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Determine the capacitance of a parallel-plate capacitor having plates \(100 \mathrm{~mm}\) by \(0.3 \mathrm{~m}\) separated by \(0.04 \mathrm{~mm}\). The dielectric constant \(\epsilon=100 \mathrm{pF} / \mathrm{m}\).

Short Answer

Expert verified
The capacitance of the capacitor is 75,000 pF.

Step by step solution

01

Understanding the Problem

The capacitance of a parallel-plate capacitor is given by the formula:\[C = \epsilon \frac{A}{d}\]where \( C \) is the capacitance, \( \epsilon \) is the permittivity (dielectric constant), \( A \) is the area of the plates, and \( d \) is the separation between the plates. We need to calculate \( C \) given \( \epsilon = 100 \text{ pF/m} \), plate dimensions, and separation.
02

Calculate the Area of the Plates

First, convert the given plate dimensions into meters:- Length = 0.3 m- Width = 0.1 m (since 100 mm = 0.1 m)Calculate the area \( A \) of one plate:\[A = \text{Length} \times \text{Width} = 0.3 \times 0.1 = 0.03 \text{ m}^2\]
03

Convert Separation to Meters

Given the plate separation is 0.04 mm, convert this to meters:\[0.04 \text{ mm} = 0.04 \times 10^{-3} \text{ m} = 0.00004 \text{ m}\]This will be used directly in the calculation for capacitance.
04

Calculate the Capacitance

Now, substitute the values into the capacitance formula:\[C = \epsilon \frac{A}{d} = 100 \frac{\text{pF/m}}{1} \frac{0.03}{0.00004}\]This simplifies to:\[C = 100 \times \frac{0.03}{0.00004} = 100 \times 750 = 75000 \text{ pF}\]Thus, the capacitance is \( 75000 \text{ pF} \) (picoFarads).

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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 device that stores electrical energy by using two flat, conductive plates. These plates face each other and are separated by a small distance. Think of it as two thin pancakes stacked with a tiny air gap. This design allows for a simple and efficient way to store charge.
The electric field generated between the plates causes one plate to gain a positive charge and the other to fill with a negative charge. This separation is crucial, as this is where the energy is stored. By applying a voltage across the plates, it enhances the potential energy difference, thus storing more energy.
Key factors include:
  • Plate Area: Larger plate areas can store more charge. A bigger pancake, in this analogy, holds more syrup.
  • Plate Separation: A smaller gap between plates increases capacitance, just like squishing those pancakes closer together.
This design is simple yet very effective in various electronic applications, from tiny circuits to large electrical systems.
Dielectric Constant
The dielectric constant, often represented by the symbol \( \epsilon \), is a measure of a material's ability to store electrical energy in an electric field. It's an important factor in capacitors, as it impacts how much electric charge the capacitor can store.
When a dielectric material is placed between the plates of a capacitor, it increases the overall capacitance by reducing the electric field's strength between the plates. This allows the capacitor to store more charge at the same voltage.
Here's why this matters:
  • Material Impact: Different materials have different dielectric constants. Materials with higher dielectric constants enable greater storage of energy.
  • Capacitance Boost: The presence of a dielectric material like plastic, ceramic, or liquid increases the storage capacity of a capacitor without altering the physical space it occupies.
The dielectric constant plays a central role in designing capacitors for various functions, tailoring them for maximum efficiency with the available material.
Capacitor Formula
The capacitor formula is a handy equation used to compute the capacitance of a parallel-plate capacitor, which is given by: \[ C = \epsilon \frac{A}{d} \]
Let’s break down what each component represents:
  • \( C \): The capacitance, typically measured in Farads, is the capacity of a capacitor to store charge.
  • \( \epsilon \): The dielectric constant, which determines how effectively the material between the plates can store and maintain charge.
  • \( A \): The area of one of the plates. Larger areas permit more charge to accumulate.
  • \( d \): The distance between the two plates. Smaller separations can lead to higher capacitance because the plates' electric fields influence each other more effectively.
For instance, in the exercise given, converting measurements to meters ensures precision. Each physical attribute directly factors into increasing or optimizing capacitance. Understanding this formula allows for practical applications in creating capacitors suited to different electrical and electronic needs.

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Most popular questions from this chapter

Before \(t=0\), the current in a 2 -H inductance is zero. Starting at \(t=0\), the current is increased linearly with time to \(5 \mathrm{~A}\) in \(1 \mathrm{~s}\). Then, the current remains constant at \(5 \mathrm{~A}\). Sketch the voltage, current, power, and stored energy to scale versus time.

Consider two initially uncharged capacitors \(C_{1}=15 \mu \mathrm{F}\) and \(C_{2}=10 \mu \mathrm{F}\) connected in series. Then, a \(50-\mathrm{V}\) source is connected to the series combination, as shown in Figure P3.28. Find the voltages \(v_{1}\) and \(v_{2}\) after the source is applied. (Hint: The charges stored on the two capacitors must be equal, because the current is the same for both the capacitors.)

Suppose that a parallel-plate capacitor has a dielectric that breaks down if the electric field exceeds \(K \mathrm{~V} / \mathrm{m}\). Thus, the maximum voltage rating of the capacitor is \(V_{\max }=K d\), where \(d\) is the thickness of the dielectric. In working Problem P3.34, we find that the maximum energy that can be stored before breakdown is \(w_{\max }=1 / 2 \epsilon_{r} \epsilon_{0} K^{2}(\mathrm{Vol})\), in which Vol is the volume of the dielectric. Air has approximately \(K=32 \times 10^{5} \mathrm{~V} / \mathrm{m}\) and \(\epsilon_{r}=1\). Find the minimum volume of air (as a dielectric in a parallel-plate capacitor) needed to store the energy content of one U.S. gallon of gasoline, which is approximately \(132 \mathrm{MJ}\). What thickness should the air dielectric have if we want the voltage for maximum energy storage to be \(1000 \mathrm{~V}\) ?

Suppose that we have a \(1000-\mathrm{pF}\) parallelplate capacitor with air dielectric charged to \(1000 \mathrm{~V}\). The capacitor terminals are open circuited. Find the stored energy. If the plates are moved farther apart so that \(d\) is doubled, determine the new voltage on the capacitor and the new stored energy. Where did the extra energy come from?

At \(t=t_{0}\) the voltage across a certain capacitance \(C\) is zero. A pulse of current flows through the capacitance between \(t_{0}\) and \(t_{0}+\) \(\Delta t\), and the voltage across the capacitance increases to \(V_{f}\). What can you say about the peak amplitude \(I_{m}\) and area under the pulse waveform (i.e., current versus time)? What are the units and physical significance of the area under the pulse? What must happen to the peak amplitude and area under the pulse as \(\Delta t\) approaches zero, assuming that \(V_{f}\) remains the same?

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