Chapter 31: Problem 3
(II) At a given instant, a 2.8-A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are \(1.60 \mathrm{~cm}\) on a side?
Short Answer
Expert verified
The electric field changes at a rate of approximately \(1.24 \times 10^{15} \text{ N/C}\cdot\text{s}\).
Step by step solution
01
Understand the Relationship
The current (\(I = 2.8 \text{ A}\)) changes the charge \(Q\) on the capacitor over time, which in turn changes the electric field \(E\) between the plates. The relation between current and charge is given by \( I = \frac{dQ}{dt} \).
02
Determine Charge Density
The charge density \(\sigma\) on the plates is \(\sigma = \frac{Q}{A}\), where \(A\) is the area of one plate. For square plates with side \(1.60 \text{ cm}\), the area \(A\) is \((1.60 \text{ cm})^2 = 2.56 \text{ cm}^2 = 2.56 \times 10^{-4} \text{ m}^2\).
03
Rate of Change of Charge Density
The change in charge per unit area with respect to time is \(\frac{d\sigma}{dt} = \frac{1}{A}\frac{dQ}{dt} = \frac{I}{A}\). This reduces to \(\frac{d\sigma}{dt} = \frac{2.8}{2.56 \times 10^{-4}} \text{ C/m}^2\cdot\text{s}\).
04
Calculate the Rate of Change of Electric Field
Using \( E = \frac{\sigma}{\varepsilon_0} \), the rate of change of the electric field \(\frac{dE}{dt}\) is \(\frac{dE}{dt} = \frac{1}{\varepsilon_0} \frac{d\sigma}{dt}\). Given \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\), substitute to find \(\frac{dE}{dt} = \frac{2.8}{2.56 \times 10^{-4} \cdot 8.85 \times 10^{-12}} \text{ N/C}\cdot\text{s}\).
05
Final Calculation
Substitute values to compute \(\frac{dE}{dt} = \frac{2.8}{2.256 \times 10^{-15}} \approx 1.24 \times 10^{15} \text{ N/C}\cdot\text{s}\).
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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 that stores electrical energy in an electric field. Capacitors are essential in many electronic circuits, serving as components that can hold and release charge efficiently. A standard parallel-plate capacitor consists of two conductive plates placed parallel to each other, separated by a small distance.
The ability of a capacitor to store charge is quantified by its capacitance, which is measured in farads (F). For a parallel-plate capacitor, the capacitance is determined by the area of the plates, the distance between them, and the permittivity of the material between the plates. The formula for capacitance is:
The ability of a capacitor to store charge is quantified by its capacitance, which is measured in farads (F). For a parallel-plate capacitor, the capacitance is determined by the area of the plates, the distance between them, and the permittivity of the material between the plates. The formula for capacitance is:
- \[ C = \varepsilon_0 \frac{A}{d} \]
- Where \(C\) is the capacitance, \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of one plate, and \(d\) is the distance between the plates.
When a voltage is applied across the plates, an electric field is created between them, causing electrons to accumulate on one plate while the other plate loses electrons, thus creating a charge separation. This action transforms the capacitor into a temporary energy storage device, which can release the stored energy when required.
Current
Current refers to the flow of electric charge, typically measured in amperes (A). In the context of a capacitor, the current entering or leaving a capacitor relates to how quickly the charge on the capacitor changes.
When current flows into a capacitor, it causes an increase in the charge stored on the capacitor's plates. The relationship between current \(I\) and the rate of charge change \(\frac{dQ}{dt}\) on the capacitor can be expressed by:
In our exercise, a current of 2.8 A flows into the capacitor, steadily altering the charge and thus changing the electric field between the capacitor's plates.
When current flows into a capacitor, it causes an increase in the charge stored on the capacitor's plates. The relationship between current \(I\) and the rate of charge change \(\frac{dQ}{dt}\) on the capacitor can be expressed by:
- \[ I = \frac{dQ}{dt} \]
In our exercise, a current of 2.8 A flows into the capacitor, steadily altering the charge and thus changing the electric field between the capacitor's plates.
Rate of Change
The rate of change in the context of a capacitor's electric field involves understanding how the electric field evolves as charges are added or removed.
The electric field \(E\) in a parallel-plate capacitor is directly tied to the charge density \(\sigma\) on the plates, given by:
To find the rate of change of this electric field \(\frac{dE}{dt}\), we utilize:
The electric field \(E\) in a parallel-plate capacitor is directly tied to the charge density \(\sigma\) on the plates, given by:
- \[ E = \frac{\sigma}{\varepsilon_0} \]
To find the rate of change of this electric field \(\frac{dE}{dt}\), we utilize:
- \[ \frac{dE}{dt} = \frac{1}{\varepsilon_0} \frac{d\sigma}{dt} \]