Question

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

Short answer

(a) Charge density: $6.195\times {10}^{-7}{\text{ C/m}}^2$ (b) Dielectric constant: $1.28$

Step-by-step solution

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.

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, $E_0=3.20\times {10}^5\text{V/m}$ and $E=2.50\times {10}^5\text{V/m}$, we have:$$\kappa =\frac{3.20\times {10}^5}{2.50\times {10}^5}\approx 1.28$$

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 $\sigma ={ϵ}_0{E}_0$ and ${\sigma }_0={ϵ}_{0}E$, substituting $\kappa$ in terms of $E$ and $E_0$, we find that the charge density on the dielectric surface becomes zero, as charge density $\sigma ={ϵ}_{0}E(\kappa -1)$.

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 $\sigma ={ϵ}_0(E_0-E)$. Using ${ϵ}_0=8.85\times {10}^{-12}{\text{ C/m}}^2$, we substitute the values:$$\sigma =8.85\times {10}^{-12}(3.20\times {10}^5-2.50\times {10}^5)\sigma =8.85\times {10}^{-12}\times 0.70\times {10}^5\sigma \approx 6.195\times {10}^{-7}{\text{ C/m}}^2$$

Key concepts

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 $\kappa =\frac{3.20\times {10}^5}{2.50\times {10}^5}\approx 1.28$, 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 $\sigma ={ϵ}_0(E_0-E)$. Here, ${ϵ}_0$ is the permittivity of free space, a constant value $8.85\times {10}^{-12}ext{C/m}^2$. ### Application:- **With Dielectric**: $\sigma ={ϵ}_0(E_0-E)$, 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 $E_0=3.20\times {10}^{5}extV/m$. When filled with a dielectric, the weakened field was $E=2.50\times {10}^{5}extV/m$, illustrating the role of dielectrics in modifying field characteristics.