Question

When a 360-nF air capacitor (11 nF = 10\(^{-9}\) F) is connected to a power supply, the energy stored in the capacitor is 1.85 \(\times\) 10\(^{-5}\) J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 \(\times\) 10\(^{-5}\) J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

Short answer

(a) 10.138 V, (b) 2.25

Step-by-step solution

Calculate Initial Potential Energy

The energy stored in a capacitor can be calculated using the formula: \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference. We know the energy before the dielectric is inserted is \( 1.85 \times 10^{-5} \ \text{J} \) and the initial capacitance \( C = 360 \times 10^{-9} \ \text{F} \). Plugging in the values, we solve for \( V \).

Solve for Potential Difference

Using the equation \( E = \frac{1}{2} C V^2 \), substitute \( E = 1.85 \times 10^{-5} \ \text{J} \) and \( C = 360 \times 10^{-9} \ \text{F} \). Thus, \( \frac{1}{2} \times 360 \times 10^{-9} \times V^2 = 1.85 \times 10^{-5} \). Solving for \( V \), we get \( V = \sqrt{\frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}} \approx 10.138 \ \text{V} \).

Calculate New Energy with Dielectric

The total energy stored becomes \( 1.85 \times 10^{-5} + 2.32 \times 10^{-5} = 4.17 \times 10^{-5} \ \text{J} \) after the dielectric slab is inserted.

Find the Dielectric Constant

When the dielectric is inserted, the capacitance increases by a factor \( k \), the dielectric constant. The new energy stored is given by \( E_{\text{new}} = \frac{1}{2} k C V^2 \). Using the new total energy \( 4.17 \times 10^{-5} \ \text{J} \), we solve \( 4.17 \times 10^{-5} = \frac{1}{2} k \times 360 \times 10^{-9} \times (10.138)^2 \). Solving for \( k \), we get \( k = \frac{2 \times 4.17 \times 10^{-5}}{360 \times 10^{-9} \times (10.138)^2} \approx 2.25 \).

Key concepts

Capacitance

Capacitance is a measure of a capacitor's ability to store charge. It determines how much electrical charge a capacitor can hold for a given voltage. The unit of capacitance is the farad (F), where 1 farad is equivalent to 1 coulomb of charge per volt. Capacitance is influenced by the physical characteristics of the capacitor, such as the surface area of the plates, the distance between them, and the material between the plates, known as the dielectric.
When a dielectric material is inserted between the plates of a capacitor, it increases the capacitance by reducing the electric field inside the capacitor, allowing it to hold more charge for the same potential difference. The factor by which the capacitance increases is called the dielectric constant, denoted by the symbol \(k\). The relationship between the initial capacitance \(C\) and the new capacitance \(C'\) when a dielectric is added can be expressed as \(C' = kC\).

Energy Stored in Capacitors

The energy stored in a capacitor is a key property that makes capacitors useful in various applications, such as energy storage, filtering, and signal processing. The stored energy can be calculated using the formula:

• \( E = \frac{1}{2} CV^2 \)

where \(E\) is the energy in joules, \(C\) is the capacitance in farads, and \(V\) is the potential difference across the capacitor in volts.
Adding a dielectric increases the total energy stored in a capacitor. In the given exercise, we observe the energy increasing from \(1.85 \times 10^{-5}\) J to \(4.17 \times 10^{-5}\) J after a dielectric slab fills the space between the plates. This increase is because the dielectric allows the capacitor to store more electric field energy between its plates due to the increased capacitance. We can say that with the dielectric, the capacitor can hold more energy for the same voltage because the dielectric constant \(k\) enhances the capacitance.

Potential Difference

Potential difference, or voltage, across the plates of a capacitor is the measure of electrical energy per unit charge between the plates. It is represented by the symbol \(V\) and measured in volts (V). In the context of capacitors, potential difference indicates the force with which the electric field in the capacitor can push charge onto the plates.
In the provided exercise, the potential difference is calculated using the energy formula. Initially, with a capacitance of 360 nF and stored energy of \(1.85 \times 10^{-5}\) J, the potential difference can be calculated using:

• \( V = \sqrt{\frac{2E}{C}} \)

By substituting the given values, the initial potential difference is found to be approximately 10.138 V. This voltage remains constant even after the dielectric is inserted, assuming the power supply still provides the same potential difference across the capacitor. This constant voltage across the capacitor underlines the importance of the potential difference in maintaining consistent energy storage characteristics in the presence of a dielectric.