91Ó°ÊÓ

Two capacitor plates with area \(A\) are separated by a distance \(d\). The space between the plates is filled with dielectric material with dielectric constant \(K\) and resistivity \(\rho\). This capacitor is attached to a battery that supplies a constant potential \(\mathcal{E},\) as shown in Fig. \(\mathbf{P} 26.69,\) and is fully charged. At time \(t=0\) the switch \(S\) is opened. Owing to the nonzero value of \(\rho,\) this capacitor discharges by leaking. We can model this device as a capacitor and a resistor in parallel. (a) Analyze this circuit and determine the time constant, characterizing the discharge in terms of the parameters given above. Now assume the capacitor has plates with area \(A=1.00 \mathrm{~cm}^{2}\) separated by \(d=30.0 \mu \mathrm{m}\) and is filled with a ceramic of dielectric constant \(K=8.70\) and resistivity \(\rho=3.10 \times 10^{12} \Omega \cdot \mathrm{m}\) (b) If \(\mathcal{E}=5.00 \mathrm{~V},\) what is the charge on the capacitor at \(t=0 ?\) (c) At what time will the capacitor have half of its original charge? (d) What is the magnitude of the leaking current at that point?

Step by step solution

01

Calculate the Time Constant

Since the capacitance \(C=\frac{K \varepsilon_0 A}{d}\) and resistance \(R=\frac{d}{\rho A}\) can be expressed in terms of the parameters, the time constant \(\tau = RC\) can be calculated as \(\tau = \frac{K \varepsilon_0 A d}{\rho} \) where \( \varepsilon_0 \) is the vacuum permittivity.

02

Calculate the Initial Charge

For a capacitor, the charge \(Q\) is given by \(Q = C\mathcal{E}\) where \(C\) is capacitance and \(\mathcal{E}\) is the supplied potential. Given the values of \(A\), \(d\), \(K\) and \(\mathcal{E}\), the initial charge can be computed as \(Q = \frac{K \varepsilon_0 A}{d} \cdot \mathcal{E}\)

03

Determine the Time of Half Charge

The charge discharging from a capacitor follows the decay equation \(Q(t) = Q_0e^{-t/\tau} \) where \(Q_0\) is the initial charge and \(t\) is the time. To find the time \(T_{1/2}\) when the capacitor has half its original charge, we equate \(Q(T_{1/2}) = \frac{Q_0}{2}\), so \(T_{1/2} = -\tau \ln\left(\frac{1}{2}\right)\)

04

Calculate the Leaking Current

The leaking current \(I\) at any time \(t\) can be computed using the formula \(I = I_0e^{-t/\tau}\) where \(I_0 = \mathcal{E}/R\). At time \(T_{1/2}\), the leaking current will be \(I(T_{1/2}) = I_0e^{-T_{1/2}/\tau}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Dielectric constant

The dielectric constant, often represented as \( K \), measures how well a material can store electrical energy in an electric field compared to a vacuum. This constant is key in capacitors, as it signifies the factor by which capacitance increases when a dielectric material is inserted between the plates instead of air. Imagine a scenario where you have two plates of a capacitor. The dielectric constant helps determine how much charge these plates can store for a given potential difference across them. A higher dielectric constant means the material can hold more charge at the same potential. In our exercise, the dielectric constant is pivotal in calculating the capacitance, as the formula \[C = \frac{K \varepsilon_0 A}{d}\] shows. Here, \( \varepsilon_0 \) is the vacuum permittivity, \( A \) is the plate area, and \( d \) is the separation between the plates. A dielectric with a high \( K \) improves the capacitor's efficiency in storing charge.

Time constant

The time constant, \( \tau \), is a crucial parameter that characterizes how quickly a capacitor discharges in a circuit. It is a measure of time taken for the charge to reduce to just over 36% of its initial value after the discharging process begins. In circuits modeled by a resistor-capacitor (RC) situation, the time constant is given by the product of resistance \( R \) and capacitance \( C \) as \[\tau = RC\]This formula implies that the greater the values for \( R \) or \( C \), the longer the capacitor will take to discharge. In the exercise at hand, the time constant is influenced by the dielectric properties and geometry of the capacitor, being calculated as:\[\tau = \frac{K \varepsilon_0 A d}{\rho}\]where \( \rho \) is the resistivity of the dielectric material. Hence, the time constant not only depends on the electrical properties but also on the physical layout of the system.

Leaking current

Leaking current in a capacitor refers to the gradual loss of charge over time due to imperfections or resistive pathways in the dielectric material, which allow the charge to escape. This is modeled using a parallel resistor in the circuit, opposite the effects of a perfect capacitor. Leaking current, \( I \), is given by the exponential decay equation:\[I = I_0 e^{-t/\tau}\]where \( I_0 = \frac{\mathcal{E}}{R} \) is the initial current (at \( t = 0 \)) determined by the supply voltage \( \mathcal{E} \), and \( R \) is the resistance offered by the dielectric. As time progresses, the current decreases exponentially due to the time constant \( \tau \). In practical scenarios, reducing the leaking current involves selecting materials with higher resistivity, as this minimizes charge movement through the dielectric over time.

Permittivity

Permittivity is a fundamental property that quantifies how a material reacts to an electric field, essentially determining how much electric flux it can support within it. The vacuum permittivity, \( \varepsilon_0 \), is a constant that represents this property in a vacuum. In the context of capacitors, permittivity directly impacts the capacitance value. A material with higher permittivity, \( \varepsilon \), increases the capacitance, allowing more charge to be stored for a given electric field. The relation is expressed in the formula for capacitance:\[C = \frac{\varepsilon A}{d} = \frac{K \varepsilon_0 A}{d}\]where \( K \) is the dielectric constant, \( A \) is the plate area, and \( d \) is the distance between them. Thus, permittivity, particularly of the dielectric material, plays an essential role in determining how efficient the capacitor is at storing and retaining charge. In practical applications, materials with higher permittivity are often chosen to enhance capacitance values.