Question

Question: (I) A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to 5.0 kV and stores 1200 J of energy. What is its capacitance?

Short answer

The capacitance of the capacitor is \(9.6 \times {10^{ - 5}}\;{\rm{F}}\).

Step-by-step solution

Understanding the energy stored in a capacitor

A capacitor is a charge and energy storage device. The stored energy is proportional to the square of the potential difference between the plates.

The expression for stored energy in the capacitor is given as:

\({\rm{PE}} = \frac{1}{2}C{V^2}\) … (i)

Here, C is the capacitance and V is the potential difference.

Given data

The potential difference across the capacitor is,\(V = 5.0\;{\rm{kV}} = 5.0 \times {10^3}\;{\rm{V}}\).

The stored energy in the capacitor is, \({\rm{PE}} = 1200\;{\rm{J}}\).

Determination of the capacitance

From equation (i), the capacitance is given as:

\(\begin{aligned}{c}{\rm{PE}} &= \frac{1}{2}C{V^2}\\C &= \frac{{2\left( {{\rm{PE}}} \right)}}{{{V^2}}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}C &= \frac{{2 \times \left( {1200\;J} \right)}}{{{{\left( {5.0 \times {{10}^3}\;{\rm{V}}} \right)}^2}}}\\C &= 9.6 \times {10^{ - 5}}\;{\rm{F}}\end{aligned}\)

Thus, the capacitance of the capacitor is \(9.6 \times {10^{ - 5}}\;{\rm{F}}\).