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Ch 24: Capacitance and Dielectrics
Chapter 24, Problem 24

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0x10^7 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

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1
Step 1: Calculate the electric field (E) when it is 80% of the dielectric strength. Use the formula E = 0.8 \times \text{dielectric strength}.
Step 2: Use the formula for energy density (u) of a capacitor with a dielectric, which is u = \frac{1}{2} \epsilon_0 \kappa E^2, where \epsilon_0 is the permittivity of free space (8.85 \times 10^{-12} F/m) and \kappa is the dielectric constant.
Step 3: For part (b), calculate the electric field (E) again using the same method as in Step 1 since the conditions are the same.
Step 4: Use the formula for the capacitance of a parallel-plate capacitor with a dielectric, C = \kappa \epsilon_0 \frac{A}{d}, where A is the area of the plates and d is the separation between them. Solve for A using the relationship between capacitance, stored energy (U), and voltage (V): U = \frac{1}{2} C V^2.
Step 5: Rearrange the formula from Step 4 to solve for the area (A) of each plate, using the given values for the stored energy, voltage, and calculated capacitance.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Dielectric Constant

The dielectric constant, also known as relative permittivity, is a measure of a material's ability to store electrical energy in an electric field. It is defined as the ratio of the capacitance of a capacitor with the dielectric material to the capacitance of the same capacitor in a vacuum. A higher dielectric constant indicates a greater ability to store charge, which is crucial for understanding how materials affect the performance of capacitors.
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Energy Density in a Capacitor

Energy density in a capacitor refers to the amount of energy stored per unit volume of the dielectric material. It can be calculated using the formula U = 1/2 ε E², where U is the energy density, ε is the permittivity of the dielectric, and E is the electric field strength. This concept is essential for determining how much energy can be stored in a capacitor under specific electric field conditions.
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Capacitance and Plate Area

The capacitance of a parallel-plate capacitor is directly related to the area of the plates and the distance between them, described by the formula C = ε(A/d), where C is capacitance, A is the area of the plates, d is the separation, and ε is the permittivity of the dielectric. Understanding this relationship is vital for calculating the required plate area when given energy storage and voltage conditions, as it allows for the design of capacitors to meet specific energy requirements.
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Related Practice
Textbook Question
You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?
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Textbook Question
A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00x10^4 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00x10^4 V/m?
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Textbook Question
A constant potential difference of 12 V is maintained between the terminals of a 0.25-uF, parallel-plate, air capacitor. (a) A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor (see Table 24.1) ? (b) What is the total induced charge on either face of the Mylar sheet? (c) What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increase the electric field.
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Textbook Question
When a 360-nF air capacitor (1 nF = 10^-9 F) is connected to a power supply, the energy stored in the capacitor is 1.85x10^-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 2.32x10^-5 J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?
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Textbook Question
A 5.00-uF parallel-plate capacitor is connected to a 12.0@V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?
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Textbook Question
A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm^2, what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?
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