Question

In open heart surgery, a much smaller amount of energy will defibrillate the heart.

(a) What voltage is applied to the \(8.00{\rm{ }}\mu F\)capacitor of a heart defibrillator that stores \(40.0{\rm{ }}J\) of energy?

(b) Find the amount of stored charge.

Short answer

(a) The amount of voltage applied to the capacitor of a heartdefibrillator is\(\Delta V = 3.16{\rm{ }}kV\).

(b) The amount of charge stored is \(Q = 25.3{\rm{ }}mC\).

Step-by-step solution

Given Information

• Energy stored in the Capacitor –\(4.40{\rm{ }}J\)
• Capacitance value for the Capacitor – \(8.00{\rm{ }}\mu F\)

Concept Introduction

A capacitor with capacitance\(C\)that is charged to a potential difference\(\Delta V\)contains the following amount of energy –

\({U_E} = \frac{1}{2}C{(\Delta V)^2}...(1)\)

Capacitors: Any pair of conductors divided by an insulating substance is referred to as a capacitor. The potential\(\Delta V\)of the positively charged conductor with respect to the negatively charged conductor is proportional to\(Q\)when the capacitor is charged, and there are charges on the two conductors of equal magnitude and opposite sign. The ratio of\(Q\)to\(\Delta V\)is what determines the capacitance\(C\).

\(C = \frac{Q}{{\Delta V}}...(2)\)

Voltage Calculation

(a)

The potential difference between the plates of the capacitor is found by solving Equation\((1)\)for\(\Delta V\):

\(\begin{array}{c}{U_E} = \frac{1}{2}C{(\Delta V)^2}\\2{U_E} = C{(\Delta V)^2}\\{(\Delta V)^2} = \frac{{2{U_E}}}{C}\\\Delta V = \sqrt {\frac{{2{U_E}}}{C}} \end{array}\)

Entering the values for\({U_E}\)and\(C\), it is obtained:

\(\begin{array}{c}\Delta V = \sqrt {\frac{{2(40.0\;J)}}{{8.00 \times {{10}^{ - 6}}\;F}}} \\ = 3.16 \times {10^3}\;V\\ = \left( {3.16 \times {{10}^3}\;V} \right)\left( {\frac{{1{\rm{ }}kV}}{{1000\;V}}} \right)\\ = 3.16{\rm{ }}kV\end{array}\)

Calculation of charge stored

(b)

The charge accumulated on the capacitor plate is found by solving Equation\((2)\)for\(Q\):

\(Q = C\Delta V\)

Substitute numerical values:

\(\begin{array}{c}Q = \left( {8.00 \times {{10}^{ - 6}}\;F} \right)\left( {3.16 \times {{10}^3}\;V} \right)\\ = \left( {25.3 \times {{10}^{ - 3}}{\rm{ }}C} \right)\left( {\frac{{1000{\rm{ }}mC}}{{1{\rm{ }}C}}} \right)\\ = 25.3{\rm{ }}mC\end{array}\)

Therefore, the value of voltage and charge stored are \(\Delta V = 3.16{\rm{ }}kV\)and \(Q = 25.3{\rm{ }}mC\)respectively.