Electrostatic Potential and Capacitance
Capacitors and dielectrics
Calculate capacitance with and without a dielectric.
A capacitor stores charge and energy. It is two conductors (plates) close together, separated by an insulator. When connected to a battery, one plate gets +Q and the other –Q.
The charge is proportional to the voltage. The ratio is the capacitance: C = Q / V. Unit: farad (F). One farad is huge, so we use µF (10⁻⁶ F) and pF (10⁻¹² F).
For parallel plates of area A, distance d apart, in air: C = ε₀A / d. Bigger plates hold more; a smaller gap holds more.
Example: plates of 1 m², 1 mm apart: C = 8.854 × 10⁻¹² / 10⁻³ ≈ 8.9 nF. Even a square metre of plate gives only nanofarads, which shows how big a farad is.
Dielectrics: fill the gap with an insulator such as glass or mica, of dielectric constant K. Its molecules line up (it is polarised), and their charges partly cancel the field. The field inside drops to E₀/K, so for the same charge V drops too, and:
C = Kε₀A / d, K times bigger. With K = 5, the 8.9 nF capacitor becomes about 44 nF.
Inserting a dielectric, in two cases:
| Battery still connected | Battery removed first | |
|---|---|---|
| Fixed quantity | V | Q |
| C | × K | × K |
| Q | × K | same |
| V | same | ÷ K |
| E between plates | same | ÷ K |
If the slab fills only part of the gap (thickness t), C = ε₀A / (d – t + t/K).
Practise this lesson
Questions that get harder as you improve, a Tutor that helps without giving the answer away, and progress you can see. Free for every learner.
No account, no email address. Opens the real product.