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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.90 mm. If a 23.0-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates magnitude kV/m direction --Select--- (b) the capacitance pF (c) the charge on each plate pC

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.90 mm. If a 23.0-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates magnitude kV/m direction --Select--- (b) the capacitance pF (c) the charge on each plate pC

College Physics
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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An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \, \text{cm}^2\), separated by a distance of \(1.90 \, \text{mm}\). If a \(23.0 \, \text{V}\) potential difference is applied to these plates, calculate the following: (a) The electric field between the plates - Magnitude: \(\_\_\_\_\_\) kV/m - Direction: [Dropdown menu for selection] (b) The capacitance - \(\_\_\_\_\_\) pF (c) The charge on each plate - \(\_\_\_\_\_\) pC
Transcribed Image Text:An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \, \text{cm}^2\), separated by a distance of \(1.90 \, \text{mm}\). If a \(23.0 \, \text{V}\) potential difference is applied to these plates, calculate the following: (a) The electric field between the plates - Magnitude: \(\_\_\_\_\_\) kV/m - Direction: [Dropdown menu for selection] (b) The capacitance - \(\_\_\_\_\_\) pF (c) The charge on each plate - \(\_\_\_\_\_\) pC
Expert Solution
Step 1

Assume that the air filled capacitor has two parallel plates.

The area of the plate is A = 7.60 cm2 = 7.60x10-4 m2.

The distance between the plates is d = 1.90 mm = 1.90x10-3 m.

The potential difference applied to the plates is V = 23.0 V.

Assume the permittivity of free space as ε0 = 8.85x10-12 m-3kg-1s4A2.

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