Capacitors & Electric Fields and how they change relative to each other

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Homework Help Overview

The discussion revolves around the behavior of a charged capacitor when the distance between its plates is increased. Participants explore the relationship between electric field strength, voltage, and plate separation in the context of capacitors, particularly focusing on a scenario involving a 10V source that has been removed.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the implications of maintaining constant charge on the plates while altering the distance between them. Questions arise about how this affects the electric field and voltage, as well as concerns about potential discharges when the plates are pulled apart.

Discussion Status

There is an ongoing exploration of the concepts involved, with some participants affirming that the electric field remains constant while the voltage increases as the distance increases. Concerns about safety and the physical limitations of pulling the plates apart are also noted, indicating a productive exchange of ideas.

Contextual Notes

Participants reference the ideal conditions of a parallel plate capacitor and acknowledge that real-world limitations may affect the scenario being discussed. The conversation hints at safety considerations regarding high voltage and stored energy in capacitors.

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1. I have a capacitor that has been fully charged with a 10v source, then that source is removed. If I increase the distance between the plates, does the electric field stay the same, and the voltage increase, or does the voltage stay the same, and the field decrease? Or is it something that depends on other factors?

If I continue to pull the plates apart, will the voltage increase until the plates are discharged via a spark? Or will the electric field just become weaker and weaker?

The answer given for this on my practice exam says the voltage will increase, but I don't understand why this is the case, unless the field cannot change under these circumstances.




Homework Equations



E = ΔV/d

The Attempt at a Solution


 
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You have to go back to the derivation of an ideal parallel plate capacitor. The plates have a certain amount of charge on them after being charged. Electric field is created by charge, and so if the charge remains the same (no reason for it to change in this case) the field will remain the same. As you said, if the field remains the same then increasing distance means the voltage also has to increase.
 
Thanks for the quick response. :-)

So the key point is since charge is not changed, the field must remain the same, thus leaving only the voltage to react to the change in distance. Correct?

How about me pulling them apart? Does nothing spectacular happen? Do I just end up two plates with an opposite charge? Or is the reason why students aren't left alone in Physics Lab since it does discharge via a spark once the voltage reaches a level where it can jump the gap? :-)
 
Yeah, as you pull apart the potential difference compensates to maintain a constant electric field, and the electric field is constant because the charges on the plates have nowhere to go so they'll be constant. As I said earlier though, this is assuming an infinite parallel plate capacitor. In reality you would not be able to pull apart the plates forever and ever. In any case, no, nothing spectacular happens.

Don't touch a capacitor that stores a lot of voltage and energy, not because you'll perturb any fields or voltages in the capacitor, but because it can shock you (pass current into your body) and make your heart fibrillate.
 
Thank you. :-)
 

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