Sure, let's go through the problem step-by-step.
Given data:
- Area of the capacitor plates, [tex]\( A = 100 \, \text{cm}^2 \)[/tex]
- Potential difference, [tex]\( V = 50 \, \text{V} \)[/tex]
- Initial capacitance without dielectric, [tex]\( C_0 = 13 \times 10^{-4} \, \mu \text{F} \)[/tex]
- Dielectric constant of mica, [tex]\( \kappa = 5.4 \)[/tex]
### Step 1: Convert Units
Convert the area into square meters:
[tex]\[ A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \][/tex]
Convert the initial capacitance into farads:
[tex]\[ C_0 = 13 \times 10^{-4} \, \mu \text{F} = 13 \times 10^{-4} \times 10^{-6} \, \text{F} = 13 \times 10^{-10} \, \text{F} \][/tex]
### Step 2: Calculate the Distance Between Plates
Use the formula for capacitance without dielectric:
[tex]\[ C_0 = \epsilon_0 \frac{A}{d} \][/tex]
where:
- [tex]\(\epsilon_0\)[/tex] is the vacuum permittivity ([tex]\(8.854 \times 10^{-12} \, \text{F/m}\)[/tex])
- [tex]\(d\)[/tex] is the separation between the plates
Rearrange the formula to solve for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{\epsilon_0 \cdot A}{C_0} \][/tex]
[tex]\[ d = \frac{8.854 \times 10^{-12} \, \text{F/m} \times 0.01 \, \text{m}^2}{13 \times 10^{-10} \, \text{F}} \][/tex]
After substituting in the values, we find [tex]\( d \approx 6.81 \times 10^{-14} \, \text{m} \)[/tex].
### Step 3: Electric Field in Mica
The electric field [tex]\( E \)[/tex] in the mica is given by:
[tex]\[ E = \frac{V}{d} \][/tex]
Using the calculated value of [tex]\( d \)[/tex]:
[tex]\[ E \approx \frac{50 \, \text{V}}{6.81 \times 10^{-14} \, \text{m}} \approx 734131.466 \, \text{V/m} \][/tex]
Therefore, [tex]\( E \approx 734131.466 \, \text{V/m} \)[/tex].
### Step 4: Displacement Vector
The displacement vector [tex]\( D \)[/tex] in the mica is given by:
[tex]\[ D = \epsilon_0 \kappa E \][/tex]
Using the given dielectric constant [tex]\(\kappa = 5.4\)[/tex]:
[tex]\[ D = (8.854 \times 10^{-12} \, \text{F/m}) \times 5.4 \times 734131.466 \, \text{V/m} \][/tex]
[tex]\[ D \approx 3.51 \times 10^{-5} \, \text{C/m}^2 \][/tex]
### Step 5: Polarization Vector
The polarization vector [tex]\( P \)[/tex] is given by:
[tex]\[ P = D - \epsilon_0 E \][/tex]
[tex]\[ P \approx 3.51 \times 10^{-5} \, \text{C/m}^2 - (8.854 \times 10^{-12} \, \text{F/m}) \times 734131.466 \, \text{V/m} \][/tex]
[tex]\[ P \approx 2.86 \times 10^{-5} \, \text{C/m}^2 \][/tex]
### Final Answers
(A) The electric field in mica: [tex]\( 734131.466 \, \text{V/m} \)[/tex]
(B) The displacement vector: [tex]\( 3.51 \times 10^{-5} \, \text{C/m}^2 \)[/tex]
(C) The polarization vector: [tex]\( 2.86 \times 10^{-5} \, \text{C/m}^2 \)[/tex]
These computed values match the results of our solved numerical calculation.
