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(1) Determine the rate at which the electric field changes between the round plates of a capacitor, 6.0 \(\mathrm{cm}\) in diameter, if the plates are spaced 1.1 \(\mathrm{mm}\) apart and the voltage across them is changing at a rate of 120 \(\mathrm{V} / \mathrm{s}\) .

Short Answer

Expert verified
The electric field changes at a rate of approximately 109091 V/m/s.

Step by step solution

01

Understand the Relationship

Identify the relationship between the electric field (E) and voltage (V) for a parallel plate capacitor. The formula to use is:\[ E = \frac{V}{d} \]where \(E\) is the electric field, \(V\) is the voltage across the plates, and \(d\) is the distance between the plates.
02

Substitute Given Values

We know the voltage \(V\) is changing at a rate of 120 \(\text{V/s}\), and the distance \(d\) between the plates is 1.1 mm or 0.0011 meters. The rate of change of the electric field can be found using the relation from Step 1:\[ \frac{dE}{dt} = \frac{dV}{dt} \cdot \frac{1}{d} \]
03

Calculate the Rate of Change

Substitute \(\frac{dV}{dt} = 120 \text{ V/s}\) and \(d = 0.0011 \text{ m}\) into the equation:\[ \frac{dE}{dt} = \frac{120}{0.0011} \]Perform the division to find the rate of change of the electric field.
04

Solve for the Rate

Calculate the expression:\[ \frac{120}{0.0011} = 109090.91 \text{ V/m/s} \]Thus, the electric field changes at a rate of approximately 109091 \(\text{V/m/s}\).

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Key Concepts

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

Capacitor
A capacitor is a device used in electrical circuits to store energy in an electric field temporarily. It consists of two conductive plates separated by an insulating material known as a dielectric. When a voltage is applied across the plates, they store electrical charge of opposite sign, creating an electric field in the space between. Capacitors have many applications, such as smoothing fluctuations in power supply lines or storing energy for quick releases. Powerful yet simple, they are crucial components in many electronic devices.

In our given exercise, the capacitor in question forms a specific type known as a parallel plate capacitor, which enhances understanding of how charge and potential relate in a simple geometry.
Voltage
Voltage, often referred to as electric potential difference, is a measure of the potential energy per unit charge between two points in an electric field. For capacitors, the voltage relates directly to the amount of charge stored on the plates: higher voltage means more stored energy.

The formula for a capacitor is defined as: \( V = E imes d \), where \( V \) is the voltage, \( E \) is the electric field, and \( d \) is the distance between plates. A change in voltage over time, as in this exercise, indicates how rapidly the energy storage capability of the capacitor adjusts, affecting the electric field strength.
Electric Field Change Rate
The rate at which the electric field changes between the capacitor plates is crucial for understanding dynamic behaviors in circuits. This rate, \( \frac{dE}{dt} \), tells us how swiftly the field's strength increases or decreases as the voltage changes.

The given exercise utilizes the relation \( \frac{dE}{dt} = \frac{dV}{dt} \cdot \frac{1}{d} \). Here, \( \frac{dV}{dt} \) is the known rate of change of voltage (120 V/s), and \( d \) is the plate separation in meters. By understanding this relation, students can predict responses to fast-paced voltage variations, crucial for designing responsive electronic systems.
Parallel Plate Capacitor
Parallel plate capacitors are a common form, famed for their simplicity and effectiveness. Made from two parallel conductive plates, these capacitors produce an electric field of uniform strength between the plates when charged.

The distance between the plates (denoted \( d \)) significantly impacts the electric field. A larger distance lowers the field for the same voltage, while a smaller distance increases it. With a known rate of voltage change, the exercise demonstrates this relationship through a capable calculation of the electric field change rate. This reveals how direct electrical parameters interplay for efficient energy storage.

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Most popular questions from this chapter

(II) How practical is solar power for various devices? Assume that on a sunny day, sunlight has an intensity of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) at the surface of Earth and that, when illuminated by that sunlight, a solar-cell panel can convert \(10 \%\) of the sunlight's energy into electric power. For each device given below, calculate the area \(A\) of solar panel needed to power it. \((a)\) A calculator consumes \(50 \mathrm{~mW}\). Find \(A\) in \(\mathrm{cm}^{2}\). Is \(A\) small enough so that the solar panel can be mounted directly on the calculator that it is powering? (b) A hair dryer consumes \(1500 \mathrm{~W}\). Find \(A\) in \(\mathrm{m}^{2}\), Assuming no other electronic devices are operating within a house at the same time, is \(A\) small enough so that the hair dryer can be powered by a solar panel mounted on the house's roof? (c) A car requires 20 hp for highway driving at constant velocity (this car would perform poorly in situations requiring acceleration). Find \(A\) in \(\mathrm{m}^{2}\). Is \(A\) small enough so that this solar panel can be mounted directly on the car and power it in "real time"?

Imagine that a steady current \(I\) flows in a straight cylindrical wire of radius \(R_{0}\) and resistivity \(\rho .(a)\) If the current is then changed at a rate \(d I / d t,\) show that a displacement current \(I_{\mathrm{D}}\) exists in the wire of magnitude \(\epsilon_{0} \rho(d I / d t) .(b)\) If the current in a copper wire is changed at the rate of \(1.0 \mathrm{~A} / \mathrm{ms}\), determine the magnitude of \(I_{\mathrm{D}}\). \((c)\) Determine the magnitude of the magnetic field \(B_{\mathrm{D}}\) (T) created by \(I_{\mathrm{D}}\) at the surface of a copper wire with \(R_{0}=1.0 \mathrm{~mm}\). Compare (as a ratio) \(B_{\mathrm{D}}\) with the field created at the surface of the wire by a steady current of \(1.0 \mathrm{~A}\).

A radio voice signal from the Apollo crew on the Moon (Fig. \(31-25\) ) was beamed to a listening crowd from a radio speaker. If you were standing \(25 \mathrm{~m}\) from the loudspeaker, what was the total time lag between when you heard the sound and when the sound entered a microphone on the Moon and traveled to Earth?

(II) Suppose an air-gap capacitor has circular plates of radius \(R=2.5 \mathrm{~cm}\) and separation \(d=1.6 \mathrm{~mm} .\) A \(76.0-\mathrm{Hz}\) emf, \(\mathscr{E}=\mathscr{E}_{0} \cos \omega t,\) is applied to the capacitor. The maximum displacement current is \(35 \mu \mathrm{A}\). Determine \((a)\) the maximum conduction current \(I,(b)\) the value of \(\varepsilon_{0},(c)\) the maximum value of \(d \Phi_{E} / d t\) between the plates. Neglect fringing,

(1) Calculate the displacement current \(I_{\mathrm{D}}\) between the square plates, 5.8 \(\mathrm{cm}\) on a side, of a capacitor if the electric field is changing at a rate of \(2.0 \times 10^{6} \mathrm{V} / \mathrm{m} \cdot \mathrm{s}\) .

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