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Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E=3.20 \times 10^{5} \mathrm{V} / \mathrm{m}\) . When the space is filled with dielectric, the electric field is \(E=2.50 \times 10^{5} \mathrm{V} / \mathrm{m}\) . (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Short Answer

Expert verified
Dielectric constant: 1.28; Charge density: 0 C/m^2.

Step by step solution

01

Understand the Problem

We need to determine two things: the charge density on each surface of the dielectric and the dielectric constant of the material between the plates. We are given the electric fields with and without the dielectric.
02

Identify Given Values

The electric field without the dielectric is given as \( E_0 = 3.20 \times 10^5 \text{ V/m} \) and with the dielectric as \( E = 2.50 \times 10^5 \text{ V/m} \).
03

Use Formula for Electric Field with Dielectric

The electric field inside a dielectric is given by \( E = \frac{E_0}{K} \), where \( K \) is the dielectric constant. Rearrange to find \( K \).
04

Calculate Dielectric Constant

Rearranging the previous formula gives \( K = \frac{E_0}{E} = \frac{3.20 \times 10^5}{2.50 \times 10^5} \). Compute \( K \).
05

Calculate Dielectric Constant Numerically

After performing the division, \( K = 1.28 \).
06

Relate Charge Density to Electric Field

The electric displacement field \( D \) is related to the surface charge density \( \sigma \) by \( D = \epsilon E \), where \( \epsilon = \epsilon_0 K \) is the permittivity of the dielectric. Since \( D = \sigma_{f} + \sigma \), with \( \sigma_{f} \) being the free surface charge density and \( \sigma \) the polarization surface charge density, calculate \( \sigma \).
07

Compute Free Surface Charge Density

Since \( \sigma_{f} = \epsilon_0 E_0 \), find \( \sigma_{f} = 8.85 \times 10^{-12} \times 3.20 \times 10^5 \).
08

Calculation of Free Surface Charge Density

\( \sigma_f = 2.832 \times 10^{-6} \text{ C/m}^2 \).
09

Calculate Total Surface Charge Density

Now calculate \( D = \epsilon E = \epsilon_0 K E \) to find \( D = 8.85 \times 10^{-12} \times 1.28 \times 2.50 \times 10^5 \).
10

Compute Total Surface Charge Density

After computation, \( D = 2.832 \times 10^{-6} \text{ C/m}^2 \), the same as \( \sigma_f \), so the polarization charge density is zero: \( \sigma = 0 \text{ C/m}^2 \).
11

Conclusion

The dielectric constant is \( 1.28 \), and the charge density on each surface of the dielectric is zero \( 0 \text{ C/m}^2 \).

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Key Concepts

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

Dielectric Constant
A dielectric constant, also known as relative permittivity, is a dimensionless measure that describes how easily a material can become polarized by an electric field. In other words, it shows how well a material can store electrical energy when placed in an electric field. The dielectric constant is typically denoted by the symbol \( K \) or \( \varepsilon_r \).

One important formula used in problems involving dielectrics is \( E = \frac{E_0}{K} \), where \( E_0 \) is the electric field without the dielectric, and \( E \) is the electric field with the dielectric present. By rearranging this formula, you can find the dielectric constant \( K \) by calculating:
  • \( K = \frac{E_0}{E} \)
In our example exercise, the electric field without and with the dielectric is given, allowing us to compute the dielectric constant as \( K = \frac{3.20 \times 10^5}{2.50 \times 10^5} = 1.28 \).

This value indicates how much the dielectric material reduces the electric field compared to a vacuum.
Charge Density
Charge density represents how the electric charge is distributed over a surface, a volume, or along a line. When a dielectric is present, the charge density can include contributions from both the free charges and those induced by polarization.

For dielectric materials, the total surface charge density \( \sigma \) is often considered to be the sum of the free surface charge density \( \sigma_f \) and the polarization surface charge density \( \sigma_p \).

In the provided exercise, to find the free surface charge density, the equation \( \sigma_f = \epsilon_0 E_0 \) is used, where \( \epsilon_0 \) is the permittivity of free space (approximately \( 8.85 \times 10^{-12} \text{ F/m} \)). Calculating, we find that \( \sigma_f = 2.832 \times 10^{-6} \text{ C/m}^2 \).

However, because the total surface charge density calculated through the electric displacement field \( D \) matches \( \sigma_f \), the polarization charge density \( \sigma_p \) is zero.
Electric Displacement Field
The electric displacement field, denoted as \( D \), is a vector quantity that accounts for both free and bound charges within a dielectric. It's used in Maxwell's equations and is crucial when analyzing electric fields in materials.

The displacement field \( D \) is defined as:
  • \( D = \epsilon E \)
where \( \epsilon = \epsilon_0 K \) is the permittivity of the dielectric, \( E \) is the electric field with dielectric, and \( \epsilon_0 \) is the permittivity of free space. For free surface charges, \( D \) equates to the surface charge density \( \sigma_f \).

In the scenario provided, we computed \( D = \epsilon E = \epsilon_0 K E \) and obtained \( D = 2.832 \times 10^{-6} \text{ C/m}^2 \), reinforcing the result that the non-polarized charge density is entirely free surface charge, ensuring \( \sigma_p = 0 \text{ C/m}^2 \).

Understanding \( D \) helps clarify how electric fields interact with dielectrics and exhibits the behavior of materials under electric influence.

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Most popular questions from this chapter

A 12.5 -\muF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{677}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

A parallel-plate capacitor has capacitance \(C=12.5 \mathrm{pF}\) when the volume between the plates is filled with air. The plates are circular, with radius 3.00 \(\mathrm{cm}\) . The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 \(\mathrm{pC}\) (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times\) \(10^{7} \mathrm{V} / \mathrm{m}\) . The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{F}\) and must be able to withstand a maximum potential difference of 5500 \(\mathrm{V} .\) What is the minimum area the plates of the capacitor may have?

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric- field magnitude between the plates? (d) What is the surface charge density on each plate?

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