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Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Short Answer

Expert verified
(a) Charge density: \(6.195\times10^{-7} \text{ C/m}^2\) (b) Dielectric constant: \(1.28\)

Step by step solution

01

Understanding the Problem

We need to determine two things: (a) the charge density on each surface of the dielectric and (b) the dielectric constant. We are given the electric fields for both the evacuated space and the space filled with dielectric.
02

Determine the Dielectric Constant (\(\kappa\))

The dielectric constant \(\kappa\) is the ratio of the electric field in the absence of the dielectric (\(E_0\)) to the electric field with the dielectric (\(E\)). This is given by the equation \(\kappa = \frac{E_0}{E}\). Substituting the given values, \(E_0 = 3.20\times10^5\space \text{V/m}\) and \(E = 2.50\times10^5\space \text{V/m}\), we have:\[ \kappa = \frac{3.20\times10^5}{2.50\times10^5} \approx 1.28 \]
03

Calculate Charge Density on Dielectric Surface

The charge density \(\sigma\) on each surface of the dielectric can be determined using \(\sigma = \sigma_0 (1 - 1/\kappa)\), where \(\sigma_0\) is the charge density without the dielectric. Given that \(\sigma = \epsilon_0 E_0\) and \(\sigma_0 = \epsilon_0 E\), substituting \(\kappa\) in terms of \(E\) and \(E_0\), we find that the charge density on the dielectric surface becomes zero, as charge density \(\sigma = \epsilon_0 E (\kappa - 1)\).
04

Apply Formula for Charge Density

Using the electric field and the dielectric constant, the formula for the induced charge density on the dielectric is given by \(\sigma = \epsilon_0 (E_0 - E)\). Using \(\epsilon_0 = 8.85 \times 10^{-12} \text{ C/m}^2\), we substitute the values:\[ \sigma = 8.85 \times 10^{-12} (3.20 \times 10^5 - 2.50 \times 10^5) \ \sigma = 8.85 \times 10^{-12} \times 0.70 \times 10^5 \ \sigma \approx 6.195 \times 10^{-7} \text{ C/m}^2 \]

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Key Concepts

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

Dielectric Constant
The dielectric constant, often represented by the Greek letter "kappa" (\( \kappa \)), is a measure of a material's ability to reduce the electric field within it. When a dielectric material is placed between two charged plates, it affects the electric field by becoming polarized. This polarization effectively reduces the field’s strength inside the dielectric compared to a vacuum. To calculate the dielectric constant, we use the formula \( \kappa = \frac{E_0}{E} \), where \( E_0 \) is the electric field without the dielectric and \( E \) is the field with the dielectric in place. ### Key Points:- **Reduces Electric Field**: The dielectric constant indicates how much a material decreases the electric field inside.- **Unitless Measure**: Since it is a ratio of two electric fields, the dielectric constant has no units.- **Polarization**: Dielectric materials polarize in response to an electric field, aligning their internal charges to oppose the field.In our example, the dielectric constant was found by calculating \( \kappa = \frac{3.20\times10^5}{2.50\times10^5} \approx 1.28 \), showcasing the material's effect on the field strength.
Charge Density
Charge density is a measure of the electric charge per unit area on a surface. It's usually represented by the Greek letter "sigma" (\( \sigma \)). In the context of parallel plates with a dielectric, charge density helps us understand how charges distribute themselves on the dielectric surfaces.The charge density on the dielectric surface can be influenced by the presence of a dielectric material, due to the modified electric field between the plates. When the dielectric constant (\( \kappa \)) is introduced, we calculate the induced charge density using \( \sigma = \epsilon_0 (E_0 - E) \). Here, \( \epsilon_0 \) is the permittivity of free space, a constant value \( 8.85 \times 10^{-12} ext{ C/m}^2 \). ### Application:- **With Dielectric**: \( \sigma = \epsilon_0(E_0 - E) \), showing the effect of the dielectric on charge distribution.- **Units**: Charge density is expressed in coulombs per square meter (C/m²).In our solved problem, \( \sigma \approx 6.195 \times 10^{-7} \text{ C/m}^2 \), indicating a redistribution of charges due to the dielectric.
Parallel Plates
In physics, parallel plates are often used to create uniform electric fields. This setup involves two flat, conductive plates facing each other, with equal but opposite electric charges on their surfaces. When analyzing electric fields and dielectrics, the parallel plate model simplifies the study of electrostatic phenomena.### Characteristics of Parallel Plates:- **Uniform Field**: Inside, the electric field is uniform and constant due to equal charge distribution.- **Simple Analysis**: It’s easier to calculate properties like charge density, capacitance, and electric fields.When a dielectric is placed between them, it reduces the effective electric field due to its polarization effect. The field strength between the plates diminishes in accordance with the dielectric constant, as previously discussed in the Dielectric Constant section.For the given problem, the empty space generated an electric field of \( E_0 = 3.20\times10^5 ext{ V/m} \). When filled with a dielectric, the weakened field was \( E = 2.50\times10^5 ext{ V/m} \), illustrating the role of dielectrics in modifying field characteristics.

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Most popular questions from this chapter

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

A 10.0- \(\mu\)F parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?

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\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 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 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?

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