181k views
5 votes
A parallel-plate air-filled capacitor has a capacitance of 388 pf.

if each of its plates has an area of 0.050 m², what is the separation?

User Mckenzm
by
7.7k points

1 Answer

3 votes

Final answer:

To determine the separation of plates in a capacitor with a capacitance of 388 pf and plate area of 0.050 m², the formula C = ε0 * (A/d) is used to find that the plate separation is 1.14 mm.

Step-by-step explanation:

Calculating the Separation of Plates in a Capacitor

A parallel-plate air-filled capacitor with a capacitance of 388 pf and plate area of 0.050 m² is given, and we're asked to determine the plate separation. To solve the mathematical problem completely, we use the formula for the capacitance of a parallel-plate capacitor, which is C = ε0 * (A/d), where C is the capacitance, ε0 is the permittivity of free space (ε0 = 8.854 x 10-12 F/m), A is the area of the plates, and d is the separation between the plates.

Rearranging the formula to solve for d gives us d = ε0 * (A/C). Plugging in the given values we get:

d = (8.854 x 10-12 F/m) * (0.050 m² / 388 x 10-12 F) = 1.14 x 10-3 meters, or 1.14 mm. This is the plate separation for the given capacitor.

This question falls under the subject of Physics, specifically dealing with the topic of capacitance and electrostatics, which is generally covered in High School grade curriculum. The problem demonstrates how to give me a 500 word answer, however, brevity and clarity are equally important, ensuring the response is not too verbose while maintaining complete information.

User Ghassen Sultana
by
7.3k points