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

In summary, the electric field will remain constant, and the voltage will increase as the distance between the plates is increased.
  • #1
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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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  • #2
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.
 
  • #3
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? :-)
 
  • #4
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.
 
  • #5
Thank you. :-)
 

1. How do capacitors store electric charge?

Capacitors store electric charge by using two parallel plates separated by an insulating material, known as a dielectric. When a voltage is applied to the plates, an electric field is created between them, causing one plate to accumulate positive charge and the other to accumulate negative charge.

2. How does the electric field change in a capacitor during charging and discharging?

During charging, the electric field in a capacitor increases as more charge is stored on the plates. During discharging, the electric field decreases as the charge is released from the plates.

3. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by the surface area of the plates, the distance between the plates, and the type of dielectric material used.

4. How do capacitors and electric fields change relative to each other in a series circuit?

In a series circuit, the capacitors are connected end-to-end, with the same amount of charge on each capacitor. The electric field remains constant throughout the circuit, but the voltage drops across each capacitor in proportion to its capacitance.

5. How do capacitors and electric fields change relative to each other in a parallel circuit?

In a parallel circuit, the capacitors are connected side-by-side, with the same voltage applied across each capacitor. The electric field between the plates is divided among the capacitors, but the total charge on each capacitor remains the same.

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