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 25

Chapter 24, Problem 25

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?

Verified step by step guidance

To find the capacitance (C), use the formula C = Q/V, where Q is the charge and V is the potential difference. Substitute Q = 0.0180 μC and V = 200 V into the formula.

To find the area (A) of each plate, use the formula for capacitance of a parallel plate capacitor: C = ε₀ * (A/d), where ε₀ is the permittivity of free space (8.85 x 10^-12 F/m) and d is the separation between the plates (1.50 mm). Rearrange the formula to solve for A.

To find the maximum voltage (V_max) that can be applied without dielectric breakdown, use the formula V_max = E_max * d, where E_max is the maximum electric field strength (3.0 x 10^6 V/m) and d is the separation between the plates.

To find the total energy (U) stored in the capacitor when the charge is 0.0180 μC, use the formula U = (1/2) * C * V^2. Substitute the values of C and V obtained from previous steps into the formula.

Ensure all units are consistent when performing calculations, converting μC to C and mm to m where necessary, to maintain accuracy in the results.

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Key Concepts

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

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined by the equation C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage. In this problem, knowing the charge and voltage allows us to calculate the capacitance of the air capacitor.

Electric Field and Dielectric Breakdown

The electric field (E) between parallel plates is given by E = V/d, where V is the voltage and d is the separation between the plates. Dielectric breakdown occurs when the electric field exceeds a material's breakdown strength, causing it to become conductive. For air, this threshold is 3.0x10^6 V/m, which limits the maximum voltage that can be applied without causing breakdown.

Energy Stored in a Capacitor

The energy (U) stored in a capacitor is given by the formula U = 0.5 * C * V^2, where C is the capacitance and V is the voltage across the plates. Alternatively, it can be expressed as U = Q^2/(2C) using the charge. This concept is crucial for determining the total energy stored when the capacitor is charged to a specific charge or voltage.

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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?

In Fig. E $24.20$ , $C sub 1 equals 6.00$ $μ$ F, $C sub 2 equals 3.00$ $μ$ F, and $C sub 3 equals 5.00$ $μ$ F. The capacitor network is connected to an applied potential $V_{a b}$ .

(a) After the charges on the capacitors have reached their final values, the charge on $C sub 2$ is $30.0$ mC. What are the charges on capacitors $C sub 1$ and $C sub 3$ ?

(b) What is the applied voltage $V_{a b}$ ?

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Textbook Question

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

A $5.80$ - $μ$ F, parallel-plate, air capacitor has a plate separation of $5.00$ mm and is charged to a potential difference of $400$ V. Calculate the energy density in the region between the plates, in units of J/m³.

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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

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 uC. (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?

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