12 min readConcept sheetSection B very short answer (2 marks) or long answer (5 marks), often with a choice: capacitance of a parallel plate capacitor with and without a dielectric; Section A objective parts on units and types; also a NEET favourite

Capacitance, spherical and parallel plate capacitors, and types of capacitors

Capacitance is C = Q/V, in farad. A sphere has C = 4πε₀r; a parallel plate capacitor has C = ε₀A/d, rising K times with a dielectric.

CapacitanceC = Q/V; SI unit farad (F), 1 F = 1 C V⁻¹; 1 μF = 10⁻⁶ F, 1 pF = 10⁻¹² F

Isolated sphereC = 4πε₀r = r/(9 × 10⁹) farad, with r in metres

Parallel plate capacitorC₀ = ε₀A/d in air; C = Kε₀A/d = KC₀ with a dielectric filling the gap

What C depends onThe size, shape and spacing of the conductors and the medium between them; not Q or V

What is a capacitor?

A capacitor is a pair of conductors carrying equal and opposite charges, +Q and −Q, separated by an insulator. The potential difference V between them is proportional to Q. The ratio C = Q/V is the capacitance: the charge stored per volt. Its unit is the farad (F), where 1 F = 1 C V⁻¹. A farad is a very large unit, so we use μF (10⁻⁶ F) and pF (10⁻¹² F). Principle: bring an earthed metal plate close to a charged plate. The charge induced on it lowers the potential of the charged plate. So the charged plate can now hold more charge at the same potential, and its capacitance rises. In a circuit, a capacitor is drawn as two short parallel lines.

Derivation: capacitance of an isolated sphere

A metal sphere of radius r, alone in air, carries a charge q.

1
Its charge sits on its surface, and outside it acts as if all the charge were at the centre.
2
So the potential of the sphere is V = (1/4πε₀) q/r.
3★
Capacitance: C = q/V = 4πε₀r. With 1/4πε₀ = 9 × 10⁹, C = r/(9 × 10⁹) farad.
4
So C ∝ r. A sphere of radius 0.18 m has C = 0.18 ÷ (9 × 10⁹) = 2 × 10⁻¹¹ F = 20 pF.
5
Even the Earth (r ≈ 6.4 × 10⁶ m) has only C ≈ 7.1 × 10⁻⁴ F, about 711 μF. The farad really is huge.

★ key step

Derivation: parallel plate capacitor

Two parallel metal plates, each of area A, are a small distance d apart in air. They carry charges +q and −q.

1
Surface charge density σ = q/A. As d is small compared with the plates, ignore the bending of the field at the edges.
2
The field between the plates is uniform: E = σ/ε₀ = q/(ε₀A).
3
Potential difference between the plates: V = Ed = qd/(ε₀A).
4★
Capacitance: C₀ = q/V = ε₀A/d.
5
With a dielectric of dielectric constant K filling the gap, the field and V fall K times, so C = Kε₀A/d = KC₀.
6
So, for a large capacitance use large plates, a small gap and a dielectric of high K.

★ key step

How to draw a parallel plate capacitor

Draw two vertical parallel lines for the plates. Label the gap d and the plate area A.
Write + signs along the left plate (+Q) and − signs along the right plate (−Q).
Draw equally spaced arrows from left to right between the plates for the uniform field E. Draw no arrows outside.
With a dielectric: shade a slab between the plates. Put − signs on its left face and + signs on its right face, draw fewer arrows inside it, and label E = E₀/K.

Types of capacitors and their uses

Type	How it is made	Dielectric	Where it is used
Paper capacitor	Long strips of metal foil and thin waxed paper (or mylar) are laid one over the other in turn, then rolled into a small cylinder.	Paper soaked in paraffin wax, or mylar	Fixed capacitors in radios and many household circuits
Metal plate (variable) capacitor	Two sets of metal plates sit between each other. In a variable one, one set is fixed and the other turns, changing the facing area and so C.	Air, or silicone oil in high-voltage types	Tuning a radio to different stations; high-voltage work
Electrolytic capacitor	A rolled metal foil sits in an electrolyte. A voltage forms a very thin insulating oxide layer on the foil.	The thin metal oxide layer	Storing large charges at low voltage, as in power supplies. Its + and − ends are marked and must not be swapped.

Worked example: capacitance and charge

Problem

Two plates, each 20 cm × 10 cm, are 1 mm apart in air. Find (a) the capacitance, (b) the charge when it is joined to a 50 V supply and (c) the capacitance when the gap is filled with a dielectric of K = 5. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)

Working

A = 0.20 m × 0.10 m = 2 × 10⁻² m²; d = 1 × 10⁻³ m.
(a) C₀ = ε₀A/d = 8.85 × 10⁻¹² × 2 × 10⁻² ÷ (1 × 10⁻³) = 1.77 × 10⁻¹⁰ F = 177 pF.
(b) Q = C₀V = 1.77 × 10⁻¹⁰ × 50 = 8.85 × 10⁻⁹ C.
(c) C = KC₀ = 5 × 177 pF = 885 pF.

Answer

(a) 177 pF; (b) 8.85 × 10⁻⁹ C (about 8.9 nC); (c) 885 pF

In real life

The tuning knob of an old radio turns a variable air capacitor. Ceiling fans and water pumps use a capacitor to start their motors, and phone chargers use electrolytic capacitors to smooth the supply.

Exam angle

1 marks: “What is the SI unit of capacitance?” The farad (F): 1 F = 1 C V⁻¹.

2 marks: “Find the capacitance of an isolated metal sphere of radius 27 cm in air. (1/4πε₀ = 9 × 10⁹ N m² C⁻²)” C = 4πε₀r = r/(9 × 10⁹) = 0.27 ÷ (9 × 10⁹) = 3 × 10⁻¹¹ F = 30 pF.

5 marks: “Derive an expression for the capacitance of a parallel plate capacitor. How does it change when a dielectric of constant K fills the gap?” σ = q/A, E = σ/ε₀ = q/(ε₀A), V = Ed = qd/(ε₀A), so C₀ = q/V = ε₀A/d. A dielectric weakens the field and V K times, so C = Kε₀A/d = KC₀. Draw the plates with A, d and the field lines.

Remember it

Big plates, small gap, strong dielectric: C = Kε₀A/d grows with A and K and shrinks with d.