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A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

Short Answer

Expert verified
(a) 28.64 pF; (b) 343.68 pC; (c) 3.24 kV/m; (d) 2.06 μJ; (e) (a) 14.32 pF, (b) 343.68 pC, (c) 6.48 kV/m, (d) 4.12 μJ.

Step by step solution

01

Calculating Capacitance (Initial Setup)

First, we need to find the capacitance of the initial setup where the separation is 3.7 mm. The formula for the capacitance of a parallel-plate capacitor is \( C = \frac{\varepsilon_0 \cdot A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{F/m}) \), \( A \) is the area of one plate, and \( d \) is the separation between the plates. Here, \( A = (0.12)^2 \) m\(^2\), and \( d = 0.0037 \) m. Substituting these values in, we calculate \( C \).
02

Calculating Initial Charge on Plates

Using the capacitance \( C \) found in the previous step, we can find the charge \( Q \) on each plate using the relation \( Q = C \cdot V \), where \( V = 12 \) V is the battery voltage. Substitute the values to find \( Q \).
03

Calculating Electric Field Between Plates (Initial Setup)

The electric field \( E \) between the plates can be calculated using \( E = \frac{V}{d} \), where \( V = 12 \) V is the voltage and \( d = 0.0037 \) m is the initial separation between the plates. Compute \( E \) using these values.
04

Calculating Initial Energy Stored

The energy \( U \) stored in the capacitor is given by \( U = \frac{1}{2} C V^2 \). Using the capacitance from step 1 and the voltage \( V = 12 \) V, calculate \( U \).
05

Capacitance after Battery Disconnection and Plate Separation

Now, with the battery disconnected and the plates pulled to \( d = 0.0074 \) m, recalculate capacitance \( C = \frac{\varepsilon_0 \, A}{d} \) using the new separation \( d = 0.0074 \) m. The area \( A \) remains the same.
06

Charge After Changing Plate Separation

Since the battery is disconnected, the charge \( Q \) remains the same as initially calculated. Note that \( Q \) does not change after the separation.
07

Electric Field After Changing Plate Separation

With the charge constant, recalculate the electric field using \( E = \frac{Q}{\varepsilon_0 \, A} \). Substitute the unchanged charge \( Q \), \( \varepsilon_0 \), and the area \( A \) to find the new \( E \).
08

Energy Stored After Changing Plate Separation

The energy in the capacitor changes because the capacitance changes. Calculate the new energy using \( U = \frac{Q^2}{2C} \) with the new capacitance from step 5 and the unchanged charge \( Q \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel-plate capacitor
A parallel-plate capacitor is one of the simplest types of capacitors, consisting of two conductive plates separated by a thin insulation layer known as the dielectric. In our exercise, the plates are separated by air, which serves as the dielectric medium. The function of this setup is to store electric energy when connected to a power source such as a battery. The stored energy depends on the area of the plates, the separation distance between them, and the dielectric material used.
The idea is that when a voltage is applied, one plate accumulates a positive charge and the other a negative charge, creating an electric field between them. With a parallel-plate capacitor, the larger the surface area of the plates and the smaller the distance between them, the more energy it can store. This makes the design quite efficient for several applications, including energy storage, filtering, and in various electronic circuits.
Capacitance calculation
Capacitance is a measure of a capacitor's ability to store charge. For a parallel-plate capacitor, the capacitance is calculated using the formula: \[ C = \frac{\varepsilon_0 \cdot A}{d} \] where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \, \text{F/m})\), \( A \) is the area of one plate, and \( d \) is the separation between the plates.
For the given problem, the area \( A \) is \((0.12)^2\) square meters and the initial separation \( d \) is 0.0037 meters. By substituting these values into the formula, you can calculate the initial capacitance.
This capacity reflects how much charge the plates can hold for every volt of electric potential difference applied, establishing a core function of capacitors in storing and releasing charge as needed in circuits.
Electric field between plates
The electric field \( E \) between the plates of a parallel-plate capacitor is a uniform field, calculated using the formula: \[ E = \frac{V}{d} \] where \( V \) is the voltage applied across the plates, and \( d \) is the distance between them.
For the initial setup with a 12-V battery and a separation of 0.0037 meters, this formula helps compute the strength of the electric field. The electric field represents the force per unit charge that would be experienced by a charge placed within it.
When the separation is increased while keeping the charge constant (as when the battery is disconnected), the voltage changes and consequently, the electric field can be recalibrated. This variability is essential in understanding capacitor behavior under different physical configurations.
Energy stored in capacitor
The energy stored in a capacitor can be calculated by the equation: \[ U = \frac{1}{2} C V^2 \] where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. In our scenario, this formula allows us to calculate how much energy is stored initially with the given capacitance and voltage.
Upon disconnecting the battery and altering the plate separation, the voltage across the capacitor changes, although the charge remains the same. The energy stored is then recalculated with the modified capacitance using the alternative expression: \[ U = \frac{Q^2}{2C} \] This highlights how energy is influenced by changes in configuration and the potential energy shift due to altered geometric or material properties. Understanding these concepts emphasizes the dynamic nature of energy storage in capacitors, vital for managing power in various applications.

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Most popular questions from this chapter

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

A 5.00\(\mu\)F 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?

Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

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