Electric charge and its properties
A plastic comb rubbed on dry hair picks up tiny bits of paper. Link to earlier classes : charging by friction and the idea of the atom from Class 8 and Class 9.
Electric charge is a basic property of matter that makes it feel electric and magnetic forces. It is of two kinds, positive and negative; like charges repel and unlike charges attract. Its SI unit is the coulomb (C). Charge is additive (total charge is the algebraic sum), conserved (it can be moved from one body to another but never created or destroyed) and quantised (it comes only in whole-number multiples of the electronic charge).
Rule 1: Quantisation of charge — q = n e (n = a whole number, e = 1.6 × 10⁻¹⁹ C, the charge of one proton). Charges smaller than e are not found in free form.
- Easy : n = 5 × 10¹² electrons carry a charge of 5 × 10¹² × 1.6 × 10⁻¹⁹ = 8 × 10⁻⁷ C in magnitude.
- Medium : A body has a charge of −3.2 μC. The number of excess electrons is n = 3.2 × 10⁻⁶ ÷ 1.6 × 10⁻¹⁹ = 2 × 10¹³.
- Exam : A glass rod loses 4 × 10¹¹ electrons to a silk cloth. The rod gets +6.4 × 10⁻⁸ C, the silk gets −6.4 × 10⁻⁸ C, and the total charge of the pair remains zero.
Rule 2: Conductors and insulators — A conductor (metal, salt water, the human body) has free charges that move easily; an insulator (glass, plastic, dry wood) has almost none, so charge put on one spot stays there. Charge given to a conductor spreads over its surface.
- Easy : A copper wire carries current because it has free electrons.
- Medium : A charged plastic rod keeps its charge on the touched spot, but a charged metal rod loses it quickly when held by hand, because the body conducts it to the ground.
- Exam : A charged body attracts uncharged paper bits because it makes the near side of each bit oppositely charged (induction) and the attraction beats the repulsion from the far side.
q = ne; total charge of an isolated system never changes; e = 1.6 × 10⁻¹⁹ C.
Saying that a rubbed rod "makes" charge. Friction only moves electrons from one body to the other.
Memory clue : "Charge comes in packets of e : count the packets, n times e."
Spot the difference
| Property | Meaning | Example |
|---|---|---|
| Additivity | charges add like numbers with signs | +5 μC and −2 μC give +3 μC |
| Conservation | total stays constant | glass and silk after rubbing |
| Quantisation | q = ne | 3.2 × 10⁻¹⁹ C is 2e |
Practice questions
- 1How many electrons together carry a charge of 1 C?
- 2Can a body have a charge of 2.4 × 10⁻¹⁹ C? Give the reason.
- 3A glass rod rubbed with silk becomes positive. Which particles were transferred, and in which direction?
- 4Two identical metal spheres carry +6 μC and −2 μC. They are touched together and separated. What is the charge on each?
- 5A body gains 5 × 10¹⁰ electrons. Find its charge.
- 6Why does a charged comb attract small neutral paper bits?
- 7A neutral sphere loses 2 × 10⁹ electrons. Find its charge and the mass it loses (mass of an electron = 9.1 × 10⁻³¹ kg).
- 8Spheres A (+9 μC) and B (−3 μC) are identical, and C is identical but neutral. C touches A, then C touches B. Find the final charge on A, B and C.
Show answers (try them yourself first)
- 1n = 1 ÷ 1.6 × 10⁻¹⁹ = 6.25 × 10¹⁸ electrons.
- 2No. 2.4 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 1.5 is not a whole number.
- 3Electrons moved from the glass rod to the silk.
- 4Total = +4 μC shared equally, so +2 μC each.
- 5q = −(5 × 10¹⁰ × 1.6 × 10⁻¹⁹) = −8 × 10⁻⁹ C.
- 6The comb polarises the paper: the near side becomes oppositely charged and is pulled harder than the far side is pushed.
- 7q = +2 × 10⁹ × 1.6 × 10⁻¹⁹ = +3.2 × 10⁻¹⁰ C; mass lost = 2 × 10⁹ × 9.1 × 10⁻³¹ = 1.82 × 10⁻²¹ kg.
- 8After A touches C : A = C = +4.5 μC. Then C (+4.5) touches B (−3) : each = +0.75 μC. Final : A = +4.5 μC, B = +0.75 μC, C = +0.75 μC (sum = +6 μC, as at the start).
Coulomb's law and forces between many charges
Two balloons rubbed on hair push each other apart, and the push grows when they are brought closer. Link to earlier classes : the inverse-square law of gravitation from Class 11.
Coulomb's law says that the force between two point charges at rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, and acts along the line joining them.
Rule 1: Coulomb's force — F = k q₁ q₂ ÷ r², where k = 1 ÷ (4π ε₀) = 9 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² (q₁, q₂ = charges, r = distance). In a medium of dielectric constant K the force becomes F ÷ K.
- Easy : q₁ = 2 μC, q₂ = 3 μC, r = 0.30 m ⇒ F = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶ ÷ 0.09 = 0.60 N.
- Medium : If one charge is doubled and the distance is also doubled, the force becomes 2 ÷ 4 = ½ of the old value.
- Exam : The same two charges placed in a liquid with K = 4 feel 0.60 ÷ 4 = 0.15 N.
Rule 2: Superposition principle — The force on one charge due to many charges is the vector sum of the forces due to each of them taken alone: F = F₁ + F₂ + F₃ + …
- Easy : Three equal charges q at the corners of an equilateral triangle of side a : each pair gives a force k q² ÷ a², and the net force on one charge is √3 k q² ÷ a² along the outward bisector.
- Medium : For q = 2 μC and a = 0.20 m, k q² ÷ a² = 0.90 N, so the net force is 0.90 × 1.732 = 1.56 N.
- Exam : q₁ = +2 μC at x = 0, q₂ = −3 μC at x = 0.30 m and q₃ = +1 μC at x = 0.60 m. On q₂, q₁ pulls with 0.60 N towards −x and q₃ pulls with 0.30 N towards +x, so the net force is 0.30 N towards −x.
F = kq₁q₂ ÷ r²; force on q₁ and on q₂ are equal and opposite; add forces as vectors.
Using the formula with charges in μC or distances in cm without converting to C and m.
Memory clue : "Charges on top, distance squared below, nine billion in front."
Spot the difference
| Law | Formula | Nature |
|---|---|---|
| Coulomb | F = kq₁q₂ ÷ r² | attractive or repulsive |
| Gravitation | F = Gm₁m₂ ÷ r² | always attractive |
Practice questions
- 1Find the force between two 1 C charges placed 1 m apart in vacuum.
- 2Write the SI unit of ε₀.
- 3By what factor does the force change if the distance is made three times?
- 4Two point charges of 5 nC each are 3 cm apart. Find the force between them.
- 5Find the force between the electron and proton of a hydrogen atom, 5.3 × 10⁻¹¹ m apart (e = 1.6 × 10⁻¹⁹ C).
- 6Charges +q and +4q are d apart. Where, between them, can a third charge stay in equilibrium?
- 7Find the ratio of the electric force to the gravitational force between two protons (m = 1.67 × 10⁻²⁷ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻²).
- 8Equal charges of 2 μC sit at the corners of an equilateral triangle of side 0.20 m. Find the net force on one of them.
Show answers (try them yourself first)
- 1F = 9 × 10⁹ N.
- 2C² N⁻¹ m⁻².
- 3The force becomes 1/9 of the old value.
- 4F = 9 × 10⁹ × (5 × 10⁻⁹)² ÷ (0.03)² = 2.5 × 10⁻⁴ N (repulsive).
- 5F = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² ÷ (5.3 × 10⁻¹¹)² = 8.2 × 10⁻⁸ N (attractive).
- 6Let x be the distance from q. Then q ÷ x² = 4q ÷ (d − x)², so d − x = 2x and x = d ÷ 3 from the charge q.
- 7k e² ÷ (G m²) = (9 × 10⁹ × 2.56 × 10⁻³⁸) ÷ (6.67 × 10⁻¹¹ × 2.79 × 10⁻⁵⁴) ≈ 1.2 × 10³⁶.
- 8√3 × 0.90 N ≈ 1.56 N, directed away from the other two charges along the bisector.
Electric field and field lines
Iron filings around a magnet show a pattern; in the same way a charge fills the space around it with an electric "influence". Link to earlier classes : force and the field idea of gravity in Class 11.
The electric field E at a point is the force per unit positive test charge placed there. It is a vector, measured in N C⁻¹ (or V m⁻¹), and it points away from a positive source charge and towards a negative one. Electric field lines are curves whose tangent at each point gives the direction of E.
Rule 1: Field of a point charge — E = F ÷ q and E = k Q ÷ r² (Q = source charge, r = distance from it). The force on a charge q in the field is F = q E.
- Easy : Q = 5 μC, r = 0.50 m ⇒ E = 9 × 10⁹ × 5 × 10⁻⁶ ÷ 0.25 = 1.8 × 10⁵ N C⁻¹.
- Medium : An electron in E = 2 × 10³ N C⁻¹ feels F = 1.6 × 10⁻¹⁹ × 2 × 10³ = 3.2 × 10⁻¹⁶ N, so a = 3.2 × 10⁻¹⁶ ÷ 9.1 × 10⁻³¹ = 3.5 × 10¹⁴ m s⁻².
- Exam : At the midpoint of +4 μC and −4 μC placed 0.40 m apart, each gives 9 × 10⁵ N C⁻¹ in the same direction (towards the negative charge), so E = 1.8 × 10⁶ N C⁻¹.
Rule 2: Properties of field lines — They start on positive charges and end on negative charges; they never cross; they are closer where the field is stronger; for static charges they never form closed loops; and they are always perpendicular to the surface of a conductor.
- Easy : Near a point charge the lines are crowded, far away they spread out.
- Medium : If two lines crossed, E would have two directions at the crossing point, which is impossible.
- Exam : The lines of a uniform field are parallel, equally spaced straight lines.
In the figure : Field lines of a positive charge (outward), a negative charge (inward) and a dipole (from + to −).
E = kQ ÷ r²; F = qE; lines start on +, end on −, never cross.
Thinking that field lines are paths followed by a charge. They only show the direction of the force at each point.
Memory clue : "Field lines flow from Plus to Minus, and never cross."
Spot the difference
| Quantity | Symbol | Unit |
|---|---|---|
| Electric field | E | N C⁻¹ |
| Force on a test charge | F = qE | N |
Practice questions
- 1Write the SI unit of the electric field.
- 2What is the direction of E near a negative point charge?
- 3Can two electric field lines intersect?
- 4A 2 μC charge is placed where E = 300 N C⁻¹. Find the force on it.
- 5Find the field at 3 cm from a 6 nC point charge.
- 6An oil drop of mass 1.63 × 10⁻¹⁵ kg hangs still in an upward field of 1.0 × 10⁴ N C⁻¹. Find its charge and the number of electrons it carries (g = 9.8 m s⁻²).
- 7Charges +9 μC and +1 μC are 0.40 m apart. At what point on the line joining them is the field zero?
- 8An electron starts from rest in a uniform field of 5 × 10³ N C⁻¹. Find its speed after 2 ns.
Show answers (try them yourself first)
- 1N C⁻¹ (equal to V m⁻¹).
- 2Towards the charge.
- 3No; at the crossing point the field would have two directions.
- 4F = qE = 2 × 10⁻⁶ × 300 = 6 × 10⁻⁴ N.
- 5E = 9 × 10⁹ × 6 × 10⁻⁹ ÷ (0.03)² = 6 × 10⁴ N C⁻¹.
- 6q = mg ÷ E = 1.63 × 10⁻¹⁵ × 9.8 ÷ 10⁴ ≈ 1.6 × 10⁻¹⁸ C, which is 10 electrons.
- 79 ÷ x² = 1 ÷ (0.40 − x)² gives x = 3(0.40 − x), so x = 0.30 m from the 9 μC charge.
- 8a = eE ÷ m = 8.8 × 10¹⁴ m s⁻²; v = a t = 8.8 × 10¹⁴ × 2 × 10⁻⁹ ≈ 1.8 × 10⁶ m s⁻¹.
Electric dipole
A water molecule has a slightly positive end and a slightly negative end, so it behaves like a tiny pair of opposite charges. Link to earlier classes : torque and couples from Class 11.
An electric dipole is a pair of equal and opposite charges +q and −q separated by a small distance 2a. Its dipole moment is a vector p pointing from −q to +q, with magnitude p = q × 2a, measured in C m.
Rule 1: Field of a dipole (far away, r much larger than a) — On the axis, E = 2 k p ÷ r³ along p. On the equatorial line, E = k p ÷ r³ opposite to p. A dipole field falls as 1 ÷ r³, faster than the 1 ÷ r² of a single charge.
- Easy : Charges ±2 μC held 1 cm apart have p = 2 × 10⁻⁶ × 0.01 = 2 × 10⁻⁸ C m.
- Medium : For p = 2 × 10⁻⁸ C m at r = 0.50 m, the axial field is 2 × 9 × 10⁹ × 2 × 10⁻⁸ ÷ 0.125 = 2880 N C⁻¹, and the equatorial field is half of it, 1440 N C⁻¹.
- Exam : For p = 8 × 10⁻⁹ C m, the equatorial field at 0.40 m is 9 × 10⁹ × 8 × 10⁻⁹ ÷ 0.064 = 1125 N C⁻¹, directed opposite to p.
Rule 2: Dipole in a uniform field — The net force is zero, but there is a torque τ = p E sin θ (vector form τ = p × E), where θ is the angle between p and E. It tries to turn the dipole along the field.
- Easy : With p = 2 × 10⁻⁸ C m and E = 4 × 10⁴ N C⁻¹, the greatest torque (θ = 90°) is 8 × 10⁻⁴ N m.
- Medium : At θ = 30° the torque is 8 × 10⁻⁴ × ½ = 4 × 10⁻⁴ N m.
- Exam : When θ = 0° or 180°, the torque is zero; θ = 0° is the stable position and θ = 180° is unstable.
In the figure : A dipole of moment p in a uniform field E; the two equal and opposite forces form a couple.
p = q(2a), from − to +; axial field = 2 × equatorial field; τ = pE sin θ.
Using the point-charge formula E = kq ÷ r² for a dipole. The dipole field falls as 1 ÷ r³.
Memory clue : "On the Axis, double (2kp÷r³); on the Equator, single, and backwards."
Spot the difference
| Position | Field | Direction |
|---|---|---|
| Axial | 2kp ÷ r³ | along p |
| Equatorial | kp ÷ r³ | opposite to p |
Practice questions
- 1Write the SI unit of the dipole moment.
- 2What is the direction of p?
- 3What is the net force on a dipole in a uniform field?
- 4Charges ±5 nC are 4 mm apart. Find p.
- 5A dipole with p = 3 × 10⁻¹⁰ C m is at 90° to a field of 2 × 10⁵ N C⁻¹. Find the torque.
- 6What is the ratio of the axial field to the equatorial field at equal distances?
- 7A dipole of p = 1.5 × 10⁻⁹ C m lies at 60° to a uniform field of 2 × 10⁴ N C⁻¹. Find the torque.
- 8Why does the dipole field fall faster with distance than the field of a single charge?
Show answers (try them yourself first)
- 1C m.
- 2From the negative charge to the positive charge.
- 3Zero.
- 4p = 5 × 10⁻⁹ × 4 × 10⁻³ = 2 × 10⁻¹¹ C m.
- 5τ = pE = 3 × 10⁻¹⁰ × 2 × 10⁵ = 6 × 10⁻⁵ N m.
- 62 : 1.
- 7τ = 1.5 × 10⁻⁹ × 2 × 10⁴ × sin 60° = 2.6 × 10⁻⁵ N m.
- 8The two opposite fields partly cancel at a distance, leaving only a small difference that falls as 1 ÷ r³.
Electric flux and Gauss's law
The amount of rain collected by a tilted bucket depends on its opening and its tilt; flux is the electric version of this idea. Link to earlier classes : the cosine of an angle and area vectors.
Electric flux measures how much electric field passes through a surface. Gauss's law states that the total flux through any closed surface equals the net charge enclosed divided by ε₀.
Rule 1: Flux — Φ = E A cos θ (E = field, A = area, θ = angle between E and the normal to the surface). Unit : N m² C⁻¹.
- Easy : E = 500 N C⁻¹ perpendicular to A = 0.20 m² ⇒ Φ = 100 N m² C⁻¹.
- Medium : The same surface turned so that the normal makes 60° with E ⇒ Φ = 500 × 0.20 × 0.5 = 50 N m² C⁻¹.
- Exam : For a surface parallel to the field, θ = 90° and Φ = 0.
Rule 2: Gauss's law — Φ = q ÷ ε₀ (q = net charge inside the closed surface). Charges outside contribute zero net flux, because every line that enters also leaves.
- Easy : A charge of 8.85 μC inside any closed surface gives Φ = 8.85 × 10⁻⁶ ÷ 8.85 × 10⁻¹² = 10⁶ N m² C⁻¹.
- Medium : A 2 μC charge at the centre of a cube sends Φ = 2.26 × 10⁵ N m² C⁻¹ through the cube, so each of the six faces gets 3.8 × 10⁴.
- Exam : A closed surface enclosing +3 μC and −1 μC has Φ = 2 × 10⁻⁶ ÷ 8.85 × 10⁻¹² = 2.26 × 10⁵ N m² C⁻¹.
Φ = EA cos θ; Φ(closed surface) = q(inside) ÷ ε₀.
Counting the charges outside the closed surface in Gauss's law.
Memory clue : "Gauss counts only what is inside the bag."
Spot the difference
| Case | Net flux |
|---|---|
| Charge inside | q ÷ ε₀ |
| Charge outside only | zero |
| Surface parallel to the field | zero |
Practice questions
- 1What is the flux through a surface kept parallel to the field?
- 2What is the net flux through a closed surface that encloses no net charge?
- 3Write the unit of electric flux.
- 4A charge of 2 μC is at the centre of a cube. Find the flux through one face.
- 5A closed surface encloses +3 μC and −1 μC. Find the net flux through it.
- 6Why does a charge outside a closed surface not change the net flux through it?
- 7The net outward flux through a sphere is 1.2 × 10⁴ N m² C⁻¹. Find the charge inside.
- 8A hemispherical bowl of radius 0.10 m is placed in a uniform field of 300 N C⁻¹ parallel to its axis. Find the flux through its curved surface.
Show answers (try them yourself first)
- 1Zero.
- 2Zero.
- 3N m² C⁻¹.
- 4(2 × 10⁻⁶ ÷ 8.85 × 10⁻¹²) ÷ 6 ≈ 3.8 × 10⁴ N m² C⁻¹.
- 5Φ = (3 − 1) × 10⁻⁶ ÷ 8.85 × 10⁻¹² ≈ 2.3 × 10⁵ N m² C⁻¹.
- 6Each field line from the outside charge enters the surface and leaves it again, so inward and outward fluxes cancel.
- 7q = Φ ε₀ = 1.2 × 10⁴ × 8.85 × 10⁻¹² ≈ 1.06 × 10⁻⁷ C.
- 8The flux through the curved surface equals the flux through its circular opening: Φ = E π R² = 300 × π × 0.01 ≈ 9.4 N m² C⁻¹.
Applications of Gauss's law
The wire of a high-voltage line and the plates of a camera flash capacitor are long lines and flat sheets of charge. Link to earlier classes : symmetry and Gauss's law of the previous section.
When a charge distribution is symmetric (a long line, a large flat sheet or a sphere), Gauss's law gives the field quickly by choosing a closed surface on which E has a constant value and a known direction.
Rule 1: Infinite line and infinite sheet — For a long line of charge density λ : E = λ ÷ (2π ε₀ r) (r = distance from the line). For a large flat sheet of surface density σ : E = σ ÷ (2 ε₀), independent of the distance.
- Easy : λ = 2 μC m⁻¹ at r = 0.50 m ⇒ E = 2 × 9 × 10⁹ × 2 × 10⁻⁶ ÷ 0.50 = 7.2 × 10⁴ N C⁻¹.
- Medium : σ = 4.425 × 10⁻⁶ C m⁻² ⇒ E = σ ÷ 2ε₀ = 2.5 × 10⁵ N C⁻¹ at any distance.
- Exam : Two parallel sheets with +σ and −σ give E = σ ÷ ε₀ between them and zero outside. For σ = 2.0 × 10⁻⁸ C m⁻², E = 2.26 × 10³ N C⁻¹.
Rule 2: Thin spherical shell — Outside (r ≥ R) the shell acts like a point charge at its centre, E = k Q ÷ r². Inside (r < R) the field is zero.
- Easy : A shell with Q = 3 μC and R = 0.10 m gives E = 9 × 10⁹ × 3 × 10⁻⁶ ÷ 0.09 = 3 × 10⁵ N C⁻¹ at r = 0.30 m.
- Medium : At the surface (r = 0.10 m) the field is 2.7 × 10⁶ N C⁻¹.
- Exam : Anywhere inside the shell, E = 0, so a sensitive instrument inside a charged metal shell is protected from outside fields.
Line : E ∝ 1/r; sheet : E is constant; shell : E = kQ/r² outside, 0 inside.
Using σ ÷ ε₀ for a single sheet. For one sheet it is σ ÷ 2ε₀.
Memory clue : "Line: one over r. Sheet: no r at all. Shell: nothing inside."
Spot the difference
| Source | Field | Depends on distance? |
|---|---|---|
| Infinite line | λ ÷ 2πε₀r | yes, as 1/r |
| Infinite sheet | σ ÷ 2ε₀ | no |
| Thin shell, outside | kQ ÷ r² | yes, as 1/r² |
| Thin shell, inside | 0 | — |
Practice questions
- 1What is the electric field inside a uniformly charged thin spherical shell?
- 2Does the field of a large flat sheet change with distance?
- 3How does the field of a long line charge vary with distance?
- 4A large sheet has σ = 8.85 μC m⁻². Find the field near it.
- 5A shell of Q = 5 nC and radius 0.20 m is given. Find the field at 0.50 m from its centre.
- 6Two parallel sheets carry +σ and −σ with σ = 2.0 × 10⁻⁸ C m⁻². Find the field between them.
- 7A sphere of radius 0.05 m holds a total charge of 2 μC on its surface. Find the surface charge density.
- 8A long line charge gives a field of 9 × 10³ N C⁻¹ at 2 m. Find the field at 6 m.
Show answers (try them yourself first)
- 1Zero.
- 2No; for a large sheet it is constant.
- 3It falls as 1 ÷ r.
- 4E = σ ÷ 2ε₀ = 8.85 × 10⁻⁶ ÷ (2 × 8.85 × 10⁻¹²) = 5 × 10⁵ N C⁻¹.
- 5E = 9 × 10⁹ × 5 × 10⁻⁹ ÷ 0.25 = 180 N C⁻¹.
- 6E = σ ÷ ε₀ = 2.0 × 10⁻⁸ ÷ 8.85 × 10⁻¹² ≈ 2.3 × 10³ N C⁻¹.
- 7σ = Q ÷ 4πR² = 2 × 10⁻⁶ ÷ (4π × 0.0025) ≈ 6.4 × 10⁻⁵ C m⁻².
- 8E ∝ 1/r, so at three times the distance E = 9 × 10³ ÷ 3 = 3 × 10³ N C⁻¹.