Question

A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery remains attached to the capacitor, the distance between the parallel plates increases by \(25 \% .\) What happens to the energy stored in the capacitor?

Short answer

The energy stored in the capacitor decreases to 80% of its original value.

Step-by-step solution

Understand Capacitor Energy Formula

The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. Since the voltage remains constant as per the problem statement, we will focus on the capacitance \( C \).

Capacitance Dependence on Plate Distance

For a parallel plate capacitor, capacitance \( C \) is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. As the plate distance \( d \) increases by \( 25\% \), \( d \) is replaced by \( 1.25d \). This changes capacitance to \( C' = \frac{\varepsilon_0 A}{1.25d} \).

Calculate the Change in Capacitance

Substitute the new distance into the capacitance formula: \( C' = \frac{C}{1.25} \). The new capacitance is \( 0.8C \), indicating that the capacitance decreases. This decreased capacitance affects the energy stored in the capacitor.

Apply Change to Energy Formula

Substitute the new capacitance into the energy formula: \( E' = \frac{1}{2} C' V^2 = \frac{1}{2} \times 0.8C \times V^2 = 0.8 \times \frac{1}{2} CV^2 = 0.8E \). The new energy stored in the capacitor is \( 80\% \) of the original energy.

Conclusion: Energy Decreases

The energy stored in the capacitor decreases to \( 80\% \) of its original value because the capacitance decreases when the plate distance is increased by \( 25\% \), while the voltage remains constant.

Key concepts

Capacitance

Capacitance is a fundamental concept in electronics that refers to a component's ability to store electrical charge. It is typically measured in farads (F), a unit derived from the International System of Units. The capacitance of a system depends on the physical characteristics of the system. For a parallel plate capacitor, the capacitance is calculated using the formula \( C = \frac{\varepsilon_0 A}{d} \).
\(\varepsilon_0\) represents the permittivity of free space, \(A\) is the area of the capacitor plates, and \(d\) is the distance between these plates.

Several factors can affect the capacitance of a device:

• Area of the plates: Larger plates can store more charge, increasing capacitance.
• Distance between plates: A smaller distance leads to higher capacitance because the charged plates are closer.
• Dielectric constant: Inserting a dielectric material between the plates can enhance capacitance significantly.

Understanding these dependencies helps explain changes in energy storage in capacitors when physical parameters are altered.

Parallel Plate Capacitor

A parallel plate capacitor is a specific type of capacitor formed by two conductive plates separated by a distance, typically with air or a dielectric material in between. This setup is one of the simplest and most widely studied capacitor designs due to its straightforward geometry and predictable behavior.

Key features of a parallel plate capacitor include:

• Uniform electric field: The electric field between the plates is uniform, meaning it has constant strength and direction.
• Predictability: The simple geometry allows easy calculation of capacitance with the formula \( C = \frac{\varepsilon_0 A}{d} \).

When connected to a power source, such as a battery, a constant potential difference is maintained, leading to the storage of electric energy. The energy stored is dependent on the capacitance and the square of the voltage according to the formula \( E = \frac{1}{2} C V^2 \).

This straightforward relationship allows for easy predictions of energy changes when dimensions such as plate distance are altered.

Electric Potential Energy

Electric potential energy in capacitors reflects the work done to charge the plates. Specifically, it is the energy stored in the electric field between the plates of a capacitor. This energy can be released when needed, which is why capacitors are valuable in circuits.

For a capacitor, the energy stored is given by the equation \( E = \frac{1}{2} C V^2 \), where \( C \) indicates the capacitance and \( V \) is the voltage.

Important points about electric potential energy in capacitors:

• Dependence on Capacitance: As capacitance decreases with increased plate distance, so does the stored energy.
• Voltage Constant: In a situation where voltage remains constant, any change in capacitance directly impacts the stored energy.

This connection explains why, in the given problem, the increase in plate distance reduces energy to 80% of its original value. The decreased capacitance results in less potential energy stored, despite unchanged voltage levels.