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At what rate must the potential difference between the plates of a parallel- plate capacitor with a \(2.0 \mu \mathrm{F}\) capacitance be changed to produce a displacement current of \(1.5 \mathrm{~A}\) ?

Short Answer

Expert verified
The potential difference must change at a rate of 750,000 V/s.

Step by step solution

01

Understand the relationship

To solve this problem, we need to understand the relationship between displacement current, potential difference, and capacitance. Displacement current \(I_d\) in a capacitor is given by the formula \(I_d = C \cdot \frac{dV}{dt}\), where \(C\) is the capacitance and \(\frac{dV}{dt}\) is the rate of change of potential difference (voltage).
02

Rearrange the formula for rate of change of voltage

Rearrange the formula \(I_d = C \cdot \frac{dV}{dt}\) to solve for \(\frac{dV}{dt}\). This gives us \(\frac{dV}{dt} = \frac{I_d}{C}\).
03

Substitute the given values

Substitute the given values into the formula: \(I_d = 1.5 \, \text{A} \) and \(C = 2.0 \, \mu\text{F} = 2.0 \, \times 10^{-6} \, \text{F}\). The equation becomes \(\frac{dV}{dt} = \frac{1.5}{2.0 \times 10^{-6}}\).
04

Calculate \(\frac{dV}{dt}\)

Calculate the result: \(\frac{1.5}{2.0 \times 10^{-6}} = 750,000 \, \text{V/s}\). Thus, the rate at which the potential difference must be changed is \(750,000 \, \text{V/s}\).

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Key Concepts

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

Understanding Parallel-Plate Capacitors
A parallel-plate capacitor consists of two conductive plates separated by a dielectric material.
This simple structure is key in various electronic applications. Unlike typical conductors, capacitors do not allow a continuous flow of current.
Instead, they store energy in the electric field between the plates.
When a voltage is applied across the plates, positive charge accumulates on one plate, while an equal negative charge gathers on the other.
This results in an electric field, influencing how much energy the capacitor can store. Here are key points about parallel-plate capacitors:
  • They store energy as an electric field.
  • The larger the plate area and smaller the separation, the greater the capacitance.
  • The insulating material affects how much voltage the capacitor can handle.
Parallel-plate capacitors serve as the foundation for understanding more complex dynamic scenarios, such as displacement current.
Potential Difference in Capacitors
Potential difference, or voltage, across the plates of a capacitor, determines the amount of charge that gets stored.
Higher voltage means more charge is stored.
In essence, potential difference measures the energy per unit charge that causes the charges to accumulate on the plates. In the scenario of our exercise, changing the potential difference at a specific rate is crucial for generating the desired displacement current.
This concept is vital when dealing with altering electric fields in capacitive systems.
Key aspects about potential difference in capacitors include:
  • It is directly proportional to the accumulation of charges on the plates.
  • Changing it affects both the electric field and the displacement current.
  • It is essential in determining the energy storage efficiency of a capacitor.
Recognizing how potential difference interacts with charge storage helps in understanding energy dynamics in circuits involving capacitors.
Capacitance and Its Role
Capacitance is a measure of a capacitor's ability to store charge per unit of voltage.
It is a fundamental characteristic of capacitors that influences how they respond to changes in voltage.
The SI unit of capacitance is the farad (F), representing the capacity to store one coulomb of charge per volt. For parallel-plate capacitors, capacitance depends on several factors:
  • The area of the plates: Larger plate areas lead to higher capacitance.
  • The distance between plates: Smaller separations increase capacitance.
  • The material between the plates: Different dielectrics can significantly alter capacitance.
In our exercise, the given capacitance value is crucial for determining the rate of change of voltage necessary to achieve a specific displacement current.
Capacitance essentially dictates how much charge will accumulate for a given potential difference.
Rate of Change of Voltage and Its Significance
The rate of change of voltage, \(\frac{dV}{dt}\), is key in controlling displacement current in capacitors.
As voltage across the capacitor changes, so does the electric field, which in turn affects the displacement current.
In our example, \(\frac{dV}{dt}\) needs to be manipulated to produce a displacement current of 1.5 A.
The relationship is given by the equation \(I_d = C \cdot \frac{dV}{dt}\).
By rearranging and calculating, we derive the precise rate of voltage change needed.This concept is significant because:
  • It demonstrates how rapidly we need to change voltage to achieve desired electrical effects.
  • It provides insight into how capacitors behave in dynamic circuits.
  • It highlights the interplay between voltage change and displacement current generation.
Understanding how to manipulate the rate of change of voltage effectively allows for precise control in applications involving capacitive components and transient voltages.

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Most popular questions from this chapter

A magnetic rod with length \(6.00 \mathrm{~cm}\), radius \(3.00 \mathrm{~mm}\), and (uniform) magnetization \(2.70 \times 10^{3} \mathrm{~A} / \mathrm{m}\) can turn about its center like a compass needle. It is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(35.0 \mathrm{mT}\), such that the directions of its dipole moment and \(\vec{B}\) make an angle of \(68.0^{\circ}\). (a) What is the magnitude of the torque on the rod due to \(\vec{B} ?\) (b) What is the change in the orientation energy of the rod if the angle changes to \(34.0^{\circ} ?\)

A magnetic compass has its needle, of mass \(0.050 \mathrm{~kg}\) and length \(4.0 \mathrm{~cm}\), aligned with the horizontal component of Earth's magnetic field at a place where that component has the value \(B_{h}=\) \(16 \mu \mathrm{T}\). After the compass is given a momentary gentle shake, the needle oscillates with angular frequency \(\omega=45 \mathrm{rad} / \mathrm{s}\). Assuming that the needle is a uniform thin rod mounted at its center, find the magnitude of its magnetic dipole moment.

Assume the average value of the vertical component of Earth's magnetic field is \(43 \mu \mathrm{T}\) (downward) for all of Arizona, which has an area of \(2.95 \times 10^{5} \mathrm{~km}^{2} .\) What then are the (a) magnitude and (b) direction (inward or outward) of the net magnetic flux through the rest of Earth's surface (the entire surface excluding Arizona)?

A Rowland ring is formed of ferromagnetic material. It is circular in cross section, with an inner radius of \(5.0 \mathrm{~cm}\) and an outer radius of \(6.0 \mathrm{~cm}\), and is wound with 400 turns of wire. (a) What current must be set up in the windings to attain a toroidal field of magnitude \(B_{0}=0.20 \mathrm{mT} ?\) (b) A secondary coil wound around the toroid has 50 turns and resistance \(8.0 \Omega\). If, for this value of \(B_{0}\), we have \(B_{M}=800 B_{0}\), how much charge moves through the secondary coil when the current in the toroid windings is turned on?

You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is \(0.312 \mathrm{~Hz}\). Earth's magnetic field at the location of the compass has a horizontal component of \(18.0 \mu \mathrm{T}\). The needle has a magnetic moment of \(0.680 \mathrm{~mJ} / \mathrm{T}\). What is the needle's rotational inertia about its (vertical) axis of rotation?

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