Question

A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to 5.0?kV and stores 1300J of energy.

What is its capacitance?

Answer 1

The capacitance of a defibrillator's capacitor charged to 5.0 kV and storing 1300J of energy is 104 microfarads (μF).

Step-by-step explanation:

The student asked about the capacitance of a defibrillator's capacitor that is charged to 5.0 kV and stores 1300J of energy. The amount of energy (E) stored in a capacitor is given by the formula E = 1/2 C V^2, where E is the energy in joules, C is the capacitance in farads, and V is the potential difference in volts. To find the capacitance (C), we rearrange the formula to C = 2E / V^2.

Using the given values, E = 1300 J and V = 5000 V (5.0 kV), we can calculate the capacitance:

C = 2 * 1300 J / (5000 V)^2
C = 2600 J / 25000000 V^2
C = 0.000104 farads or 104 microfarads (μF)

Therefore, the capacitance of the capacitor in the defibrillator is 104 μF.