Q69PE - College Physics (Urone)

Question

Construct Your Own Problem 

Consider a heart defibrillator similar to that discussed in Example 19.11 . Construct a problem in which you examine the charge stored in the capacitor of a defibrillator as a function of stored energy. Among the things to be considered are the applied voltage and whether it should vary with energy to be delivered, the range of energies involved, and the capacitance of the defibrillator. You may also wish to consider the much smaller energy needed for defibrillation during open-heart surgery as a variation on this problem.

Step-by-Step Solution

Answer

The charge stored in the capacitor of a defibrillator as a function of stored energy is $\mathrm{Q}=\sqrt{2{\mathrm{CE}}_{\mathrm{cap}}}$ and it ranges from 28.8 mC to 80.0 mC.

Step 1: Concept Introduction

When the capacitor is charged, there are charges of equal magnitude Q and opposite sign on the two conductors, and the potential difference $\mathbf{∆}\mathbf{V}$ is proportional to Q . The formula for the charge on the capacitor is given as –

 $\mathbf{Q}\mathbf{=}\mathbf{CV}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Step 2: Calculation for stored charge

The applied voltage depends on the energy to be stored in the capacitor.

The applied voltage can be calculated as –

 $$\begin{gathered} {\mathbf{E}}_{\mathbf{cap}}\mathbf{=}\frac{{\mathbf{CV}}^{\mathbf{2}}}{\mathbf{2}} \\ \mathbf{V}\mathbf{=}\sqrt{\frac{\mathbf{2}{\mathbf{E}}_{\mathbf{ap}}}{\mathbf{C}}} \end{gathered}$$

The charge stored on the capacitor as a function of stored energy can be calculated as –

 $\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \mathbf{CV}\\ & \mathbf{=}& \mathbf{C}\mathbf{\times }\sqrt{\frac{\mathbf{2}{\mathbf{E}}_{\mathbf{cap}}}{\mathbf{C}}}\\ & \mathbf{=}& \sqrt{\mathbf{2}{\mathbf{CE}}_{\mathbf{cap}}}\\ & & \end{array}$

Step 3: Range Calculation

The range of energy stored in the capacitor is 50J to 400J .

For ${\mathrm{E}}_{\mathrm{cap}}=50 \mathrm{J},$  the charge on the capacitor can be calculated as –

 $\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \sqrt{\mathbf{2}\mathbf{\times }\left(8\times {10}^{-6} F\right)\mathbf{(}\mathbf{50}\mathbf{ }\mathbf{J}\mathbf{)}}\\ & \mathbf{=}& \mathbf{28}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{C}\\ & \mathbf{=}& \mathbf{28}\mathbf{.}\mathbf{28}\mathbf{ }\mathbf{mC}\end{array}$

For ${\mathrm{E}}_{\mathrm{cap}}=400 \mathrm{J},$ , the charge on the capacitor can be calculated as –

 $\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \sqrt{\mathbf{2}\mathbf{\times }\left(8\times {10}^{-6} F\right)\mathbf{(}\mathbf{400}\mathbf{ }\mathbf{J}\mathbf{)}}\\ & \mathbf{=}& \mathbf{80}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{C}\\ & \mathbf{=}& \mathbf{80}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mC}\end{array}$

 

Thus, the charge stored in the capacitor of a defibrillator as a function of stored energy is $\mathrm{Q}=\sqrt{2{\mathrm{CE}}_{\mathrm{cap}}}$ and it ranges from 28.8 mC to 80.0 mC.