Questions in the attachment **Question:**An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is ΔV 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*,- ○ ΔV increases, E remains the same- ○ ΔV remains the same, E remains the same- ○ ΔV increases, E increases- ○ ΔV decreases, E decreases- ○ ΔV decreases, E remains the same- ○ ΔV remains the same, E increases- ○ ΔV decreases, E increases- ○ ΔV increases, E decreases- ○ ΔV remains the same, E decreases**Explanation and Analysis:**This question explores the behavior of an air-filled parallel plate capacitor when the capacitance is altered by changing the distance between the plates after disconnecting the battery. - **ΔV (Potential Difference):** When the plates are moved apart after disconnecting the battery, the capacitance decreases, hence the potential difference across the plates *increases* to maintain the charge (Q remains constant since no current flows).- **E (Electric Field):** The electric field (E) is related to the potential difference and the distance (d) between the plates by the equation: $ E = \frac{\Delta V}{d} $. As distance increases, to keep E constant with increasing ΔV, E would decrease.In conclusion:When the distance is increased after disconnecting the battery:- ΔV increases and E decreases. The correct choice is: - ○ ΔV increases, E decreases

Answered: par plate Capacitor plates is AV and…

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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Step 1

Given

The potential difference is △V.

The electric field is E.

The gap between the capacitor plates is d, which is increased.

The mathematical formula by which the electric field is calculated between the plates is,

$E = \frac{Q}{A ε_{0}}$

Here, Q is a charge, A is area and $ε_{0}$ is the primitivity of vacuum (constant value)

From the mathematical formula of the electric field, it is clear that the electric field is independent of the gap between the plates (d). There is no change in Q and A. Therefore, the magnitude of the electric field remains unchanged.

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