Energy Stored by Capacitor: Videos & Practice Problems

Capacitance relates charge and voltage, storing potential energy in capacitors. The energy stored is expressed as ${\frac{1}{2}}^c{v}^2$, with alternative forms $\frac{1}{2}qv$ and $\frac{1}{2}{q}^2/c$. Energy density, $u$, is energy per volume, calculated as $\frac{U}{V}$, leading to $\frac{1}{2}\epsilon E^2$ for parallel plate capacitors.

Energy Stored by Capacitor

Understanding capacitance is crucial for grasping how capacitors store energy. A capacitor consists of two plates that separate charges, leading to stored potential energy. The energy stored in a capacitor can be expressed using the formula:

$$U = \frac{1}{2} C V^2$$

where \(U\) is the potential energy, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor. This equation can be rearranged into two other forms based on the relationship between charge (\(Q\)), capacitance, and voltage:

$$U = \frac{1}{2} Q V$$

$$U = \frac{1}{2} \frac{Q^2}{C}$$

These variations allow flexibility depending on the known variables in a problem. The energy density (\(u\)), which is the energy per unit volume, is defined as:

$$u = \frac{U}{V}$$

where \(V\) is the volume between the plates, calculated as the area of the plates multiplied by the distance between them:

$$V = A \cdot d$$

For a parallel plate capacitor, the energy density can also be expressed in terms of the electric field (\(E\)) as:

$$u = \frac{1}{2} \epsilon_0 E^2$$

Here, \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{F/m}\). This relationship indicates that the energy density is directly related to the electric field strength between the plates.

To calculate the energy stored in a capacitor, one can use the capacitance formula for parallel plates:

$$C = \frac{\epsilon_0 A}{d}$$

By substituting this into the energy formula, one can derive the energy stored based on the area and distance between the plates, along with the voltage applied.

For example, if a capacitor has an area of \(50 \, \text{cm}^2\) (converted to \(0.005 \, \text{m}^2\)) and a separation distance of \(10 \, \text{mm}\) (or \(0.01 \, \text{m}\)), with a voltage of \(20 \, \text{V}\), the energy stored can be calculated as:

$$U = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) V^2$$

After substituting the values, the potential energy stored is found to be approximately \(8.85 \times 10^{-10} \, \text{J}\).

To find the energy density, divide the total energy by the volume:

$$u = \frac{U}{A \cdot d}$$

Using the previously calculated energy, the energy density can be determined, yielding a result of approximately \(1.77 \times 10^{-5} \, \text{J/m}^3\).

In another scenario, if the energy density is given as \(2.5 \, \text{mJ/cm}^3\), it can be converted to the appropriate units and used to find the electric field strength using the rearranged energy density formula:

$$E = \sqrt{\frac{2u}{\epsilon_0}}$$

After performing the necessary conversions and calculations, the electric field strength can be determined, illustrating the relationship between energy density and electric field in capacitors.

A cardiac defibrillator can be modeled as a parallel plate capacitor. When it is charged to a voltage of 2 kV, it has a stored energy of 1 kJ. What is the capacitance of the defibrillator?

• A. 2.5×10⁻⁴ F
• B. 5×10⁻⁴ F
• C. 2.5×10⁻¹ F
• D. 5×10⁻¹ F
• E. 1.0 F

Answer: (B)

Energy Released by Flashbulb

In this scenario, we analyze the energy dynamics of a flashlight that operates using a capacitor. The flashlight has a specified capacitance and voltage, and when it is activated, it releases energy as the bulb lights up. The key concept here is to determine the energy released when the flashlight discharges a certain amount of charge.

The energy stored in a capacitor can be expressed using the formula for potential energy, which is given by:

$$ U = \frac{1}{2} C V^2 $$

or alternatively, in terms of charge:

$$ U = \frac{1}{2} Q V $$

and

$$ U = \frac{1}{2} \frac{Q^2}{C} $$

In this case, we need to find the change in energy (ΔU) when the flashlight loses 80% of its charge. This means that the final charge (Q_final) is 20% of the initial charge (Q_initial). Therefore, we can express the final charge as:

$$ Q_{final} = 0.2 \times Q_{initial} $$

Given that the capacitance (C) remains constant, we can use the equation:

$$ \Delta U = U_{final} - U_{initial} = \frac{1}{2} \frac{Q_{final}^2}{C} - \frac{1}{2} \frac{Q_{initial}^2}{C} $$

To find the initial charge, we use the relationship between charge, capacitance, and voltage:

$$ Q_{initial} = C \times V $$

For example, if the capacitance is 1,000 millifarads (or 1 farad) and the voltage is 500 volts, the initial charge would be:

$$ Q_{initial} = 1 \, \text{F} \times 500 \, \text{V} = 500 \, \text{C} $$

Consequently, the final charge is:

$$ Q_{final} = 0.2 \times 500 \, \text{C} = 100 \, \text{C} $$

Now, substituting these values into the energy change equation gives:

$$ \Delta U = \frac{1}{2} \frac{(100 \, \text{C})^2}{C} - \frac{1}{2} \frac{(500 \, \text{C})^2}{C} $$

After calculating, the energy released is found to be:

$$ \Delta U = -120,000 \, \text{J} $$

This negative value indicates a decrease in potential energy, which aligns with the understanding that the energy is converted into light when the flashlight is activated. Thus, the energy released during the discharge process is significant, demonstrating the relationship between charge, capacitance, and energy in capacitive systems.

Here’s what students ask on this topic:

What is the formula for the energy stored in a capacitor?

The energy stored in a capacitor is given by the formula:

$U_{}=\frac{1}{2}{C}_{}{V}^2$

where $U_{}$ is the energy stored, $C_{}$ is the capacitance, and $V^2$ is the voltage squared. Alternative forms of this equation include:

$U_{}=\frac{1}{2}{Q}_{}{V}_{}$

and

$U_{}=\frac{1}{2}\frac{{Q_{}}^2}{C_{}}$

These forms are useful depending on the given variables in a problem.

How do you calculate the energy density of a parallel plate capacitor?

The energy density $u$ of a parallel plate capacitor is the energy stored per unit volume. It is given by:

$u=\frac{U_{}}{V}$

where $U_{}$ is the potential energy and $V$ is the volume between the plates. For a parallel plate capacitor, this can be simplified to:

$u=\frac{1}{2}{\epsilon }_{\epsilon }$

where ${\epsilon }_{}$ is the permittivity of free space and $E^2$ is the electric field squared.

What are the different forms of the energy stored equation for capacitors?

The energy stored in a capacitor can be expressed in three common forms:

1. $U_{}=\frac{1}{2}{C}_{}{V}^2$

2. $U_{}=\frac{1}{2}{Q}_{}{V}_{}$

3. $U_{}=\frac{1}{2}\frac{{Q_{}}^2}{C_{}}$

These forms are derived from the relationships between charge ($Q_{}$), capacitance ($C_{}$), and voltage ($V_{}$), and can be used depending on the given variables in a problem.

How do you find the electric field strength in a capacitor storing a specific energy density?

To find the electric field strength $E$ in a capacitor storing a specific energy density $u$, use the formula:

$u=\frac{1}{2}{\epsilon }_{}{E}^2$

Rearrange to solve for $E$:

$E=\sqrt{\frac{2}{u_{}}}$

where ${\epsilon }_{}$ is the permittivity of free space. Ensure the energy density $u$ is in the correct units (joules per cubic meter) before substituting into the equation.

What is the significance of the energy density in a capacitor?

The energy density $u$ in a capacitor is significant because it represents the amount of energy stored per unit volume. This is particularly important in applications where space is limited, and maximizing energy storage efficiency is crucial. The energy density is given by:

$u=\frac{1}{2}{\epsilon }_{}{E}^2$

where ${\epsilon }_{}$ is the permittivity of free space and $E^2$ is the electric field squared. Higher energy density means more energy can be stored in a smaller volume, which is beneficial for compact electronic devices.