Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E=3.20×105 V/m. When the space is filled with dielectric, the electric field is E=2.50×105 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: 6.195×107 C/m2 (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 (κ)

The dielectric constant κ is the ratio of the electric field in the absence of the dielectric (E0) to the electric field with the dielectric (E). This is given by the equation κ=E0E. Substituting the given values, E0=3.20×105 V/m and E=2.50×105 V/m, we have:κ=3.20×1052.50×1051.28
03

Calculate Charge Density on Dielectric Surface

The charge density σ on each surface of the dielectric can be determined using σ=σ0(11/κ), where σ0 is the charge density without the dielectric. Given that σ=ϵ0E0 and σ0=ϵ0E, substituting κ in terms of E and E0, we find that the charge density on the dielectric surface becomes zero, as charge density σ=ϵ0E(κ1).
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 σ=ϵ0(E0E). Using ϵ0=8.85×1012 C/m2, we substitute the values:σ=8.85×1012(3.20×1052.50×105) σ=8.85×1012×0.70×105 σ6.195×107 C/m2

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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" (κ), 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 κ=E0E, where E0 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 κ=3.20×1052.50×1051.28, 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" (σ). 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 (κ) is introduced, we calculate the induced charge density using σ=ϵ0(E0E). Here, ϵ0 is the permittivity of free space, a constant value 8.85×1012extC/m2. ### Application:- **With Dielectric**: σ=ϵ0(E0E), 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, σ6.195×107 C/m2, 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 E0=3.20×105extV/m. When filled with a dielectric, the weakened field was E=2.50×105extV/m, illustrating the role of dielectrics in modifying field characteristics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An air capacitor is made by using two flat plates, each with area A, separated by a distance d. Then a metal slab having thickness a (less than d) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits a 0 and a d.

A 20.0-μF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-μF capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00× 106 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

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 1675 s with an average light power output of 2.70 × 105 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 V when the stored energy equals the value calculated in part (a). What is the capacitance?

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free