Question

You have been hired as an expert witness by the defense attorneys for a company that builds defibrillators, which are discussed in the opening storyline of the chapter on capacitance and dielectrics. One of the company's defibrillator units was used by paramedics to attempt to stop the ventricular fibrillation of a patient. Despite several attempts in the ambulance on the way to the emergency room, the fibrillation could not be stopped until a different defibrillator was used in the ER. Although the patient survived, the failure to stop the fibrillation early caused some damage to the heart. The company is being sued by attorneys representing the patient, who claim that the design of the unit is faulty and does not deliver enough energy to the patient. The company claims that its unit delivers an energy of 360 J, in compliance with a safety directive from the government. You open the device and find a 32.6 µF capacitor. You also find a battery and charging circuit that, according to tests you perform, can apply a voltage of 4.20 ✕ 103 V to the capacitor. Determine the energy stored in the capacitor (in J) so that you can give advice to the attorneys.

__________J

Answer 1

The energy stored in a capacitor with capacitance connected to potential difference is

$E=\frac{1}{2}CV^{2}$

For
C= 32.6μ F-32.6 × 10-6F
,
4.2 × 103V

We get,

0-6 × (4.2 × 103)2 E =-× 32.6 × 1

E = 287 ,5 J