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 39

Chapter 24, Problem 39

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.

Verified step by step guidance

Step 1: Understand the initial setup of the problem. We have a parallel-plate capacitor with a capacitance of 0.25 μF and a potential difference of 12 V. The initial charge on the capacitor can be calculated using the formula Q = C * V, where Q is the charge, C is the capacitance, and V is the potential difference.

Step 2: When a dielectric material like Mylar is inserted between the plates, the capacitance of the capacitor changes. The new capacitance can be calculated using the formula C' = κ * C, where κ is the dielectric constant of Mylar. Refer to Table 24.1 for the value of κ for Mylar.

Step 3: Calculate the new charge on the capacitor with the Mylar inserted using the formula Q' = C' * V. The additional charge that flows onto the positive plate is the difference between the new charge and the initial charge, ΔQ = Q' - Q.

Step 4: To find the total induced charge on either face of the Mylar sheet, use the concept that the induced charge is equal to the additional charge that flows onto the plates due to the presence of the dielectric. This is because the dielectric reduces the effective electric field, allowing more charge to accumulate on the plates.

Step 5: The Mylar sheet affects the electric field between the plates by reducing it. The dielectric constant κ reduces the electric field by a factor of 1/κ. This reduction in the electric field is reconciled with the increase in charge because the dielectric allows more charge to be stored for the same potential difference, effectively increasing the capacitance and thus the charge storage capability of the capacitor.

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Key Concepts

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

Capacitance and Dielectrics

Capacitance is the ability of a system to store charge per unit voltage, and is given by C = Q/V. When a dielectric material like Mylar is inserted between the plates of a capacitor, it increases the capacitance by a factor known as the dielectric constant (k). This is because the dielectric reduces the electric field within the capacitor, allowing more charge to be stored for the same potential difference.

Electric Field in a Capacitor

The electric field (E) between the plates of a parallel-plate capacitor is uniform and is given by E = V/d, where V is the potential difference and d is the separation between the plates. Inserting a dielectric reduces the effective electric field because the dielectric becomes polarized, creating an opposing field. This allows the capacitor to store more charge without increasing the voltage.

Induced Charge and Polarization

When a dielectric like Mylar is placed in an electric field, its molecules align such that their positive and negative charges are separated, a process known as polarization. This results in induced charges on the surfaces of the dielectric, which partially cancel the field within the dielectric. The total induced charge on the dielectric surfaces is equal to the additional charge stored on the capacitor plates due to the increased capacitance.

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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?

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

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?

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Textbook Question

An air capacitor is made from two flat parallel plates $1.50$ mm apart. The magnitude of charge on each plate is $0.0180$ $μ$ 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 $0.0180$ $μ$ C, what total energy is stored?

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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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