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An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is AV and the magnitude of the electric field between the plates is E. If the battery is disconnected, and then the distance between the plates is increased, O AV increases, E remains the same O AV remains the same, E remains the same O AV increases, E increases O AV decreases, E decreases O AV decreases, E remains the same O AV remains the same, E increases O AV decreases, E increases O AV increases, E decreases O AV remains the same, E decreases

An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is AV and the magnitude of the electric field between the plates is E. If the battery is disconnected, and then the distance between the plates is increased, O AV increases, E remains the same O AV remains the same, E remains the same O AV increases, E increases O AV decreases, E decreases O AV decreases, E remains the same O AV remains the same, E increases O AV decreases, E increases O AV increases, E decreases O AV remains the same, E decreases

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 parallel plate capacitor is connected to a battery so the difference in potential between the plates is \( \Delta V \) and the magnitude of the electric field between the plates is \( E \). If the battery is disconnected, and then the distance between the plates is increased, - \( \Delta V \) increases, \( E \) remains the same - \( \Delta V \) remains the same, \( E \) remains the same - \( \Delta V \) increases, \( E \) increases - \( \Delta V \) decreases, \( E \) decreases - \( \Delta V \) decreases, \( E \) remains the same - \( \Delta V \) remains the same, \( E \) increases - \( \Delta V \) decreases, \( E \) increases - \( \Delta V \) increases, \( E \) decreases - \( \Delta V \) remains the same, \( E \) decreases
Transcribed Image Text:An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is \( \Delta V \) and the magnitude of the electric field between the plates is \( E \). If the battery is disconnected, and then the distance between the plates is increased, - \( \Delta V \) increases, \( E \) remains the same - \( \Delta V \) remains the same, \( E \) remains the same - \( \Delta V \) increases, \( E \) increases - \( \Delta V \) decreases, \( E \) decreases - \( \Delta V \) decreases, \( E \) remains the same - \( \Delta V \) remains the same, \( E \) increases - \( \Delta V \) decreases, \( E \) increases - \( \Delta V \) increases, \( E \) decreases - \( \Delta V \) remains the same, \( E \) decreases
Expert Solution
Step 1

After the disconnection of the battery from the parallel plate capacitor, the charge stored in the capacitor remains the same.

 

Write the expression for the electric field between the given plate capacitor.

E=σε0E=QAε0E=Qε0A

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