A 5.805.80-μ\muF, parallel-plate, air capacitor has a plate separation of 5.005.00 mm and is charged to a potential difference of 400400 V. Calculate the energy density in the region between the plates, in units of J/m3.

Ch 24: Capacitance and Dielectrics

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 24: Capacitance and Dielectrics

Problem 23

Chapter 24, Problem 23

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.

Verified step by step guidance

Start by understanding the concept of energy density in a capacitor. Energy density is the energy stored per unit volume between the plates of the capacitor.

Recall the formula for the energy stored in a capacitor: $ U = \frac{1}{2} C V^2 $, where $ U $ is the energy, $ C $ is the capacitance, and $ V $ is the potential difference.

Calculate the energy $ U $ using the given values: $ C = 5.80 \times 10^{-6} $ F and $ V = 400 $ V. Substitute these values into the formula to find $ U $.

Determine the volume between the plates. The volume $ V_{\text{volume}} $ can be calculated using the formula $ V_{\text{volume}} = A \times d $, where $ A $ is the area of the plates and $ d $ is the separation between the plates. Since $ A $ is not given, assume it cancels out in the final calculation.

Finally, calculate the energy density $ u $ using the formula $ u = \frac{U}{V_{\text{volume}}} $. Substitute the values of $ U $ and $ V_{\text{volume}} $ to find the energy density in $ \text{J/m}^3 $.

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Key Concepts

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

Capacitance

Capacitance is the ability of a system to store charge per unit voltage. For a parallel-plate capacitor, it is determined by the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between them. This concept helps understand how the capacitor stores energy.

Energy Stored in a Capacitor

The energy stored in a capacitor is given by the formula U = 0.5 * C * V², where C is the capacitance and V is the voltage across the plates. This energy is stored in the electric field between the plates, and understanding this concept is crucial for calculating the energy density.

Energy Density

Energy density is the amount of energy stored per unit volume in a field. For a capacitor, it is calculated using the formula u = U/V, where U is the energy stored and V is the volume between the plates. This concept is essential for determining how much energy is concentrated in the space between the capacitor plates.

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Intro to Density

Related Practice

Textbook Question

For the system of capacitors shown in Fig. E $24.16$ , find the equivalent capacitance between $b$ and $c$ .

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Textbook Question

Figure E $24.14$ shows a system of four capacitors, where the potential difference across ab is $50.0$ V. How much charge is stored in each of the $10.0$ - $μ$ F and the $9.0$ - $μ$ F capacitors?

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Textbook Question

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

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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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?

A 5.805.80-μ\(\mu\)F, parallel-plate, air capacitor has a plate separation of 5.005.00 mm and is charged to a potential difference of 400400 V. Calculate the energy density in the region between the plates, in units of J/m3.

Verified video answer for a similar problem:

Key Concepts

Capacitance

Energy Stored in a Capacitor

Energy Density

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.