Ch 24: Capacitance and Dielectrics

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 24: Capacitance and Dielectrics

Problem 40

Chapter 24, Problem 40

Polystyrene has dielectric constant $2.6$ and dielectric strength $2.0 \times 10^{7}$ 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 $80 percent sign$ of the dielectric strength. What is the area of each plate if the capacitor stores $0.200$ mJ of energy under these conditions?

Verified step by step guidance

Step 1: Understand the problem by identifying the given values: dielectric constant (κ) = 2.6, dielectric strength = 2.0 × 10^7 V/m, electric field (E) = 80% of dielectric strength, battery voltage (V) = 500.0 V, and stored energy (U) = 0.200 mJ.

Step 2: For part (a), calculate the electric field (E) when it is 80% of the dielectric strength. Use the formula E = 0.8 × dielectric strength. Then, use the formula for energy density (u) in a dielectric: u = 0.5 × κ × ε₀ × E², where ε₀ is the permittivity of free space (8.85 × 10^-12 C²/N·m²).

Step 3: For part (b), calculate the electric field (E) again using E = 0.8 × dielectric strength. Then, use the formula for the energy stored in a capacitor: U = 0.5 × C × V², where C is the capacitance. Rearrange this formula to solve for C: C = 2U / V².

Step 4: Use the relationship between capacitance, dielectric constant, area of the plates (A), and separation distance (d): C = κ × ε₀ × A / d. Since the electric field E = V/d, rearrange to find d = V/E. Substitute d into the capacitance formula to solve for A: A = C × d / (κ × ε₀).

Step 5: Substitute the values for C, d, κ, and ε₀ into the formula for A to find the area of each plate. Ensure all units are consistent when performing calculations.

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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 the ratio of the permittivity of the dielectric material to the permittivity of free space. A higher dielectric constant indicates better insulating properties, allowing the capacitor to store more energy.

Dielectric Strength

Dielectric strength is the maximum electric field a material can withstand without breaking down or losing its insulating properties. It is expressed in volts per meter (V/m). In the context of capacitors, it determines the maximum voltage that can be applied across the dielectric without causing electrical breakdown, which is crucial for safe operation.

Energy Density in Capacitors

Energy density refers to the amount of energy stored in a capacitor per unit volume. It is calculated using the formula: energy density = 0.5 * ε * E^2, where ε is the permittivity of the dielectric material and E is the electric field strength. Understanding energy density helps in determining how much energy a capacitor can store under specific conditions.

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Related Practice

Textbook Question

A parallel-plate capacitor has capacitance $C sub 0 equals 8.00$ 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 $3.00 \times 10^{4}$ V/m?

(b) A dielectric with $K equals 2.70$ 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 $3.00 \times 10^{4}$ V/m?

A constant potential difference of $12$ V is maintained between the terminals of a $0.25$ - $μ$ F, 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

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

When a $360$ -nF air capacitor ( $1$ nF = $10^{- 9}$ F) is connected to a power supply, the energy stored in the capacitor is $1.85 x 10^{- 5}$ 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?

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