Quick physics question

[archcommus]

If the electric field between two plates of a capacitor decreases, does the potential difference across the plates decrease and thus the capacitance increases?

 

[TuxDave]

If the potential difference across the plates decreases, then capacitance increases assuming everything else stays the same. However, if the electric field decreased because you simply removed charge from the two plates, then capacitance is the same.

edit: so many edits... i guess it depends on what changed.

 

[archcommus]

Originally posted by: TuxDave
If the potential difference across the plates decreases, then capacitance increases assuming everything else stays the same. However, if the electric field decreased because you simply removed charge from the two plates, then capacitance is the same.

The electric field decreased because one of the plates "moved up."

 

[Heisenberg]

Q=C*V and V = E*d (for parallel plates), so Q=C*E*d or C=Q/(E*d). So if the electric field decreases then the capacitance should increase assuming Q and d stay the same.

Edit: One of the plates moves up? Well that's a bit different then...

 

[TuxDave]

Thinking....

 

[archcommus]

Originally posted by: Heisenberg
Q=C*V and V = E*d (for parallel plates), so Q=C*E*d or C=Q/(E*d). So if the electric field decreases then the capacitance should increase assuming Q and d stay the same.

Edit: One of the plates moves up? Well that's a bit different then...

This was a theoetical problem, no numbers, so how do I know whether the total voltage went up or down? E went down, but d went up. It basically said that the e-field decreased due to the plate moving up, increasing the distance between them.

 

[Heisenberg]

Originally posted by: archcommus

Originally posted by: Heisenberg
Q=C*V and V = E*d (for parallel plates), so Q=C*E*d or C=Q/(E*d). So if the electric field decreases then the capacitance should increase assuming Q and d stay the same.

Edit: One of the plates moves up? Well that's a bit different then...

This was a theoetical problem, no numbers, so how do I know whether the total voltage went up or down? E went down, but d went up. It basically said that the e-field decreased due to the plate moving up, increasing the distance between them.

Edit: Hmm...let me think about that for a sec...

 

[archcommus]

Originally posted by: Heisenberg

Originally posted by: archcommus

Originally posted by: Heisenberg
Q=C*V and V = E*d (for parallel plates), so Q=C*E*d or C=Q/(E*d). So if the electric field decreases then the capacitance should increase assuming Q and d stay the same.

Edit: One of the plates moves up? Well that's a bit different then...

This was a theoetical problem, no numbers, so how do I know whether the total voltage went up or down? E went down, but d went up. It basically said that the e-field decreased due to the plate moving up, increasing the distance between them.

Well if the potential difference remains the same, it shouldn't change the capacitance. Changing the distance between the plates doesn't affect the voltage, just the electric field in between.

It affects the electric field in between, but doesn't that in turn affect the voltage? According to V=Ed. The first question asked how voltage changed, the second question asked how capacitance changed. It seems as if you'd need to know numbers to make a conclusion.

 

[Heisenberg]

Ok, if the capacitor is connected to a some kind of battery/power supply then changing the distance between the two plates won't change the potential difference. Charge would be added/subtracted as necessary to maintain the same potential. For the second part, I would use C=epsilon*A/d, since any dependence on electric field or potential is gone. In that case, increasing the distance would decrease the capacitance.

 

[archcommus]

Originally posted by: Heisenberg
Ok, if the capacitor is connected to a some kind of battery/power supply then changing the distance between the two plates won't change the potential difference. Charge would be added/subtracted as neccessary to maintain the same potential. For the second part, I would use C=epsilon*A/d, since any dependence on electric field or potential is gone. In that case, increasing the distance would decrease the capacitance.

Lol so in other words I goofed up this whole thing.

I'm pretty sure it said it was disconnected from any other circuit element, so I was thinking, E down, V down, C up.

 

[InlineFour]

lol, i just studied this electric field last chapter.

 

[TuxDave]

See, my problem with this answer is that if the charge density on the plates remain the same, and assuming no fringing is happening, electric field strength should be constant regardless of how far apart the plates are. That would lead me to believe that if E is constant, d increases, V increases leading to an increase in C.

 

[Heisenberg]

Originally posted by: TuxDave
See, my problem with this answer is that if the charge density on the plates remain the same, and assuming no fringing is happening, electric field strength should be constant regardless of how far apart the plates are. That would lead me to believe that if E is constant, d increases, V increases leading to an increase in C.

But that's only if the capacitor is disconnected from any external power source correct? I was looking at it the other way - the potential between the two plates remains the same via adding/subtracting charge and thus the electric field changes. I'm sure that the answer to the second question is that C decreases, as C for parallel plates is defined entirely through the geometry.

 

[Random Variable]

I'm thinking that if the potential hasn't changed, the movement of one of the plates would increase the capacitance in the same way as the placement of a diaelectric material in between the two plates. But I'm sure I'm wrong.

 

[archcommus]

Good input, guys! Like I said, I think it was disconnected from all other circuit elements, but it stated IN the problem that d definitely increases and E definitely decreases. But if d increases, then C decreases, which must mean V increases, which must mean E decreased LESS than d increased, which I find hard to believe, considering d is like .00something meters in most cases.