KinexiaLEARN PHYSICS

Capacitance — Defibrillator Capacitor (Worked Example)

Medium
Problem

A defibrillator charges a 32 μF capacitor to 5000 V before a shock. What charge is stored?

Given

  • C = 32 μF = 3.2 × 10⁻⁵ F

  • V = 5000 V

Find

The charge Q.

Principle

High voltage × modest capacitance still stores serious charge: Q = CV.

Formula

Solution

  1. Convert: C=3.2×10−5 FC = 3.2 \times 10^{-5}\ \mathrm{F}

  2. Substitute: Q=(3.2×10−5)(5000)Q = (3.2 \times 10^{-5})(5000)

  3. Multiply: Q=0.16 CQ = 0.16\ \mathrm{C}

0.16 C

Unit check

F × V = C — correct.

What this means physically

0.16 C released in ~10 ms drives ~16 A through the chest — the therapeutic jolt. This is why defibrillator capacitors are handled with extreme care.

Common mistake

Converting μF wrong: 32 μF = 32 × 10⁻⁶ = 3.2 × 10⁻⁵ F. Write the power of ten explicitly every time.