Questions: The square plates of a 3000-pF parallel-plate capacitor measure 40 mm by 40 mm and are separated by a dielectric that is 0.29 mm thick and completely fills the region between the plates. What is the dielectric constant of the dielectric? 61 56 50 45 67

Solution Steps

We are given a parallel-plate capacitor with a capacitance of 3000 pF, plate dimensions of 40 mm by 40 mm, and a dielectric thickness of 0.29 mm. We need to find the dielectric constant of the material between the plates.

The capacitance \( C \) of a parallel-plate capacitor is given by the formula:

\[ C = \frac{{\varepsilon_r \varepsilon_0 A}}{d} \]

where:

• \( C \) is the capacitance,
• \( \varepsilon_r \) is the dielectric constant,
• \( \varepsilon_0 \) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)),
• \( A \) is the area of one plate,
• \( d \) is the separation between the plates.

The area \( A \) of one plate is:

\[ A = 40 \, \text{mm} \times 40 \, \text{mm} = (0.04 \, \text{m}) \times (0.04 \, \text{m}) = 0.0016 \, \text{m}^2 \]

We know:

• \( C = 3000 \, \text{pF} = 3000 \times 10^{-12} \, \text{F} \)
• \( d = 0.29 \, \text{mm} = 0.00029 \, \text{m} \)

Substitute these values into the capacitance formula:

\[ 3000 \times 10^{-12} = \frac{{\varepsilon_r \times 8.854 \times 10^{-12} \times 0.0016}}{0.00029} \]

Rearrange the equation to solve for \( \varepsilon_r \):

\[ \varepsilon_r = \frac{{3000 \times 10^{-12} \times 0.00029}}{{8.854 \times 10^{-12} \times 0.0016}} \]

Calculate \( \varepsilon_r \):

\[ \varepsilon_r = \frac{{3000 \times 0.00029}}{{8.854 \times 0.0016}} \approx 61.36 \]

Final Answer