Voltage in Capacitors

[justadream]

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I thought that voltage across a capacitor was set by the voltage source (e.g., if battery is 5V, voltage across the capacitor is 5V). But Voltage = E*d.

So if you increase the distance between the plates, you can increase the voltage? How come voltage in a capacitor can change (I thought it was always the voltage of the battery)?

 

[The Brown Knight]

No, if you increase the distance and as long as you don't disconnect the battery, the voltage will NOT change. However, the capacitance will decrease and charge stored on the capacitor will, as a result, decrease.

 

[Cawolf]

If you increase the distance, then the E field will get smaller in magnitude.

 

[justadream]

@Cawolf
@The Brown Knight

So is this wrong? See answer choice B (and the explanation).

 

[The Brown Knight]

@Cawolf
@The Brown Knight

So is this wrong? See answer choice B (and the explanation).

No, I still think C is the answer since the dielectric creates an opposing field that slightly "cancels" the one originally present.
Also, the problem states that the plate area is much larger than plate separation; so as long as you stay close to the middle of the capacitor, you can make the approximation that E field strength is independent of distance from the plate (you can probably find the Gauss's Law derivation of the exact formula for an infinite sheet somewhere online); this is why I thought A and B were correct statements. Choice B might not make the most sense, but C is the wrongest.

 

[Cawolf]

On the bright side - choice C is clearly and blatantly wrong.

My initial response was only accurate if we are dealing with a circuit that is charging a capacitor.

If the capacitor is fully charged and disconnected then apparently the E field remains constant. That is a concept that is slightly confusing to me.

But the formula that E = Q/Aε states that field strength is independent of separation with a constant charge.

 

[justadream]

@Cawolf

So if a capacitor is fully charged and disconnected, then V and d are inversely proportional?

 

[Cawolf]

Yes - as you put work into separating the charges, it translates into a greater potential difference between the plates.