In open-heart surgery, a much smaller amount of energy will defibrillate the heart. (a) What voltage is applied to the capcitor of a heart defibrilator that 40.0J of energy? (b) Find the amount of the stored charge.

In Open Heart Surgery A Small Amount Of Energy Will Defibrillate The Heart

This question aims to understand the concept of capacitors, how the electrical charge charges the capacitor, and how to calculate the energy stored in the capacitor.

In electrical circuits, the capacitor is commonly used as an electrical component, with storing an electrical charge as the main role. Charge of opposite value and same magnitude is present on adjacent plates in standard parallel plate capacitors. The electrical potential energy is stored in the capacitor.  The conductor in the capacitor is initially uncharged and requires a potential difference V by connecting it to the battery. if at that time q is the charge on the plate, then q = CV. The product of the potential and charge is equal to the work done. Hence, W = Vq. The battery provides a little amount of charge at a stable voltage V, and the energy stored in the capacitor becomes:

\[ U = \dfrac{1}{2}CV^2\]

Applications of capacitors in microelectronics are handheld calculators, audio tools, camera flashes, uninterruptible power supplies, and pulsed loads such as magnetic coils and lasers.

Expert Answer

Part a:

In this question, we are given:

The capacitance of the capacitor that is: $C \space=\space 8 \mu F$ and that is equal to: $\space 8 \times 10^{-6}$

The energy stored in the capacitor that is: $U_c \space=\space 40J$

And we are asked to find the voltage in the capacitor.

The formula that relates the voltage in the capacitor, the capacitance of the capacitor, and the energy stored in the capacitor are given as:


Rearranging the formula to make Voltage $V$ the subject because it is an unknown parameter that we are asked to find:

\[V=\sqrt{ \dfrac{2U_c}{C}}\]

Now plugging the values of $U_c$ and $C$ and solving for $V$:

\[ V= \sqrt{ \dfrac{2 \times 40}{8 \times 10^{-6}}} \]

By solving the expression, $V$ comes out to be:

\[ V=3.162 \space KV \]

Part b:

The stored charge $Q$ is the unknown parameter.

The formula that related the energy stored in the capacitor $U_c$, Voltage $V$ and the stored charge $Q$ is given as:

\[ U_c = \dfrac{1}{2}QV \]

Making $Q$ the subject:

\[ Q = \dfrac{2U_c}{v} \]

Plugging the values and solving:

\[ Q = \dfrac{2 \times 40}{3162} \]

By solving the expression, $Q$ comes out to be:

\[Q=0.0253 \space C\]

Numerical Results

Part a: Voltage is applied to the $8.00 \mu F$ capacitor of a heart defibrillator that stores $40.0 J$ of energy is $3.16 \space KV$.

Part b:  The amount of the stored charge is $0.0253C$.


A $12pF$ capacitor is connected to a $50V$ battery. Once the capacitor is fully charged, how much electrostatic energy is stored?

The formula given for finding the amount of energy stored in the capacitor is:

\[E \space = \space \dfrac{1}{2} CV^2\]

\[E \space = \space \dfrac{1}{2} (12 \times 10^{-12})(50)^2 \]

By solving the expression, Energy $E$ comes out to be:

\[E \space = 1.5 \times 10^{-8} J \]

Once the capacitor is fully charged, electrostatic energy stored is $ 1.5 \times 10^{-8} J$

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