# capacitors

#### inaccensa

##### Full Member
10+ Year Member
I was reviewing EK and I'm a little confused.
C= Q/V. So I understand that the capacitance will decrease with an increase in voltage ina circuit. Now EK also said that the V will increase with an increase in Q stating a linear relationship for the energy in a capacitor.

I understand the formulae C= Q/V and

U = .5 QV; U=.5CV^2;U=Q^2/C

#### inaccensa

##### Full Member
10+ Year Member
can anyone please explain this? Also, if the capacitor is fully charged and then the battery is disconnected, then the V across the capacitor will not necessarily be zero rt? The charge will remain constant, but what happens to the capacitance and voltage?

#### stockraider

##### Full Member
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5+ Year Member
can anyone please explain this? Also, if the capacitor is fully charged and then the battery is disconnected, then the V across the capacitor will not necessarily be zero rt? The charge will remain constant, but what happens to the capacitance and voltage?
ok so say you have a battery connected to a capacitor, and keep in mind the equation Q=CV. when you UNPLUG the battery, you decrease V and also as a result decrease C, when you keep a battery plugged in, V remains constant while C increases and Q increases.

#### sv3

##### Full Member
10+ Year Member
ok so say you have a battery connected to a capacitor, and keep in mind the equation Q=CV. when you UNPLUG the battery, you decrease V and also as a result decrease C, when you keep a battery plugged in, V remains constant while C increases and Q increases.

How does that work? If you unplug the battery, then the Q has nowhere to go. So if V and C decrease, how does the Q=CV equation balance?

Are you assuming there's a dieelectric present or something?

I just reviewed this section so worried i ain't grasping this (felt good when i did problems)

Also, I didn't think C was dependent on V or Q - you don't need to know those to even determine C. I thought C only changed if a dieelectric was present.
thanks

#### stockraider

##### Full Member
10+ Year Member
5+ Year Member
How does that work? If you unplug the battery, then the Q has nowhere to go. So if V and C decrease, how does the Q=CV equation balance?

Are you assuming there's a dieelectric present or something?

I just reviewed this section so worried i ain't grasping this (felt good when i did problems)

Also, I didn't think C was dependent on V or Q - you don't need to know those to even determine C. I thought C only changed if a dieelectric was present.
thanks
ok sorry, i mean if u unplug the battery you decrease V and raise C by the same amount to keep Q constant. also, C is dependent on V, remember C= constant*A/d, and V=Ed, so doing some math you can see d=V/E and hence C=constant*A/(V/E). now decreasing V using this equation has what effect on C?...it increases it.

#### thebillsfan

##### Unseasoned Veteran
10+ Year Member
5+ Year Member
ok sorry, i mean if u unplug the battery you decrease V and raise C by the same amount to keep Q constant. also, C is dependent on V, remember C= constant*A/d, and V=Ed, so doing some math you can see d=V/E and hence C=constant*A/(V/E). now decreasing V using this equation has what effect on C?...it increases it.

i dont know if this is right. if you unplug the battery, the charge has nowhere to go, so there is still a potential diff between the two plates. nothing changes if you unplug the voltage but dont separate the plates

#### sv3

##### Full Member
10+ Year Member
ok sorry, i mean if u unplug the battery you decrease V and raise C by the same amount to keep Q constant. also, C is dependent on V, remember C= constant*A/d, and V=Ed, so doing some math you can see d=V/E and hence C=constant*A/(V/E). now decreasing V using this equation has what effect on C?...it increases it.

d is not a dependent. it is invariable. but i see what your saying.

#### inaccensa

##### Full Member
10+ Year Member
i dont know if this is right. if you unplug the battery, the charge has nowhere to go, so there is still a potential diff between the two plates. nothing changes if you unplug the voltage but dont separate the plates

I don't know if I agree with you, I just did a problem with kaplan. The capacitor was fully charged and the battery was unpluged. I thought the V =0 volts. i thought I saw that somewhere. Anyhow the V which was initially 10 volts went upto 20v.

Can someone please explain what exactly happens after the battery is disconnected? i thought the battery was not only responsible for supplying charge, but it also helped build and maintain the potential difference. please correct me if I'm wrong

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