Question

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

Short answer

Answer: The energy stored in the capacitor increases after the insertion of the dielectric, and the dielectric is pulled into the space between the plates.

Step-by-step solution

Part a: Energy before dielectric insertion

The energy stored in a capacitor before the dielectric was inserted can be calculated using the formula \[U = \frac{1}{2}CV^2,\] where \(U\) is the stored energy, \(C\) is the capacitance, and \(V\) is the constant potential difference. Using this formula, we can calculate the initial energy stored as \[U_{initial} = \frac{1}{2}CV^2.\]

Part b: Energy after dielectric insertion

When the dielectric is inserted between the plates, the capacitance changes. The new capacitance, \(C'\), is given by \[C' = \kappa C,\] where \(\kappa\) is the dielectric constant. The energy stored after dielectric insertion can be calculated using the same formula as before: \[U_{final} = \frac{1}{2}C'V^2,\] By substituting \(C' = \kappa C\), we get \[U_{final} = \frac{1}{2}(\kappa C)V^2.\]

Part c: Dielectric interaction with the plates

To determine if the dielectric was pulled or pushed into the space between the plates, we should analyze the forces acting on it. The force can be determined by the change in energy when the dielectric is inserted. \[F = U_{final} - U_{initial}\] Substitute the expressions for \(U_{final}\) and \(U_{initial}\): \[F = \frac{1}{2}(\kappa C)V^2 - \frac{1}{2}CV^2\] Factor out \(\frac{1}{2}CV^2\): \[F = \frac{1}{2}CV^2(\kappa-1)\] Since \(\kappa > 1\) for dielectric materials, the value of \(F\) will be positive. This means that the energy stored in the capacitor has increased after inserting the dielectric. Therefore, the dielectric was pulled into the space between the plates.

Key concepts

Dielectric Constant

A dielectric constant, denoted by \(\kappa\), is a key property of dielectric materials. It measures the material's ability to increase the capacitance of a capacitor when inserted between the plates. This occurs because dielectric materials are electrical insulators that can be polarized by an electric field. As a result, they store more electrical energy compared to the vacuum.

• \(\kappa > 1\) for all dielectric materials: This indicates their effectiveness in increasing capacitance.
• The larger the \(\kappa\), the greater the capacitance and stored energy.

Understanding dielectric constants is essential because they affect how capacitors are used in circuits and electronic devices.

Potential Difference

The potential difference, often denoted as \(V\), is the voltage applied across the capacitor plates. It is crucial in determining the amount of energy stored in the capacitor. This voltage creates an electric field between the plates that allows the capacitor to store energy.

• A higher potential difference results in more energy being stored, assuming the capacitance remains constant.
• Even when a dielectric material is inserted, the potential difference \(V\) remains constant in this scenario.

Therefore, potential difference plays a vital role in both the initial and final energy storage within the capacitor.

Stored Energy

Stored energy in a capacitor, represented by \(U\), is calculated using the formula \(U = \frac{1}{2}CV^2\), where \(C\) is the capacitance and \(V\) is the potential difference. The energy quantifies how much electric potential energy is kept within the capacitor.

• Before a dielectric is inserted, energy depends solely on the initial capacitance \(C\).
• After a dielectric is inserted, energy increases due to increased capacitance, expressed as \(C' = \kappa C\).

This change means more energy is stored after the dielectric insertion, demonstrating how dielectrics enhance a capacitor's performance.

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a small distance. Capacitance depends on the area of the plates, the distance between them, and the properties of any insulating material (dielectric) between the plates.

• Capacitance without dielectric: \(C = \frac{\varepsilon_0 A}{d}\), where \(A\) is the area, \(d\) is the distance, and \(\varepsilon_0\) is the permittivity of free space.
• With dielectric: Capacitance becomes \(C' = \kappa C\) when a dielectric is added.

Parallel plate capacitors are fundamental in storing energy and are used extensively in circuits, making understanding their operation crucial.