Battery connected to capacitor

[negation]

In connecting a battery to a capactor, doubling the spacing halve the capacitance while keeping voltage fixed.
Why?

 

[phinds]

In connecting a battery to a capactor, doubling the spacing halve the capacitance while keeping voltage fixed.
Why?

Why not? What is your understanding of how capacitors work?

 

[abitslow]

please read the first paragraph of this:http://en.wikipedia.org/wiki/Capacitance
or simply review your textbook.

 

[negation]

Why not? What is your understanding of how capacitors work?

I have managed to start with a derivation with V=Ed to show that if d=2d, for a battery connected to a capacitor, then c is halved.
But I do not know how should I show what happens when distance is doubled as capacitor is unconnected to battery.

 

[negation]

please read the first paragraph of this:http://en.wikipedia.org/wiki/Capacitance
or simply review your textbook.

It states nothing...wouldn't really be asking to obvious otherse

I want the derivation...

 

[negation]

Why not? What is your understanding of how capacitors work?

Fairly poor. I at least do know that it is a device used to store charges?

 

[phinds]

Yes, it stores a charge and the voltage on the capacitor after it reaches steady state is whatever voltage you put on it. If you hook it to a 12V battery, it will hold 12volts. This is independent of the value of the capacitor, but the value does tell you how much charge it stores, which is not the same thing as voltage.

If you hook a capacitor DIRECTLY to a battery it will most likely explode. Any idea why?

EDIT: You edited your post while I was entering this one. Going back and editing your entries makes a thread confusing. Better to just do a new post (in the thread) with the update, OR ... put EDIT: below your original post and then enter the new thing. Hm ... I now see 2 posts in a row. Maybe I just missed the first one.

 

[negation]

Yes, it stores a charge and the voltage on the capacitor after it reaches steady state is whatever voltage you put on it. If you hook it to a 12V battery, it will hold 12volts. This is independent of the value of the capacitor, but the value does tell you how much charge it stores, which is not the same thing as voltage.

If you hook a capacitor DIRECTLY to a battery it will most likely explode. Any idea why?

EDIT: You edited your post while I was entering this one. Going back and editing your entries makes a thread confusing. Better to just do a new post (in the thread) with the update, OR ... put EDIT: below your original post and then enter the new thing. Hm ... I now see 2 posts in a row. Maybe I just missed the first one.

Interesting input! It never occurred to me the practicality of a capacitor. In analogy, a capacitor stores as we know and we can view it in similarity to a tank of a certain capacity. The batter that is connected to a capacitor serves as a water tap. The capacity of the capacitor is independent of the rate at which water flows through the water tap.

Edit: It will explode because the capacitor has a finite capacitance.

 

[jbriggs444]

I want the derivation...

At a simple level...

You understand about the inverse square law, right? The electric field strength from a point charge is inversely proportional to the square of the distance from that charge.

If the charge is spread out along an infinite wire (with a fixed amount of charge per unit length), do you understand that the relationship changes so that the electric field strength is inversely proportional to the distance from the wire?

If the charge is spread out along an infinite plane (with a fixed amount of charge per unit area), do you understand that the relationship changes again so that electric field strength is a constant independent of distance from the plane?

A capacitor is approximately a pair of planes with a fixed charge per unit area. The charge on the two planes is equal and opposite. As a result, the electric field strength is constant between the two planes and is zero in their exterior. Electric potential (aka voltage) is given by the integral of electric field strength on (any) path running from the arbitrarily chosen zero point to the point where the potential is measured. If you double the distance between the two plates and keep the field strength constant, you've doubled the resulting integral. That means you've doubled the voltage difference between the two plates.

Twice the voltage holding the same charge means you've halved the capacitance. In order to hold the voltage fixed (because that's what batteries do), that means that the capacitor must now only be holding half the charge.

 

[negation]

At a simple level...

You understand about the inverse square law, right? The electric field strength from a point charge is inversely proportional to the square of the distance from that charge.

If the charge is spread out along an infinite wire (with a fixed amount of charge per unit length), do you understand that the relationship changes so that the electric field strength is inversely proportional to the distance from the wire?

If the charge is spread out along an infinite plane (with a fixed amount of charge per unit area), do you understand that the relationship changes again so that electric field strength is a constant independent of distance from the plane?

A capacitor is approximately a pair of planes with a fixed charge per unit area. The charge on the two planes is equal and opposite. As a result, the electric field strength is constant between the two planes and is zero in their exterior. Electric potential (aka voltage) is given by the integral of electric field strength on (any) path running from the arbitrarily chosen zero point to the point where the potential is measured. If you double the distance between the two plates and keep the field strength constant, you've doubled the resulting integral. That means you've doubled the voltage difference between the two plates.

Twice the voltage holding the same charge means you've halved the capacitance. In order to hold the voltage fixed (because that's what batteries do), that means that the capacitor must now only be holding half the charge.

Yes it drops off like 1/r²

For a battery connected to a capacitor and where d = 2d:

I begin with:

EA = Q/ε₀

E = Q/Aε₀

But V = E.d
∴V = Qd/Aε₀

Expressing the above as Q:

Q = VAε₀/d

But suppose d dobules. So, d= 2d
∴C =Aε₀/2d

The formula for capacitance is C = Q/V:
From this,

Q = [VAε₀/2d]/V which simplifies to Aε₀/2d
∴C = Aε₀/2d

for C = Aε₀/2d to be true, it must be the case that C/2

therefore, if d doubles, then C must be halved because if it were not then, we get an expression not equivalent to C/2 via reductio ad asurdum. (on a side note, what does this tells me about the voltage and charge?)


I really shouldn't be spending all the time doing such derivation because it's really jeopardizing my grades and it isn't really what my course focuses at the moment but it's harder to just regurgitate.

Really, a derivation of what happens when d=2d in the event a battery is disconnected would be great.

 

[cjl]

If you hook a capacitor DIRECTLY to a battery it will most likely explode. Any idea why?

Having done this a number of times (and never experienced a resulting explosion), I disagree.

 

[jbriggs444]

therefore, if d doubles, then C must be halved because if it were not then, we get an expression not equivalent to C/2 via reductio ad asurdum. (on a side note, what does this tells me about the voltage and charge?)

So you've derived that C must be halved if d is doubled. But you already know the relationship between voltage and charge. voltage = charge/capacitance.

With a battery connected, voltage is held contant. An ideal battery will source (or sink) whatever current is required to maintain a fixed voltage difference between its two terminals.

That means that as you separate the capacitor plates (which, if the charge on the plates were fixed would normally increase the voltage difference between them) the voltage increases only slightly. The increased voltage leads to a current that drains into the battery (against the grain, charging the battery). This drains the stored charge on the capacitor until equilibrium is regained.

Really, a derivation of what happens when d=2d in the event a battery is disconnected would be great.

Already provided. If the battery is disconnected, the charge on the plates is fixed and the voltage doubles.

 

[negation]

Having done this a number of times (and never experienced a resulting explosion), I disagree.

He said 'most likely' so universal violation does not applies here.

 

[phinds]

Edit: It will explode because the capacitor has a finite capacitance.

That answer is relevant but not really the reason. I suppose if a capacitor had infinite capacitance, it could handle the infinite current that is really the problem.

Now, in a real-world battery, you can't GET infinite current you just get however much the battery can provide so it may not explode but the point is that the capacitor acts like a short circuit when first connected ... THAT is the key point to understand ... and short circuits draw all the current you can feed them and too much current all at once can make violent things happen inside a physical device. It can also depend on the construction. There are LOTS of ways to make capacitors. Some have liquids inside, some don't and there are different dielectrics and so forth.

 

[negation]

So you've derived that C must be halved if d is doubled. But you already know the relationship between voltage and charge. voltage = charge/capacitance.

With a battery connected, voltage is held contant. An ideal battery will source (or sink) whatever current is required to maintain a fixed voltage difference between its two terminals.

That means that as you separate the capacitor plates (which, if the charge on the plates were fixed would normally increase the voltage difference between them) the voltage increases only slightly. The increased voltage leads to a current that drains into the battery (against the grain, charging the battery). This drains the stored charge on the capacitor until equilibrium is regained.



Already provided. If the battery is disconnected, the charge on the plates is fixed and the voltage doubles.

I see it. I overlooked the relationship C = Q/V.
Rearranging, V = Q/C and where C = C/2, then V = Q/(C/2) = 2 Q/C.
This means with a batter connected, distance is halved and V remains constant then Q must double. Am I right?

 

[negation]

I'm still confused about the part where the battery is disconnected.

 

[jbriggs444]

I see it. I overlooked the relationship C = Q/V.
Rearranging, V = Q/C and where C = C/2, then V = Q/(C/2) = 2 Q/C.
This means with a batter connected, distance is halved and V remains constant then Q must double. Am I right?

Not quite. You halved the denominator. The quotient is fixed. What has to happen to the numerator?

 

[jbriggs444]

I'm still confused about the part where the battery is disconnected.

With the battery disconnected, the charge on the plates is fixed but the voltage is no longer fixed.

Decrease capacitance and keep charge the same. What has to happen to voltage?

 

[negation]

Not quite. You halved the denominator. The quotient is fixed. What has to happen to the numerator?

It has to be halved.

Edit: wait, what equation are we referring to? V = Q/C?

 

[negation]

With the battery disconnected, the charge on the plates is fixed but the voltage is no longer fixed.

Decrease capacitance and keep charge the same. What has to happen to voltage?

Voltage has to increase.

 

[jbriggs444]

It has to be halved.

Edit: wait, what equation are we referring to? V = Q/C?

Right. Q is in the numerator. It has to be halved. Not doubled.

 

[jbriggs444]

Voltage has to increase.

Right. Double the distance with the battery disconnected and voltage doubles.

 

[DrZoidberg]

Exactly. With the battery disconnected Q stays constant.
And if you look at this equation
E = Q/Aε0
you see that E stays constant too.

 

[sophiecentaur]

It is important to bear in mind that moving the plates together and apart involves WORK. As you take them further apart (not connected to anything) the charge is constant and the Capacitance decreases - so the PD increases pro-rata (Q=CV) and the Energy will increase (=CV²/2)If you move them apart when the battery is connected, the voltage is unchanged so the charge (Q=CV again) must flow back into the battery (more work done in shifting charge back against the battery emf).

In any of these problems, the way to get things straight is to apply the Formula, whenever you can and that will give you the right answer. You can rely on that answer, so it only remains to come to terms with it. If you try to approach it form the 'intuitive' direction (as a misplaced matter of principle) then you can fall over. Elementary stuff like this is totally reliable so it it never worth looking for an excuse for it being 'wrong'. i.e. you can rely on the book work. It is not a cop out.
When you have finally reached a stage in your Science Education that you are at the frontiers of Science and breaking new ground then you will need to start to doubt that what you have learned is 'enough'. Until then, stand on the shoulders of previous Giants (as Sir Isaac did). If you just don't want to learn the book work then that's up to you - very dangerous.

 

[davenn]

by phinds View Post

If you hook a capacitor DIRECTLY to a battery it will most likely explode. Any idea why?

Having done this a number of times (and never experienced a resulting explosion), I disagree.

Yeah I have to disagree with that one too

have hooked hundreds of capacitors across batteries no one of them has even got hot, let alone exploded

The only way it may explode is if the capacitor voltage rating is much lower than the applied voltage.
And many many years ago as a young learner I did explode a cap with a 150V rating when I accidentally placed 240V AC across it oops

Dave

 

[sophiecentaur]

Yeah I have to disagree with that one too

have hooked hundreds of capacitors across batteries no one of them has even got hot, let alone exploded

The only way it may explode is if the capacitor voltage rating is much lower than the applied voltage.
And many many years ago as a young learner I did explode a cap with a 150V rating when I accidentally placed 240V AC across it oops

Dave

I endorse your disagreement. The only way you will get 'too much' charge in a capacitor will be if there is some appreciable series Inductance in the circuit and that can lead to voltage magnification due to series resonance.
You are more likely to damage a Capacitor by Shorting it out, due to the potentially high discharge current. Any battery will have an appreciable internal resistance and limit the charge current.

 

[enorbet]

In actual practice capacitors connected to a DC source eg: a battery will not explode unless the dielectric breakdown voltage or the rated use temperature is exceeded. While in theory it is a capacitance problem, all real world capacitors have ESR which is Effective Series Resistance. In most cases this will bleed off any excesses in a system where battery and capacitor voltages match. They will reach an equilibrium, much like two batteries connected together.

 

[negation]

It is important to bear in mind that moving the plates together and apart involves WORK. As you take them further apart (not connected to anything) the charge is constant and the Capacitance decreases - so the PD increases pro-rata (Q=CV) and the Energy will increase (=CV²/2)If you move them apart when the battery is connected, the voltage is unchanged so the charge (Q=CV again) must flow back into the battery (more work done in shifting charge back against the battery emf).

In any of these problems, the way to get things straight is to apply the Formula, whenever you can and that will give you the right answer. You can rely on that answer, so it only remains to come to terms with it. If you try to approach it form the 'intuitive' direction (as a misplaced matter of principle) then you can fall over. Elementary stuff like this is totally reliable so it it never worth looking for an excuse for it being 'wrong'. i.e. you can rely on the book work. It is not a cop out.
When you have finally reached a stage in your Science Education that you are at the frontiers of Science and breaking new ground then you will need to start to doubt that what you have learned is 'enough'. Until then, stand on the shoulders of previous Giants (as Sir Isaac did). If you just don't want to learn the book work then that's up to you - very dangerous.

What about derivation? When does it become important? I often find it easier to see a map of the relationship between variables in the process of deriving certain equations. While it is VERY tedious, it does suits my mental stipulation. "just" applying formula isn't quite doing physics is it?

 

[sophiecentaur]

I couldn't agree with you more. Where does "derivation" come from, if not by using previous, established formulae (the formal statements of ideas)? Starting with the simple and going into the complex, step by rigorous step is the way things get found out - not arm waving and "no but what is really happening 'Physically'?"
The 'Giants' of the past, provide us with the building blocks of formal analysis of the way things work. The formulae have shown themselves to be the most reliable way of passing this stuff from brain to brain. How we, as individuals, deal with those formulae depends on our individual ability and application. Your "mental stipulation" shows that you have the 'right attitude' in my book.