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

AI Thread Summary
When a capacitor is charged and then disconnected from a voltage source, increasing the distance between the plates results in an increase in voltage while the electric field remains constant, assuming the charge stays the same. The electric field is determined by the charge on the plates, which does not change in this scenario. As the plates are pulled apart, the potential difference adjusts to maintain the electric field, leading to a higher voltage until it potentially discharges through a spark if the voltage exceeds the breakdown voltage of the air gap. In practical terms, while the ideal model assumes infinite separation, real-world limitations prevent indefinite separation. Caution is advised when handling charged capacitors due to the risk of electric shock.
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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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