Solve the electric field strength of the capacitor's dielectric when the polarization P=2μCm2P=2μCm2, the relative permittivity of the dielectric is εr=5.5εr=5.5 and when the permittivity of the vacuum is known ε0=8.854∗10−12Fmε0=8.854∗10−12Fm.

To solve the electric field strength \(E\) of the capacitor's dielectric, we can use the relationship between the electric displacement field \(D\), the polarization \(P\), and the electric field \(E\).

Given:
- Polarization \( P = 2 \, \mu C/m^2 = 2 \times 10^{-6} \, C/m^2 \)
- Relative permittivity \( \varepsilon_r = 5.5 \)
- Permittivity of vacuum \( \varepsilon_0 = 8.854 \times 10^{-12} \, F/m \)

The total displacement field \(D\) in the dielectric is related to the electric field \(E\) and the polarization \(P\) as follows:
\[
D = \varepsilon_0 \varepsilon_r E
\]
And we know that the displacement field \(D\) is also related to the polarization \(P\) by:
\[
D = \varepsilon_0 E + P
\]

Now, equating both expressions for \(D\):
\[
\varepsilon_0 \varepsilon_r E = \varepsilon_0 E + P
\]

Solving for \(E\):
\[
E(\varepsilon_0 \varepsilon_r - \varepsilon_0) = P
\]
\[
E = \frac{P}{\varepsilon_0 (\varepsilon_r - 1)}
\]

Now substitute the given values:
\[
E = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12} \times (5.5 - 1)}
\]

Let's calculate the electric field \(E\).It seems like I can’t do more advanced data analysis right now. Please try again later. 

However, you can manually calculate the electric field \(E\) using this formula:
\[
E = \frac{P}{\varepsilon_0 (\varepsilon_r - 1)}
\]
where:
- \( P = 2 \times 10^{-6} \, C/m^2 \)
- \( \varepsilon_0 = 8.854 \times 10^{-12} \, F/m \)
- \( \varepsilon_r = 5.5 \)

This will give you the electric field strength.

If you have any other questions or need more details, feel free to ask!

Here are 5 related questions to consider:
1. How does the relative permittivity of a material affect the electric field inside a capacitor?
2. What is the physical meaning of polarization \(P\) in dielectric materials?
3. How do you compute the capacitance of a capacitor with a dielectric material?
4. How would the electric field change if the relative permittivity were doubled?
5. What role does the permittivity of vacuum (\( \varepsilon_0 \)) play in electric field calculations?

**Tip**: Always ensure that units are consistent when solving physics problems to avoid errors in your calculations!

Solve the electric field strength of the capacitor's dielectric when the polarization P = 2 μC/m², the relative permittivity of the dielectric is εr = 5.5, and the permittivity of the vacuum is known ε0 = 8.854 * 10^-12 F/m.

This tutorial explains how to solve for the electric field strength in a capacitor's dielectric when the polarization P = 2 μC/m², the relative permittivity εr = 5.5, and the permittivity of the vacuum ε0 = 8.854 * 10^-12 F/m. Learn key concepts of electromagnetism, including electric displacement and polarization, with step-by-step calculations. Suitable for undergraduate students.

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