NEET Physics · NCERT Class 12

Electric Charges and Fields notes

Electric Charge & Coulomb's Law, Electric Field & Field Lines, Electric Flux, Electric Dipole, Gauss's Law and Applications of Gauss's Law — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 7 topics
  • 64 min read
  • 30 formulas
  • 24 PYQs
1

Chapter Overview

8 min

Electric Charges and Fields introduces electrostatics, the study of charges at rest and the fields produced by them. The chapter begins with properties of electric charge, quantization, conservation, conductors, insulators and Coulomb’s law. It then develops electric field, field lines and continuous charge distribution. Electric flux prepares the foundation for Gauss’s law, which relates net electric flux through a closed surface to the charge enclosed. Electric dipole is important for torque, potential energy and fields on axial and equatorial lines. Applications of Gauss’s law give electric fields due to line charge, plane sheet, spherical shell and conductors. For NEET, this chapter is high-yield because questions are formula-based, conceptual and symmetry-based.

Key points6

  • NEET commonly asks direct numerical questions from Coulomb’s law, electric field, dipole torque and Gauss’s law.
  • Electric field is a vector, so superposition requires vector addition.
  • Electric flux is not the number of field lines exactly, but it is proportional to field lines crossing a surface.
  • Gauss’s law is always true, but it is useful for finding electric field only when symmetry is high.
  • A closed Gaussian surface counts only enclosed charge, not charges outside it.
  • Conductors in electrostatic equilibrium have zero field inside and charge on the surface.

Memory tricks2

Chapter Flow Trick: Remember the flow: Charge makes Force, Force defines Field, Field gives Flux, Flux leads to Gauss.

Gauss Shortcut: Gauss counts only what is inside: flux through a closed surface depends on enclosed charge.

Examples2

Real-Life Example

A rubbed plastic comb attracts tiny paper bits because charges are induced in paper and electrostatic force pulls them.

NEET-Style Snapshot

If distance between two charges is doubled, electrostatic force becomes one-fourth because F ∝ 1/r².

Common mistakes2

Forgetting Vector Nature

Coulomb force and electric field are vectors. Add directions carefully, not only magnitudes.

Using Gauss's Law Without Symmetry

Gauss’s law is always valid, but direct E calculation requires symmetry so E can be taken constant on the Gaussian surface.

Reference tables2

Quick revision8

Formulas4

Coulomb's Law
F=kq1q2/r2=(1/4πε0)(q1q2/r2)F = kq1q2/r² = (1/4πε0)(q1q2/r²)

Magnitude of electrostatic force between two point charges separated by distance r.

F
Electrostatic force
k
Coulomb constant, 8.99 × 10⁹ N m² C⁻²
q1, q2
Point charges
r
Separation between charges
ε0
Permittivity of free space
Electric Field
E=F/q0E = F/q0

Force experienced per unit positive test charge.

E
Electric field intensity
F
Force on test charge
q0
Small positive test charge
2 more formulas — sign up free

Diagrams3

2

Electric Charge & Coulomb's Law

9 min

Electric charge is a fundamental property of matter responsible for electrostatic force. Charges are of two types: positive and negative. Like charges repel and unlike charges attract. Charge is quantized, meaning any observable charge is an integral multiple of elementary charge e. Charge is also conserved, so total charge of an isolated system remains constant. Conductors allow charges to move freely, while insulators do not. Coulomb’s law gives the force between two stationary point charges: F = kq1q2/r². The force acts along the line joining the charges. For multiple charges, the net force is found by the superposition principle, adding individual forces vectorially.

Key points6

  • Elementary charge e = 1.6 × 10⁻¹⁹ C.
  • A body cannot have charge 0.5e or 2.3e; charge must be ne.
  • Electrostatic force can be attractive or repulsive depending on signs of charges.
  • Coulomb’s law applies accurately to point charges or spherically symmetric charge distributions at large separation.
  • In superposition, force between any pair is unaffected by presence of other charges.
  • Permittivity of medium affects Coulomb force.

Memory tricks2

Charge Properties: Remember QCI: Quantized, Conserved, Invariant.

Coulomb Law: Force grows with charge product and falls with distance square: more charge means more force, more distance means much less force.

Examples3

Given Coulomb Numerical

For q1 = 3 μC, q2 = 6 μC and r = 4 m, F = kq1q2/r² = 8.99 × 10⁹ × 3 × 10⁻⁶ × 6 × 10⁻⁶ / 4² ≈ 0.0101 N.

Quantization Example

If a body has charge -3.2 × 10⁻¹⁹ C, then n = q/e = -2, meaning it has gained 2 electrons.

NEET-Type Shortcut

If both charges are doubled and distance is doubled, force becomes same because F ∝ q1q2/r² gives factor 4/4.

Common mistakes3

Forgetting Microcoulomb Conversion

1 μC = 10⁻⁶ C. Always convert charges to coulomb before using k = 9 × 10⁹.

Ignoring Direction in Superposition

Forces must be added as vectors. Opposite directions subtract; perpendicular directions need Pythagoras.

Using Coulomb Law for Large Irregular Bodies Directly

Coulomb’s law in simple form is for point charges. Extended bodies need distribution methods or symmetry.

Reference tables2

Quick revision8

Formulas4

Quantization of Charge
q=neq = ne

Any charge is an integral multiple of elementary charge.

q
Total charge
n
Integer number of electrons gained or lost
e
Elementary charge, 1.6 × 10⁻¹⁹ C
Coulomb's Law in Vacuum
F=kq1q2/r2=(1/4πε0)(q1q2/r2)F = kq1q2/r² = (1/4πε0)(q1q2/r²)

q1 and q2 are charges; r is distance. Gives the magnitude of electrostatic force between two point charges.

F
Magnitude of electrostatic force
k
8.99 × 10⁹ N m² C⁻²
q1
First charge in coulomb
q2
Second charge in coulomb
r
Distance between charges in metre
2 more formulas — sign up free

Diagrams3

3

Electric Field & Field Lines

9 min

Electric field describes the influence of a charge in the space around it. It is defined as force per unit positive test charge, E = F/q0. The electric field due to a positive point charge is radially outward, while that due to a negative charge is radially inward. Field is a vector, so for multiple charges the net field is found by vector superposition. Electric field lines are visual tools: their tangent gives field direction and their density indicates field strength. Lines never intersect and begin on positive charges and end on negative charges. Continuous charge distributions such as line, surface and volume charges require integration or Gauss’s law, depending on symmetry.

Key points6

  • Electric field exists even if no test charge is placed.
  • A test charge should be very small so it does not disturb the source charge distribution.
  • At any point, only one electric field direction exists, so field lines cannot cross.
  • Uniform electric field has parallel equally spaced field lines.
  • Electric field inside a conductor in electrostatic equilibrium is zero.
  • For continuous charge distributions, divide charge into small elements dq.

Memory tricks2

Field Line Direction: Field lines go from Plus to Minus: P to M means Positive to Negative.

Line Density: Crowded lines mean crowded force: stronger electric field.

Examples2

Solved Example

Field due to q = 2 μC at r = 0.3 m is E = kq/r² = 9 × 10⁹ × 2 × 10⁻⁶ / 0.09 = 2 × 10⁵ N/C.

Previous NEET-Type Question

At the midpoint between equal and opposite charges, electric fields due to both charges are in the same direction, from positive to negative.

Common mistakes2

Drawing Intersecting Field Lines

Field lines never intersect because electric field cannot have two directions at one point.

Confusing Field with Force

Electric field depends on source charge only; force also depends on the test charge.

Reference tables2

Quick revision8

Formulas5

Electric Field Intensity
E=F/q0E = F/q0

Defines electric field as force per unit positive test charge.

E
Electric field intensity
F
Force on test charge
q0
Positive test charge
Electric Field due to Point Charge
E=kq/r2=q/(4πε0r2)E = kq/r² = q/(4πε0r²)

Magnitude of field at distance r from a point charge q.

E
Electric field
k
Coulomb constant
q
Source charge
r
Distance from charge
Superposition Principle for Fields
Enet=E1+E2+E3+...E_{net} = E1 + E2 + E3 + ...

Net field is vector sum of fields due to individual charges.

E_net
Resultant electric field
E1, E2, E3
Individual electric field vectors
2 more formulas — sign up free

Diagrams3

4

Electric Flux

8 min

Electric flux measures the amount of electric field passing through a surface. For a uniform electric field through a plane surface, flux is Φ = EA cosθ, where θ is the angle between the electric field and the area vector. The area vector is always normal to the surface. Flux is maximum when the surface is perpendicular to the field and zero when the surface is parallel to the field. For curved or non-uniform situations, flux is calculated using integration: Φ = ∫E·dA. Through a closed surface, outward flux is taken positive and inward flux negative. Electric flux is central to Gauss’s law, where total closed-surface flux depends only on enclosed charge.

Key points6

  • Flux is a scalar quantity because it is a dot product.
  • The angle θ is between electric field and area vector, not between field and surface.
  • Solid angle helps describe how much a surface subtends at a point charge.
  • Flux through a closed surface can be positive, negative or zero.
  • Charges outside a closed surface contribute zero net flux through it.
  • Flux gives a bridge between field-line visualization and Gauss’s law.

Memory tricks2

Angle Reminder: Flux angle is with Area vector, not with the surface. Area vector stands normal to the surface.

Maximum Flux Trick: Maximum flux occurs when field pierces the surface straight through, so E is parallel to A.

Examples2

Solved Example

A surface of area 0.2 m² is placed perpendicular to E = 100 N/C. Since area vector is parallel to E, Φ = EA = 20 N m²/C.

Closed Surface Example

If no charge is enclosed by a closed surface, net flux is zero even if electric field exists due to outside charges.

Common mistakes2

Taking Angle with Surface

If the field makes angle α with the surface, then it makes 90° - α with the area vector.

Assuming Flux Must Always Be Positive

Flux through a closed surface can be negative if more field lines enter than leave.

Reference tables2

Quick revision8

Formulas4

Flux Through Plane Surface
Φ=E⋅A= EA cosθΦ = E·A =\text{ EA cos}θ

Electric flux through a plane area in uniform electric field.

Φ
Electric flux
E
Uniform electric field
A
Area vector magnitude
θ
Angle between E and area vector
Flux Through Curved Surface
Φ=∫E⋅dAΦ = ∫E·dA

Used when field or surface direction changes from point to point.

Φ
Total electric flux
E
Electric field
dA
Small area vector
2 more formulas — sign up free

Diagrams3

5

Electric Dipole

10 min

An electric dipole consists of two equal and opposite charges separated by a small distance. Its dipole moment is p = q × 2a, directed from negative charge to positive charge. Dipoles are important because many molecules behave as electric dipoles. The electric field of a dipole is different on its axial and equatorial lines. For a short dipole, axial field is twice the equatorial field in magnitude at the same large distance. In a uniform electric field, a dipole experiences no net force but experiences torque τ = pE sinθ, which tends to align it with the field. Its potential energy is U = -pE cosθ, minimum when aligned with the field.

Key points6

  • Dipole moment is a vector directed from -q to +q.
  • Net force on a dipole in a uniform electric field is zero.
  • Torque tries to align the dipole moment with the electric field.
  • Dipole field falls as 1/r³, faster than point charge field which falls as 1/r².
  • Axial field and equatorial field directions are different.
  • Potential energy is minimum when p is parallel to E.

Memory tricks2

Dipole Moment Direction: Dipole moment goes from Minus to Plus: M to P.

Axial vs Equatorial: Axial field is double; Equatorial field is equal-half compared with axial at same r.

Examples3

Solved Numerical

A dipole has q = 2 μC and separation 4 cm. p = q(2a) = 2 × 10⁻⁶ × 0.04 = 8 × 10⁻⁸ C m.

Torque Example

If p = 5 × 10⁻⁸ C m, E = 2 × 10⁵ N/C and θ = 90°, τ = pE sinθ = 10⁻² N m.

NEET-Type Question

A dipole is in stable equilibrium when p is parallel to E because U = -pE is minimum.

Common mistakes3

Wrong Direction of Dipole Moment

Electric dipole moment is from negative to positive charge, not from positive to negative.

Assuming Net Force in Uniform Field

In a uniform electric field, forces on +q and -q are equal and opposite, so net force is zero but torque may be non-zero.

Forgetting Equatorial Field Direction

On the equatorial line of a short dipole, electric field is opposite to dipole moment.

Reference tables2

Quick revision8

Formulas5

Dipole Moment
p=q(2a)p = q(2a)

Magnitude of electric dipole moment.

p
Dipole moment
q
Magnitude of either charge
2a
Separation between charges
Axial Field of Short Dipole
Eaxial=(1/4πε0)(2p/r3)E_{\text{axial}} = (1/4πε0)(2p/r³)

Electric field on axial line of a short dipole at distance r from its centre.

E_axial
Electric field on axial line
p
Dipole moment
r
Distance from dipole centre
ε0
Permittivity of free space
Equatorial Field of Short Dipole
Eequatorial=(1/4πε0)(p/r3)E_{\text{equatorial}} = (1/4πε0)(p/r³)

Magnitude of field on equatorial line; direction is opposite to dipole moment.

E_equatorial
Electric field on equatorial line
p
Dipole moment
r
Distance from dipole centre
2 more formulas — sign up free

Diagrams3

6

Gauss's Law

9 min

Gauss’s law states that the total electric flux through any closed surface equals the net charge enclosed divided by permittivity of free space: ∮E·dA = q_enclosed/ε0. The closed surface chosen for applying the law is called a Gaussian surface. Gauss’s law is true for any closed surface, regular or irregular, but it becomes useful for calculating electric field only when symmetry allows E to be constant over parts of the surface. Common useful symmetries are spherical, cylindrical and planar. Charges outside the Gaussian surface may affect electric field at individual points, but their net contribution to total flux through the closed surface is zero.

Key points6

  • Gauss’s law is a flux law, not directly a force law.
  • A Gaussian surface must be closed.
  • Area vector for a closed surface is outward normal.
  • To calculate E easily, E should be constant in magnitude on the chosen surface.
  • If q_enclosed = 0, net flux is zero, but electric field need not be zero everywhere.
  • Gauss’s law is especially powerful for infinite or highly symmetric charge distributions.

Memory tricks2

Gauss Law Core: Gauss sees only enclosed charge for net flux.

Surface Choice Trick: Sphere for spherical, cylinder for line, pillbox for plane.

Examples2

Solved Example

If a closed surface encloses +3 μC and -1 μC, q_enclosed = +2 μC. Net flux = 2 × 10⁻⁶/ε0.

Concept Example

A charge outside a closed box may create electric field on the box surface, but net flux through the box is zero if no charge is enclosed.

Common mistakes3

Using Open Surface as Gaussian Surface

A Gaussian surface must be closed. An open sheet cannot directly enclose charge.

Thinking Zero Flux Means Zero Field

Net flux can be zero even when electric field exists, because inward and outward contributions cancel.

Counting Outside Charges in q Enclosed

Only charges inside the Gaussian surface are included in q_enclosed.

Reference tables2

Quick revision8

Formulas3

Statement of Gauss's Law
Φ=qenclosed/ε0Φ = q_{\text{enclosed}}/ε0

Net electric flux through a closed surface equals enclosed charge divided by ε0.

Φ
Net electric flux
q_enclosed
Net charge inside closed surface
ε0
Permittivity of free space
Mathematical Form
∮E⋅dA=qenclosed/ε0∮E·dA = q_{\text{enclosed}}/ε0

Integral form of Gauss’s law for any closed surface.

∮
Closed surface integral
E
Electric field
dA
Outward area vector element
q_enclosed
Charge enclosed
1 more formula — sign up free

Diagrams3

7

Applications of Gauss's Law

11 min

Applications of Gauss’s law use symmetry to calculate electric fields quickly. For an infinite line charge, cylindrical symmetry gives E = λ/(2πε0r), directed radially outward for positive charge. For an infinite plane sheet, planar symmetry gives E = σ/(2ε0), independent of distance. For a charged spherical shell, field outside behaves as if all charge were concentrated at the centre, while field inside the shell is zero. For conductors in electrostatic equilibrium, electric field inside the conducting material is zero, excess charge lies on the outer surface and field just outside is σ/ε0. Electrostatic shielding works because the field inside a closed conductor is zero.

Key points6

  • For line charge, choose a cylindrical Gaussian surface coaxial with the line.
  • For plane sheet, choose a pillbox Gaussian surface cutting through the sheet.
  • For spherical shell, choose a concentric spherical Gaussian surface.
  • The field of an infinite plane sheet does not depend on distance.
  • Inside a charged spherical shell, enclosed charge is zero, so electric field is zero.
  • A conductor shields its cavity from external electrostatic fields if no charge is placed inside the cavity.

Memory tricks3

Three Famous Results: Line falls as 1/r, sheet is constant, shell is zero inside.

Conductor Shielding: Conductor cancels field inside: charges rearrange until internal E becomes zero.

Gaussian Surface Match: Line-cylinder, Sheet-pillbox, Shell-sphere.

Examples4

Solved Example: Plane Sheet

For σ = 8.85 × 10⁻⁶ C/m², field due to an infinite non-conducting sheet is E = σ/(2ε0) = 8.85 × 10⁻⁶/(2 × 8.85 × 10⁻¹²) = 5 × 10⁵ N/C.

Previous NEET-Type Question

Electric field inside a charged spherical shell is zero because a Gaussian sphere inside encloses no charge.

Quick Revision Note

For an infinite line charge, if distance r is doubled, E becomes half because E ∝ 1/r.

Real-Life Application

A car can act like a Faraday cage during lightning because charges stay on the outer conducting surface, shielding the inside.

Common mistakes3

Using Sheet Formula for Conducting Surface

Infinite non-conducting sheet has E = σ/(2ε0), but just outside a conductor E = σ/ε0.

Thinking Spherical Shell Field Is Zero Outside

Inside a uniformly charged spherical shell E = 0; outside it behaves like a point charge at the centre.

Forgetting Direction

For positive charge distributions, field is outward; for negative distributions, field is inward.

Reference tables2

Quick revision8

Formulas5

Electric Field due to Infinite Line Charge
E=λ/(2πε0r)E = λ/(2πε0r)

Field at perpendicular distance r from an infinitely long line charge.

E
Electric field
λ
Linear charge density
ε0
Permittivity of free space
r
Distance from line charge
Electric Field due to Infinite Plane Sheet
E=σ/(2ε0)E = σ/(2ε0)

Field on either side of a uniformly charged infinite non-conducting plane sheet.

E
Electric field
σ
Surface charge density
ε0
Permittivity of free space
Field Just Outside a Conductor
E=σ/ε0E = σ/ε0

Electric field immediately outside a charged conductor surface.

E
Electric field just outside conductor
σ
Surface charge density on conductor
ε0
Permittivity of free space
2 more formulas — sign up free

Diagrams5

Videos for Electric Charges and Fields

  • Electric Charges & Fields-1⚡ | Class 12 PhysicsBest for NEET · English
  • Electric Field Due to Charges🔥 | Class 12 PhysicsVedantu NEET English · English
  • Electric dipoles & dipole moments | Electric charges & fieldsKhan Academy India - English · English
  • Electric Charges & Fields-1⚡ | Class 12 Physics Best for NEET · English
  • Electric Charges & Coulomb’s Force 🔥| NEET Class 12 PhysicsVedantu NEET English · English
  • Electric Field Due to Charges🔥 | Class 12 PhysicsVedantu NEET English · English
Watch with a free account

NEET PYQs from Electric Charges and Fields

The latest 3 of 24 previous-year questions from this chapter.

  1. NEET 2025Set 45MediumQ1

    Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radius of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :

    • A3F/5
    • B2F/3
    • CF/2
    • D3F/8
  2. NEET 2025Set 45MediumQ2

    An electric dipole with dipole moment 5 × 10⁻⁶ Cm is aligned with the direction of a uniform electric field of magnitude 4 × 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is :

    • A0.8 J
    • B1.0 J
    • C1.2 J
    • D1.5 J
  3. NEET 2024Set T1EasyQ3

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential $V$ at any axial point, at $2\,m$ distance $(r)$ from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude $4\times10^{-6}\,C\,m$, is $\pm 9\times10^{3}\,V$. (Take: $\frac{1}{4\pi\varepsilon_0}=9\times10^9\;\text{SI units)}$ Reason R: $V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}$ where r is the distance of any axial point situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

    • AA is true but R is false.
    • BA is false but R is true.
    • CBoth A and R are true and R is the correct explanation of A.
    • DBoth A and R are true and R is NOT the correct explanation of A.

Get the whole of Electric Charges and Fields

A free account opens every formula, diagram, table and quick-revision list in this chapter, and tracks the topics you finish.

  • Free to start
  • Progress saved per topic
  • PYQ practice for Electric Charges and Fields