Question 19.11: A heart defibrillator delivers 4.00×10^2 J of energy by disc...

College Physics[EXP-4825]

A heart defibrillator delivers $4.00 \times 10^{2} J$ of energy by discharging a capacitor initially at $1.00 \times 10^{4} V$ . What is its capacitance?
Strategy
We are given $E_{\text {cap }}$ and V , and we are asked to find the capacitance C . Of the three expressions in the equation for $E_{\text {cap }}$ , the most convenient relationship is

$E_{\text {cap }}=\frac{C V^{2}}{2}$ . (19.77)

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Solving this expression for C and entering the given values yields

$C=\frac{2 E_{\text {cap }}}{V^{2}}=\frac{2\left(4.00 \times 10^{2} J \right)}{\left(1.00 \times 10^{4} V \right)^{2}}=8.00 \times 10^{-6} F$ (19.78)

= 8.00 μF.

Discussion
This is a fairly large, but manageable, capacitance at $1.00 \times 10^{4} V$ .

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