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Electrostatics: Class 12 Physics
Electric charge, Coulomb's law, electric field and field lines, Gauss's law, electric potential, capacitors and dielectrics, with formulas and worked numericals.
Key points
Electric charge is a basic property of matter. There are two kinds, positive and negative. Like charges repel and unlike charges attract. Charge is quantised, so every charge is a whole number multiple of the electron charge e = 1.6 × 10⁻¹⁹ C, and charge is conserved in every isolated system.
Coulomb's law gives the force between two point charges at rest. F = k q₁q₂ / r², where k = 1/(4πε₀) = 9 × 10⁹ N m²/C². The force acts along the line joining the charges. In a medium of relative permittivity εᵣ the force becomes F = k q₁q₂ / (εᵣ r²).
The electric field E at a point is the force per unit positive test charge, E = F/q. For a point charge Q, E = kQ/r². Field is a vector and fields from several charges add as vectors. Field lines start on positive charge and end on negative charge, they never cross, and they are closer together where the field is stronger.
Electric flux is Φ = E A cos θ, where θ is the angle between the field and the normal to the area. Gauss's law states that the net flux through any closed surface equals the net enclosed charge divided by ε₀: Φ = Q/ε₀. It is used to find the field of symmetric charge distributions.
Fields from Gauss's law:
- Charged sphere or point charge, outside: E = kQ/r², as if all charge were at the centre.
- Infinite plane sheet with surface charge density σ: E = σ / (2ε₀).
- Between two oppositely charged parallel plates: E = σ / ε₀.
- Inside a conductor in equilibrium: E = 0, and the excess charge sits on the outer surface.
Electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point, V = W/q. For a point charge V = kQ/r. Potential is a scalar. The potential difference between two points is ΔV = W/q, and the work done in moving a charge q through ΔV is W = qΔV. The electric field is related to potential by E = −ΔV/Δr.
The electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt: 1 eV = 1.6 × 10⁻¹⁹ J.
A capacitor stores charge and energy. Its capacitance is C = Q/V, in farads. For parallel plates C = ε₀A/d with a vacuum between them. If a dielectric of relative permittivity εᵣ fills the space, the capacitance is multiplied by εᵣ.
Combining capacitors:
- In parallel the charges add and the potential difference is the same: C = C₁ + C₂ + C₃.
- In series the charge is the same and the potential differences add: 1/C = 1/C₁ + 1/C₂ + 1/C₃.
The energy stored in a charged capacitor is U = ½CV² = ½QV = Q²/2C. This energy is stored in the electric field between the plates.
Study tip: For a field or force question with more than one charge, draw the diagram, find the magnitude and direction of each contribution separately, and only then add them as vectors.
Definitions and flashcards
Say the meaning first, then open the card to check.
Coulomb's law
The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Electric field intensity
The force acting on a unit positive charge placed at the point.
Electric flux
The number of field lines passing normally through a surface, equal to E A cos θ.
Gauss's law
The net electric flux through a closed surface equals the enclosed charge divided by ε₀.
Electric potential
The work done per unit positive charge in bringing it from infinity to the point.
Capacitance
The ratio of the charge on a capacitor to the potential difference across it.
Dielectric
An insulating material placed between the plates that increases the capacitance.
Electron volt
The energy gained by an electron through a potential difference of one volt, 1.6 × 10⁻¹⁹ J.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Coulomb's law | F = k q₁q₂ / r² | k = 9 × 10⁹ N m²/C². F in newtons, q in coulombs, r in metres. |
| Electric field of a point charge | E = kQ / r² | E in N/C. |
| Gauss's law | Φ = Q / ε₀ | ε₀ = 8.85 × 10⁻¹² C²/N m². |
| Potential of a point charge | V = kQ / r | V in volts. |
| Work in a potential difference | W = q ΔV | Energy gained by a charge q moved through ΔV. |
| Capacitance | C = Q / V | C in farads. |
| Parallel plate capacitor | C = ε₀ A / d | With a dielectric multiply by εᵣ. |
| Energy stored | U = ½ C V² | Also ½ Q V or Q² / 2C. |
| Parallel combination | C = C₁ + C₂ + C₃ | Series: 1/C = 1/C₁ + 1/C₂ + 1/C₃. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
State Coulomb's law and write its equation. (2 marks)
The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: F = k q₁q₂ / r², with k = 9 × 10⁹ N m²/C².
Why can two electric field lines never cross? (2 marks)
The field at a point has one direction only. If two lines crossed, there would be two directions of force at the crossing point, which is impossible.
State Gauss's law. (2 marks)
The total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by the surface: Φ = Q/ε₀.
Why is the electric field inside a charged conductor zero? (3 marks)
Free charges in a conductor move until the net force on each is zero. In equilibrium there is no net field inside, so all the excess charge stays on the outer surface.
Define capacitance and state what a dielectric does to it. (3 marks)
Capacitance is the charge stored per unit potential difference, C = Q/V. A dielectric between the plates reduces the effective field, so more charge can be stored for the same potential difference, and the capacitance is multiplied by the relative permittivity of the dielectric.
Numericals with worked steps
Two charges of 4 μC and 6 μC are 0.3 m apart in air. Find the force between them.
- F = k q₁q₂ / r².
- = 9 × 10⁹ × (4 × 10⁻⁶) × (6 × 10⁻⁶) / (0.3)².
- = 9 × 10⁹ × 24 × 10⁻¹² / 0.09.
- = 0.216 / 0.09 = 2.4 N.
Answer: 2.4 N, attractive if the charges have opposite signs and repulsive if they have the same sign
Find the electric field at a distance of 0.5 m from a point charge of 2 μC.
- E = kQ / r².
- = 9 × 10⁹ × 2 × 10⁻⁶ / (0.5)².
- = 1.8 × 10⁴ / 0.25 = 7.2 × 10⁴ N/C.
Answer: 7.2 × 10⁴ N/C
A 5 μF capacitor is charged to 200 V. Find the charge and the energy stored.
- Q = CV = 5 × 10⁻⁶ × 200 = 1.0 × 10⁻³ C.
- U = ½ C V² = ½ × 5 × 10⁻⁶ × (200)² = ½ × 5 × 10⁻⁶ × 40,000 = 0.1 J.
Answer: Q = 1.0 × 10⁻³ C and U = 0.1 J
Capacitors of 2 μF and 4 μF are joined in series across 12 V. Find the equivalent capacitance and the charge on each.
- 1/C = 1/2 + 1/4 = 3/4, so C = 4/3 μF = 1.33 μF.
- Q = CV = 1.33 × 10⁻⁶ × 12 = 1.6 × 10⁻⁵ C.
- In series each capacitor carries the same charge.
Answer: C = 1.33 μF and Q = 1.6 × 10⁻⁵ C on each
Common mistakes
- Using the distance in centimetres in Coulomb's law. Convert to metres.
- Adding electric fields as plain numbers. They are vectors and must be added with directions.
- Forgetting that potential is a scalar, so potentials of several charges add as numbers with their signs.
- Using the parallel capacitor formula for a series combination, or the other way round.
- Leaving out the sign of the charge when finding the direction of the force.