The capacitor in an automatic external defibrillator is charged to \(7.5 \mathrm{kV}\) and stores \(2400 \mathrm{~J}\) of energy. What is its capacitance?

Short Answer

Expert verified
Answer: The capacitance of the capacitor is approximately \(8.53 \times 10^{-5} F\).

Step by step solution

01

Convert voltage to volts

Given voltage is 7.5 kV, which is equal to 7.5 * 1000 volts. So, \(V = 7500 V\)
02

Write down the energy stored in the capacitor

From the given information, the energy stored in the capacitor is \(W = 2400 J\).
03

Write down the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by \(W = \frac{1}{2}CV^2\).
04

Rearrange the formula to solve for capacitance

To find the capacitance, we need to rearrange the formula as follows: \(C = \frac{2W}{V^2}\)
05

Substitute the given values and calculate the capacitance

Plug in the values for energy and voltage into the formula and solve for the capacitance: \(C = \frac{2(2400)}{7500^2}\) \(C = \frac{4800}{56250000}\) \(C = 8.53333 \times 10^{-5} F\)
06

Write down the final answer

The capacitance of the capacitor in the automatic external defibrillator is approximately \(8.53 \times 10^{-5} F\).

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Most popular questions from this chapter

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

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