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24.48 . A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant \(K\) . The magnitude of the charge on each plate is \(Q\) . Each plate has area \(A\) , and the distance between the plates is \(d\) . (a) Use Gauss's law as stated in Eq. \((24.23)\) to calculate the magnitude of the electric field in the

Short Answer

Expert verified
The electric field magnitude is given by \( E = \frac{Q}{K \varepsilon_0 A} \).

Step by step solution

01

Understand Gauss's Law

Gauss's law, in the context of dielectrics, is given by the equation \( \oint \mathbf{E} \, d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), where \( Q_{enc} \) is the charge enclosed within a Gaussian surface and \( \varepsilon_0 \) is the permittivity of free space. In a dielectric, this becomes \( \varepsilon = K \varepsilon_0 \), where \( K \) is the dielectric constant of the material between the plates.
02

Define the Gaussian Surface

In this problem, we use a Gaussian pillbox between the capacitor plates. This pillbox intersects both plates, enclosing a charge \( Q \). We seek the electric field \( E \) within the plates.
03

Apply Gauss's Law

For a parallel-plate capacitor filled with a dielectric, Gauss's Law simplifies to \( E \cdot A = \frac{Q}{K \varepsilon_0} \), given that the electric field \( E \) is uniform across the area \( A \) of the plate.
04

Solve for the Electric Field

Rearranging the equation from Step 3 gives the electric field magnitude as \( E = \frac{Q}{K \varepsilon_0 A} \). This expression allows us to calculate \( E \) using the provided values of \( Q \), \( K \), \( \varepsilon_0 \), and \( A \).

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle used to describe the distribution of electric charge. It states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. This law is expressed mathematically as:
  • \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \)
where \( Q_{enc} \) is the enclosed charge and \( \varepsilon_0 \) is the permittivity of free space. When dealing with dielectrics, the equation is modified by including the dielectric constant \( K \) of the material. This changes the expression to include:
  • \( \varepsilon = K \varepsilon_0 \)
In the context of a capacitor with a dielectric material, Gauss's Law enables us to understand how the presence of the dielectric affects the electric field between the plates. The dielectric essentially reduces the effective electric field by a factor of the dielectric constant \( K \), impacting the capacitor's ability to store charge.
Electric Field
The electric field within a capacitor is crucial for understanding how the device functions. It is the force exerted by the electric charge stored on the capacitor plates and is uniform between the parallel plates of the capacitor:
  • A higher charge leads to a stronger electric field.
  • The field is uniform due to the parallel plate geometry.
The relationship between the charge, the electric field \( E \), and other factors like plate area \( A \) and dielectric constant \( K \) can be described by rearranging Gauss's Law:
  • \( E = \frac{Q}{K \varepsilon_0 A} \)
This formula tells us that:
  • \( Q \) is the charge on the plate.
  • \( A \) is the area of the plates.
  • \( \varepsilon_0 \) is the permittivity of free space.
  • \( K \) is the dielectric constant.
The dielectric constant \( K \) reduces the effective electric field within the capacitor, offering better efficiency in charge storage by allowing more charge to be held at a given electric potential.
Dielectric Constant
The dielectric constant \( K \) is a measure of a material's ability to store electrical energy in an electric field. It plays a critical role in capacitors, influencing the device's capacitance:
  • Higher \( K \) values indicate better insulating properties.
  • It reduces the effective electric field inside the capacitor.
This constant is used to adjust the permittivity of free space \( \varepsilon_0 \) to account for the material filling the space between a capacitor's plates:
  • \( \varepsilon = K \varepsilon_0 \)
In practical terms, a greater dielectric constant leads to a larger capacitance \( C \) since the formula for capacitance becomes:
  • \( C = \frac{K \varepsilon_0 A}{d} \)
where \( d \) is the distance between the plates. Thus, using a material with a higher dielectric constant allows capacitors to store more charge at the same voltage, which is highly desirable in many electronics applications.

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Most popular questions from this chapter

24.24. A parallel-plate air capacitor has a capacitance of 920 \(\mathrm{pF}\) . The charge on each plate is 2.55\(\mu \mathrm{C}\) . (a) What is the potential difference between the plates? (b) If the charge is kept constant,what will be the potential difference between the plates if the sep ration is doubled? (c) How much work is required to double the separation?

24.4. Capacitance of an Oscilloscope. Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates. Typically, they are squares 3.0 \(\mathrm{cm}\) on a side and separated by \(5.0 \mathrm{mm},\) with vacuum in between. What is the capacitance of these deflecting plates and hence of the oscilloscope? (Note: This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.)

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.28. A capacitor of capacitance \(C\) is charged to a potential difference \(V_{0}\) . The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance \(C / 2\) Compute (a) the original charge of the system; (b) the final potential difference across each capacitor; (e) the final energy of the system; (d) the decrease in energy when the capacitors are connected. (e) Where did the "lost" energy go?

24.50. A parallel-plate air capacitor is made by using two plates 16 \(\mathrm{cm}\) square, spaced 4.7 \(\mathrm{mm}\) apart. It is connected to a \(12-\mathrm{V}\) bat- tery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(9.4 \mathrm{mm},\) what are the answers to parts \((\mathrm{a})-(\mathrm{d}) ?\)

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