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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 spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

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