon_0 \) |
| Field of a spherical shell | pp. 35–36 | For \( r \lt R \), state the Gaussian surface encloses no charge, so \( E = 0 \) |
Figure Walkthrough: Reading Field Line and Gaussian Surface Diagrams
Exam questions often refer to these printed figures. Know what each one is telling you.
- Fig. 1.12 (p. 19): field of a point charge. The arrows get shorter with distance and point radially outward. When connected, their density — not their length — becomes the strength indicator.
- Fig. 1.13 (p. 20): equal solid angles subtend larger areas at greater distance, so the same number of lines crosses larger areas. Lines per unit area thus fall as \( 1/r^2 \).
- Fig. 1.14 (p. 21): the four simple configurations — a single positive charge (radial outward), a single negative charge (radial inward), two like charges (lines bend away in repulsion), and a dipole (lines run from \( + \) to \( – \), showing attraction).
- Figs. 1.26–1.28 (pp. 33–35): the Gaussian surface must match the symmetry — a coaxial cylinder for the wire (flux only through the curved face), a pillbox for the sheet (flux through the two flat faces), a concentric sphere for the shell (flux over the whole sphere).
Electric Charges and Fields Class 12 Notes: Last-Minute Recap
The whole chapter on one screen, night-before format:
- Two kinds of charge; like repel, unlike attract (Franklin’s convention).
- Three properties: Quantise, Add, Conserve (QAC).
- Coulomb’s law: \( F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2} \), inverse-square, \( \mathbf{F}_{12} = -\mathbf{F}_{21} \).
- Electric field: \( \mathbf{E} = \mathbf{F}/q \), point charge \( \propto Q/r^2 \), spherical symmetry, unit N/C.
- Four field-line rules: start at +, stop at −; continuous in free space; never cross; never loop.
- Flux: \( \Delta\phi = \mathbf{E}\cdot\Delta\mathbf{S} = E\,\Delta S\cos\theta \), outward normal convention, unit \( \text{N m}^2/\text{C} \).
- Dipole: \( \mathbf{p} = q \times 2a \), axis \( E = 2p/(4\pi\varepsilon_0 r^3) \), equator \( E = -p/(4\pi\varepsilon_0 r^3) \); falls as \( 1/r^3 \); torque \( \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E} \).
- Gauss’s law: \( \Phi = q/\varepsilon_0 \).
| Configuration | Field |
|---|---|
| Infinite wire | \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \) |
| Infinite sheet | \( E = \dfrac{\sigma}{2\varepsilon_0} \) |
| Spherical shell, outside | \( E = \dfrac{q}{4\pi\varepsilon_0 r^2} \) |
| Spherical shell, inside | \( E = 0 \) everywhere |
Constants box: \( e = 1.6 \times 10^{-19}\ \text{C} \); \( k = 9 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2} \); \( \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\ \text{N}^{-1}\ \text{m}^{-2} \).
Reference: NCERT Class 12 Physics textbook, chapter Electric Charges and Fields.
FAQs on Electric Charges and Fields
Why is 1 coulomb considered a very large unit of charge?
Because a single electron carries only \( 1.6 \times 10^{-19}\ \text{C} \), so 1 C is the charge of about \( 6 \times 10^{18} \) electrons. NCERT Example 1.1 shows it would take roughly 200 years to collect 1 C if \( 10^9 \) electrons move out every second (NCERT, p. 6).
Is a charge of \( 2.4 \times 10^{-19}\ \text{C} \) possible? Why or why not?
No. Charge is quantised: \( q = ne \) with \( n \) an integer (NCERT, p. 5). Since \( 2.4 \times 10^{-19} / 1.6 \times 10^{-19} = 1.5 \) is not a whole number, such a charge cannot exist.
Why is the electric field zero everywhere inside a uniformly charged spherical shell?
A concentric spherical Gaussian surface inside the shell encloses no charge, so Gauss’s law forces the flux, and hence \( E \), to zero at every interior point — not just the centre (NCERT, pp. 35–36).
What is the difference between electric field and electric flux?
The electric field \( \mathbf{E} \) is the force per unit positive test charge at a point (unit N/C). Electric flux \( \Delta\phi = \mathbf{E}\cdot\Delta\mathbf{S} \) is the count of field lines through a surface (unit \( \text{N m}^2/\text{C} \)).
The field is a local property at one point; flux is a sum over a surface (NCERT, pp. 21–22).
Does an electric dipole experience a net force in a uniform electric field?
No. The forces \( q\mathbf{E} \) and \( -q\mathbf{E} \) are equal and opposite, so the net force is zero; only a torque \( \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E} \) acts, rotating the dipole to align with the field (NCERT, p. 27). A net force appears only when the field is non-uniform.
Why can two electric field lines never cross each other?
At the crossing point the field would need two directions at once. The tangent to a field line gives the direction of \( \mathbf{E} \), so two tangents at one point would mean two different field directions — impossible (NCERT, p. 21).
For the wider revision set, see all Class 12 revision notes or start from the CBSE notes index.
Reference: NCERT Class 12 Physics textbook, chapter Electric Charges and Fields.
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