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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

Expert verified
(a) 10.138 V, (b) 2.25

Step by step solution

01

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 \( V \).
02

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 \( V = \sqrt{\frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}} \approx 10.138 \ \text{V} \).
03

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.
04

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 \( k = \frac{2 \times 4.17 \times 10^{-5}}{360 \times 10^{-9} \times (10.138)^2} \approx 2.25 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 \(\mu\)C on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm\(^2\). Each plate carries a charge of magnitude 4.35 \(\times\) 10\(^{-8}\) C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

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