Answered: Imagine a parallel plate capacitor with…

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)...

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Imagine a parallel plate capacitor with mica between the plates. Suppose the potential difference between the plates is 82V, and there is an electric field between the plates with a strength of 320 N/C. If each plate has an area of 0.030 square meters,

a) What is the separation between the plates?

b) Find the capacitance of the capacitor.

c) How much charge is on each plate?

d) How much energy is stored in the capacitor?

For each part, explain your reasoning.

Expert Solution

Step 1 :Introduction

The separation distance between the plates is calculated by using the formula $d = \frac{V}{E}$ , where $V = 82 V$ is the potential difference and $E = 320 N / C$ is the electric field .

The capacitance of the capacitor is calculated by using the formula $C = \frac{k ε_{0} A}{d}$ , where $k = 6$ is the dielectric constant , $ε_{0} = 8.854 \times 10^{- 12} F / m$ is the permittivity of free space , $A = 0.03 m^{2}$ is the area of the plate and d is the distance between the plates .

The charge on the capacitor is calculated by using the formula $q = C V$ and the energy stored in the capacitor is calculated by using the formula $E = \frac{C V^{2}}{2}$ .

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