Question

(a) What is the energy stored in the $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$capacitor of a heartdefibrillator charged to $\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{V}$? (b) Find the amount of storedcharge.

Short answer

1. The stored energy is 405 J.
2. The charge stored is 0.09 C

Step-by-step solution

Given Data

The capacitance of the capacitor is:

$\begin{array}{rcl}\mathrm{C}& =& (10.0\mathrm{\mu F})\left(\frac{1\mathrm{F}}{{10}^6\mathrm{\mu F}}\right)\\ & =& 10.0\times {10}^{-6}\mathrm{F}\\ & & \end{array}$

The voltage between the capacitor plates is:$\mathrm{\Delta V}=9.00\times {10}^3\mathrm{V}$

Calculating for part (a) energy of capacitor

The stored energy for a capacitor of capacitancewith a potential difference between the platesis given by:

${\mathbf{U}}_{\mathbf{E}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{\left(\mathrm{\Delta V}\right)}^{\mathbf{2}}$

a)

The capacitance C is defined as charge stored per unit potential difference

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\left(\mathrm{\Delta V}\right)}$

Thus, entering the values of C and $\left(\Delta V\right)$, we obtain:

$\begin{array}{rcl}{\mathbf{U}}_{\mathbf{E}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\left(10.0\times {10}^{-6}F\right)\left(9.00\times {10}^{3}V\right)\\ & \mathbf{=}& \mathbf{405}\mathbf{J}\\ & & \end{array}$

Therefore, stored energy isrole="math" localid="1654921309243" ${\mathit{U}}_{\mathbf{E}}\mathbf{=}\mathbf{\text{405 J}}$

Calculating for part (b), stored charge

b)

The stored charge is given as

$\mathit{Q}\mathbf{=}\mathit{C}\mathbf{\times }\mathit{\Delta }\mathit{V}$

Substitute numerical values:

$$\begin{gathered} \mathbf{\text{Q=}}\left(10.0\times {10}^{-6}F\right)\left(9.00\times {10}^{3}V\right) \\ \mathbf{=}\mathbf{0}\mathbf{.09}\mathbf{C} \end{gathered}$$

Therefore, the charge accumulated on capacitor plates islocalid="1654921503322" $\mathit{Q}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}\mathbf{}\mathbf{\text{C}}$