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Chapter 32: Problem 15

Prove that the displacement current in a parallel-plate capacitor of capacitance \(C\) can be written as \(i_{d}=C(d V / d t)\), where \(V\) is the potential difference between the plates.

Short Answer

Expert verified
The displacement current is \( i_d = C \frac{dV}{dt} \).

Step by step solution

01

Understand the Parameters

In a parallel-plate capacitor, the electric potential difference between the plates is denoted by \( V \), and the capacitance of the capacitor is \( C \). The displacement current, \( i_d \), is related to the change in electric field (and thus the potential difference) over time.
02

Recall the Definition of Capacitance

Capacitance \( C \) is defined by the relationship \( Q = CV \), where \( Q \) is the charge on the capacitor and \( V \) is the potential difference.
03

Differentiate with Respect to Time

Differentiate both sides of the equation \( Q = CV \) with respect to time \( t \). This gives \( \frac{dQ}{dt} = C \frac{dV}{dt} \).
04

Relate to Displacement Current

The rate of change of charge with respect to time is defined as the current, \( i = \frac{dQ}{dt} \). Thus, using the earlier result, the displacement current is \( i_d = C \frac{dV}{dt} \).

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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 a fundamental and straightforward electrical component. It consists of two large, flat plates that are parallel to each other. These plates are often separated by a small gap filled with an insulating material or vacuum.

  • Each plate holds opposite charges, one positive and one negative.
  • The charges build up an electric field between the plates.
  • This field allows the capacitor to store energy effectively.
Parallel-plate capacitors are widely used in various electronic devices and circuits. They are especially useful for storing and releasing energy quickly. They operate based on simple principles of charge and electric fields. This configuration makes calculations and understanding easier, as we can apply specific equations to describe their behavior.
Electric Potential Difference
The electric potential difference, often termed as voltage, is crucial in the functioning of a capacitor. It represents the difference in electric potential between two points.

  • In a capacitor, it is the voltage difference between the two plates.
  • This potential difference causes the buildup of electric charge on the plates.
  • Voltage is measured in volts (V) and denotes the energy per unit charge.
The greater the electric potential difference across the plates, the more energy the capacitor can store. Understanding voltage is vital because it's the driving force that moves electric charge. This concept is deeply tied to the energy stored in the capacitor. As voltage increases, so does the potential energy stored as electrical energy.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. It is symbolized by the letter 'C' and is measured in Farads (F).

  • A higher capacitance implies a capacitor can store more charge at the same voltage.
  • The basic formula is given by \( Q = CV \), where \( Q \) is the charge and \( V \) is the voltage.
  • Capacitance depends on factors like the area of the plates and the material between them.
Understanding capacitance helps in designing circuits that require specific energy storage capacities. Large values of capacitance are advantageous in applications requiring large bursts of energy. Conversely, small capacitance is suitable for filtering applications where brief changes in charge are needed.
Electric Field
The electric field is a core concept in understanding how capacitors work. It is a vector field around charged particles, and in a capacitor, it exists between the plates.

  • The strength of the electric field is directly related to the voltage across the plates and the distance between them.
  • A uniform electric field is established in a parallel-plate capacitor.
  • This field is crucial as it relates to the force experienced by charges.
The electric field within a capacitor can impact many factors, including the voltage distribution and the energy stored. It provides the force that moves charge and allows for energy storage. The relationship between the field and potential difference is key to solving many capacitor-based problems.

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Most popular questions from this chapter

A parallel-plate capacitor with circular plates of radius \(0.10 \mathrm{~m}\) is being discharged. A circular loop of radius \(0.20 \mathrm{~m}\) is concentric with the capacitor and halfway between the plates. The displacement current through the loop is \(2.0 \mathrm{~A}\). At what rate is the electric field between the plates changing?

A magnet in the form of a cylindrical rod has a length of \(5.00 \mathrm{~cm}\) and a diameter of \(1.00 \mathrm{~cm}\). It has a uniform magnetization of \(5.30 \times 10^{3} \mathrm{~A} / \mathrm{m}\). What is its magnetic dipole moment?

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?

A \(0.50 \mathrm{~T}\) magnetic field is applied to a paramagnetic gas whose atoms have an intrinsic magnetic dipole moment of \(1.0 \times\) \(10^{-23} \mathrm{~J} / \mathrm{T}\). At what temperature will the mean kinetic energy of translation of the atoms equal the energy required to reverse such a dipole end for end in this magnetic field?

A charge \(q\) is distributed uniformly around a thin ring of radius \(r\). The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speed \(\omega\). (a) Show that the magnetic moment due to the rotating charge has magnitude \(\mu=\frac{1}{2} q \omega r^{2}\). (b) What is the direction of this magnetic moment if the charge is positive?
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