**Capacitor Construction and Capacitance Calculation**A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. Given the following specifications:- Dimensions of aluminum sheets: 0.260 m by 0.220 m- Thickness of waxed paper: 0.0500 mm (converted to 0.000050 m for calculation)- Dielectric constant (\( \kappa \)): 2.50**Problem:**Calculate the capacitance of the capacitor.**Solution:**To find the capacitance (\( C \)), use the formula:\[ C = \frac{{\kappa \cdot \varepsilon_0 \cdot A}}{d} \]Where:- \( \kappa \) 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 of the aluminum sheets (\(0.260 \, \text{m} \times 0.220 \, \text{m}\)),- \( d \) is the separation distance, which is the thickness of the waxed paper (0.000050 m).The computed capacitance is:\[ 4.675 \, \text{nF} \]

Answered: Image /qna-images/answer/36dd0b56-158d-4d56-8ded-4aad58496531.jpg

A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are 0.26O m by 0.220 m and the waxed paper, of slightly larger dimensions, is of thickness 0.0500 mm and dielectric constant K = 2.50, what is the capacitance of the capacitor? 4.675 nF

A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are 0.26O m by 0.220 m and the waxed paper, of slightly larger dimensions, is of thickness 0.0500 mm and dielectric constant K = 2.50, what is the capacitance of the capacitor? 4.675 nF

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

An air-filled parallel-plate capacitor has plates of area 2.40 cm2 separated by 2.00 mm. The capacitor is connected to a(n) 17.0 V battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/C
An air-filled parallel-plate capacitor has plates of area 2.40 cm2 separated by 2.50 mm. The capacitor is connected to a 21.0-V battery. (a) Find the value of its capacitance. pF(b) What is the charge on the capacitor? pC(c) What is the magnitude of the uniform electric field between the plates?
An air-filled parallel-plate capacitor has plates of area 2.80 cm separated by 1.80 mm. The capacitor is connected to a(n) 18.0 v battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/C
A parallel plate capacitor with plate separation of 4.25 cm has plates whose area has dimensions of length 3.18 cm and width 2.32 cm. Calculate the capacitance of this capacitor (in picofarads) if a dielectric material with a dielectric constant of 3.24 is placed between the plates. pF
A charged air-filled parallel plate capacitor of 22 nF has a energy density of 30x10-3 J/m³. If the distance between the plates is 200 µm, calculate the magnitude of the charge, in µC, of the capacitor.
You are given a parallel plate capacitor with plates having a rectangular area of 16.7 cm2 and a separation of 2.2 mm. The space between the plates is filled with a material having a dielectric constant κ = 1.8. Find the capacitance of this system.
Consider a parallel plate capacitor having plates of area 1.4 cm2 that are separated by 0.018 mm of neoprene rubber. You may assume the rubber has a dielectric constant κ = 6.7. a) What is the capacitance in μF? b) What charge, Q, in coulombs, does it hold when 9.00 V is applied to it?
Consider a parallel-plate capacitor having an area of 2550 mm^2 and a plate separation of 4.9mm and with a material of dielectric constant 5.9 positioned between the plates. Also, the value of E0 is 8.85x10^-12F/m. a) What is the capacitance of this capacitor in pF? b) compute the electric field that must be applied for a charge pf 7.8x10^-8C to be stored on each plate in V/m.
An air-filled parallel-plate capacitor has plates of area 2.90 cm? separated by 3.00 mm. The capacitor is connected to a(n) 13.0 V battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/C
An air-filled capacitor is made from two flat parallel plates 1.3 mm apart. The inside area of each plate is 7.6 cm². (a) What is the capacitance of this set of plates (in pF)? pF (b) If the region between the plates is filled with a material whose dielectric constant is 6.2, what is the new capacitance (in pF)? pF