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The electric field between the plates of a paper-separated \((K=3.75)\) capacitor is \(9.21 \times 10^{4} \mathrm{V} / \mathrm{m} .\) The plates are 1.95 \(\mathrm{mm}\) apart and the charge on each plate is 0.675\(\mu \mathrm{C}\) . Determine the capacitance of this capacitor and the area of each plate.

Short Answer

Expert verified
The capacitance is approximately 3.76 nF and the area of each plate is approximately 0.0233 m².

Step by step solution

01

Understand the Formula for Capacitance

The capacitance is given by the formula: \[ C = \frac{Q}{V} \] where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the voltage. Here, the electric field \( E \) is related to the voltage and distance between the plates: \[ E = \frac{V}{d} \].
02

Solve for Voltage

To find the voltage \( V \), use the formula: \[ V = E \times d \] Here, \( E = 9.21 \times 10^4 \text{ V/m} \) and \( d = 1.95 \text{ mm} = 1.95 \times 10^{-3} \text{ m} \). Therefore, \[ V = 9.21 \times 10^4 \times 1.95 \times 10^{-3} \].
03

Calculate Voltage

Perform the calculation for voltage: \[ V = 9.21 \times 10^4 \times 1.95 \times 10^{-3} = 179.695 \text{ V} \].
04

Calculate the Capacitance

Now, use the charge \( Q = 0.675 \mu \text{C} = 0.675 \times 10^{-6} \text{ C} \) and the calculated voltage \( V = 179.695 \text{ V} \) to find \( C \): \[ C = \frac{0.675 \times 10^{-6}}{179.695} \text{ F} \].
05

Simplify Capacitance Calculation

Perform the division to obtain the capacitance: \[ C \approx 3.76 \times 10^{-9} \text{ F} = 3.76 \text{ nF} \].
06

Calculate the Plate Area

For the area \( A \), use the capacitance formula for a parallel plate capacitor: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) and \( \varepsilon_r = 3.75 \). Rearrange to solve for \( A \): \[ A = \frac{C \times d}{\varepsilon_0 \times \varepsilon_r} \].
07

Substitute Values and Solve for Area

Substitute the known values into the area equation: \[ A = \frac{3.76 \times 10^{-9} \times 1.95 \times 10^{-3}}{8.85 \times 10^{-12} \times 3.75} \]. Compute this to find \( A \).
08

Compute the Plate Area

Perform the arithmetic to find the plate area: \[ A \approx 2.33 \times 10^{-2} \text{ m}^2 \].

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Key Concepts

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

Electric Field
The electric field is a vital concept in physics and plays a crucial role in understanding capacitors. It represents the force per unit charge experienced by a positive test charge placed within a field. The electric field (E) between two plates in a capacitor is uniform, meaning its strength is the same at every point between the plates when edge effects are ignored.
  • The formula for electric field (E) is given by: \[ E = \frac{V}{d} \] where \( V \) is the voltage across the plates, and \( d \) is the separation between them.
  • An electric field of \( 9.21 \times 10^{4} \, \text{V/m} \) is quite strong and indicates substantial voltage over a small distance.
  • In practical terms, the field guides the movement of electric charges, crucial for capacitor operation.
Using this understanding of electric fields, you can determine the voltage drop across the plates if you know the field strength and distance.
Parallel Plate Capacitor
A parallel plate capacitor is a simple electrical component comprising two conductive plates separated by a distance. This design is effective for storing electrical energy.

Key features of a parallel plate capacitor:

  • The capacitance (C) is determined by the formula: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) is the dielectric constant, \( A \) is the area of the plates, and \( d \) is the separation distance.
  • The charge on each plate is directly proportional to capacitance and voltage.
  • This capacitor type is easily tuned by adjusting either the plate area or the distance between the plates.
The parallel plate capacitor serves as an excellent model for understanding capacitance and forms the basis of many practical applications in electronics.
Dielectric Constant
The dielectric constant, also known as the relative permittivity (\varepsilon_r), is a dimensionless number that represents how an insulating material affects the capacitance of a capacitor. This constant indicates how much the material can reduce the electric field within the capacitor compared to a vacuum.
  • For a paper-separated capacitor, the dielectric constant is \( 3.75 \) in this exercise.
  • A higher dielectric constant generally means that the material allows the capacitor to store more charge at the same voltage.
  • The presence of a dielectric modifies the capacitance by a factor of \( \varepsilon_r \).
Understanding dielectric constants aids significantly in choosing materials for capacitors, especially when designing them for specific functions or improving existing ones. This enhances the performance of the capacitor by increasing charge storage without altering the physical size.

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Most popular questions from this chapter

The Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter has a group of General Problems that are not arranged by Section and not ranked.] (1) The two plates of a capacitor hold \(+2800 \mu C\) and \(-2800 \mu C\) of charge, respectively, when the potential difference is 930 V. What is the capacitance?

(II) How much energy must a \(28-\mathrm{V}\) battery expend to charge a \(0.45-\mu \mathrm{F}\) and a \(0.20-\mu \mathrm{F}\) capacitor fully when they are placed \((a)\) parallel, (b) in series? (c) How much charge flowed from the battery in each case?

An isolated capacitor \(C_{1}\) carries a charge \(Q_{0}\). Its wires are then connected to those of a second capacitor \(C_{2},\) previously uncharged. What charge will each carry now? What will be the potential difference across each?

Capacitors can be used as "electric charge counters." Consider an initially uncharged capacitor of capacitance \(C\) with its bottom plate grounded and its top plate connected to a source of electrons. \((a)\) If \(N\) electrons flow onto the capacitor's top plate, show that the resulting potential difference \(V\) across the capacitor is directly proportional to \(N .\) (b) Assume the voltage- measuring device can accurately resolve voltage changes of about \(1 \mathrm{mV}\). What value of \(C\) would be necessary to detect each new \(\begin{array}{lllll}\text { collected electron? } & \text { (c) Using modern semiconductor }\end{array}\) technology, a micron-size capacitor can be constructed with parallel conducting plates separated by an insulating oxide of dielectric constant \(K=3\) and thickness \(d=100 \mathrm{nm}\). To resolve the arrival of an individual electron on the plate of such a capacitor, determine the required value of \(\ell(\) in \(\mu \mathrm{m}\) ) assuming square plates of side length \(\ell\)

How much charge flows from a 12.0-V battery when it is connected to a \(12.6-\mu \mathrm{F}\) capacitor?

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