Capacitance and Capacitors
High-Yield Summary
- Capacitors store energy as charge (not heat). Capacitance: C = Q/V, measured in farads (1 F = 1 C/V).
- Parallel plate capacitor: C = ε0·A/d — capacitance increases with plate area, decreases with plate separation.
- Field between plates: E = V/d. Energy stored: U = ½CV² — doubling voltage quadruples stored energy.
- A dielectric increases capacitance: C' = KC, where K is the dielectric constant.
- Series capacitors: 1/Cs = Σ1/C (capacitance decreases). Parallel capacitors: Cp = ΣC (capacitance increases) — the mirror image of resistor rules.
Capacitance, Field, and Stored Energy
C = Q/V | C = ε0A/d | E = V/d | U = ½CV²
- C = Capacitance, in farads (F)
- Q = Charge stored on one plate
- V = Voltage across the plates
- ε0 = Permittivity of free space, 8.85×10⁻¹² F/m
- A, d = Plate area and plate separation
- U = Energy stored in the capacitor's field
Capacitors vs. Resistors: Series/Parallel Mirror Image
| Configuration | Resistors vs. Capacitors |
|---|---|
| Series | Resistors: Rs=ΣR (adds). Capacitors: 1/Cs=Σ1/C (decreases) |
| Parallel | Resistors: 1/Rp=Σ1/R (decreases). Capacitors: Cp=ΣC (adds) |
Common MCAT Trap
- Capacitor series/parallel rules are the exact mirror image of resistor rules — applying the resistor formula to a capacitor circuit (or vice versa) is a classic, high-frequency mistake.
- Energy stored scales with V², not V — doubling voltage quadruples stored energy, not doubles it.
- A dielectric increases capacitance (C' = KC with K > 1); it doesn't reduce it, even though it 'blocks' some field — it works by reducing the internal field per unit charge, letting more charge be stored.
Quick Recall
Adding capacitors in series vs. parallel — which increases total capacitance?
If voltage across a capacitor doubles, what happens to its stored energy?