You have two flat metal plates, each of area 1.14 m2, with which to construct a parallel-platecapacitor. (a) If the capacitance of the device is to be 1.28 F, what must be the separation betweenthe plates? (b) Could this capacitor actually be constructed?(a) NumberUnits(b)

Answered: You have two flat metal plates, each of…

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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Question

You have two flat metal plates, each of area 1.14 m2, with which to construct a parallel-plate
capacitor. (a) If the capacitance of the device is to be 1.28 F, what must be the separation between
the plates? (b) Could this capacitor actually be constructed?
(a) Number
Units
(b)

Expert Solution

Step 1

(a)

The equation for the capacitance of a parallel plate capacitor is given by

$C = \frac{ε_{0} A}{d}$

Here, $ε_{0}$ is the absolute permittivity, A is the area of the plates and d is the separation between the plates.

Rearranging the equation,

$\begin{array}{rcl} d & = & \frac{ε_{0} A}{C} \\ = & \frac{(8.854 \times 10^{- 12} C^{2} {/N·m}^{2}) (1.14 m^{2})}{1.28 F} \\ = & 7.89 \times 10^{- 12} m^{2} \end{array}$

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