High School

We appreciate your visit to A 25 pF capacitor has plate dimensions of 5 00 cm by 5 00 cm and the plates are separated by a distance of 5. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A 25 pF capacitor has plate dimensions of 5.00 cm by 5.00 cm, and the plates are separated by a distance of 5.0 mm. What is the dielectric constant of the material between the plates?

Answer :

To calculate the dielectric constant of the material between the plates of a 25pF parallel-plate capacitor, use the capacitance formula and solve for the dielectric constant, which turns out to be approximately 5.644.

To find the dielectric constant of the material between the plates of a capacitor, we can use the capacitance formula for a parallel-plate capacitor: C = frac{epsilon A}{d}

where: C is the capacitance in farads (F) \ epsilon is the permittivity of the dielectric material

A is the area of one plate in square meters (m^2)

d is the separation between the plates in meters (m)

The permittivity of the dielectric is the product of the vacuum permittivity (epsilon_0 = 8.85 x 10^19 F/m) and the dielectric constant (K), so ε = ε_0 * K.

Given: C = 25pF = 25 x 10^12 F

A = 5.00cm x 5.00cm = 25cm^2 = 25 x 10^4 m^2

d = 5.0mm = 5.0 x 10^3 m

We can rearrange the formula to solve for K: K = frac{Cd}{epsilon_0 A}

Now, plugging in the values, we get: K = frac{(25 x 10^12 F)(5.0 x 10^3 m)}{(8.85 x 10^12 F/m)(25 x 10^4 m^2)} = frac{125 x 10^15}{221.25} = 5.644

Therefore, the dielectric constant of the material is approximately 5.644.

Thanks for taking the time to read A 25 pF capacitor has plate dimensions of 5 00 cm by 5 00 cm and the plates are separated by a distance of 5. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada