Capacitor charging itself?

[nati levia]

Heya.
I'm a student at high school, I'm over most of the school (at knowledge) and the teacher gave me a test.
The test is easy but there was one question that i don't know (hats the question that summarizes the test),
The question is: How do you recharge a capacitor by itself?.

I must pass the test in 100% so i can go to the collage that i always wanted to go.

 

[Bluejets]

I'd suggest you answer using the knowledge you have gained.

 

[Harald Kapp]

How do you recharge a capacitor by itself?

Is that truly the original question? No other information given? To what level? in which time? Available tools or material? ...

 

[nati levia]

Is that truly the original question? No other information given? To what level? in which time? Available tools or material? ...

I'm reading it right now...

It says: A 24v 3300μf capacitor witch is charged to 9v needs to be charged to 24v without any power source

 

[Hopup]

Well if there is no source of energy available its practically impossible.

 

[hevans1944]

I'm reading it right now...

It says: A 24v 3300μf capacitor witch is charged to 9v needs to be charged to 24v without any power source

Is English your native language? WTF is a "capacitor witch"? Some sort of paranormal capacitor? How do you (or your teacher) expect to add energy to any capacitor without using a power source? Is this an advanced course in practical magic, or are you actually studying electronics? Please photograph and post a picture of the actual test question (and its context) since you are "reading it right now..."

 

[Audioguru]

On another thread, this student found out that this same teacher made a mistake about capacitors.
Then maybe this teacher has a screwed up idea about charging a capacitor?

 

[ChosunOne]

Okay, took me awhile, but I think I've got it. Hold the capacitor by its negative lead and walk across a thick carpet in a low-humidiy room, then touch the positive capacitor lead to the doorknob. Touching it to one of the screws holding on the light switch panel is even better.
Repeat until voltage reads 24V. [/satire]

@nati levia: In case I'm being too subtle, this is a facetious answer, not a real one. Don't use it for the test! If it turns out to be the answer your teacher thinks is correct, then find a local electronics expert and ask him/her to back you up with your school board.

EDIT: Afterthought: If your teacher thinks you can do it the way Ben Franklin did it, s/he's not only ignorant, s/he's dangerous. And weather that produces enough static to charge a capacitor is a power source.

 

[Harald Kapp]

All wrong

Who says the capacitor is of the commercially available type so well known to all of us?
Let's step back and look at the equation for a plate capacitor:
C = (epsilon*A) /d where A is the area of one plate and d is the distance, epsilon being the permittivity of the medium (e.g. air).
We also have the equation C = Q/V where Q is the charge on the plates (on one plate as the othe rplate has the same charge with inverse sign) and V is the voltage.

We can combine both equations to Q/V = (epsilon*A)/d

Now let's assume we have a primitive capacitor made from two conductive (metal) plates at a distance d=d1 (whatever the value of d1 is doesn't matter). Let's further assume the size of the plates is constant (A=const), as well as the permittivity of the medium between the plates (epsilon = const). As there is no power source, there will be noc change in cahrge, therefore Q=const, too.
We are left with an equation with two variables: V and d. We know that V shall change from 9 V to 24 V, that leaves us with a corresponding change in d.
In plain Englisch: when the distance between the plates of a capacitor is changed, the voltae changes, too, all other parameters being the same.
I'll leave the plain math to the op.

Of course, this answer is only particularly correct, as are all other proposed solutions, as one will need a kind of power source to change the distance between the plates (or to walk across a floor or to...).
The implicit assumption here is that by "no power source" no "electrical power source was meant.

 

[hevans1944]

@Harald Kapp: That "solution" for a 24 V, 3300 μF capacitor is so impractical I didn't even mention it.

But just supposing you could construct such a capacitor with a very thin removable (plastic?) dielectric, you could "charge" it up (or, more correctly, energize it) with 9 V DC and then slide the dielectric out until the capacitance decreased to 1237.5 μF, thereby raising the voltage to 24 V DC. That, to me, seems easier than changing the separation between the plates, although either method would work... and, as you mentioned, require work (expend energy) to perform. TANSTAAFL.

 

[BobK]

A 24v 3300μf capacitor witch is charged to 9v needs to be charged to 24v

Harald's method does not solve the problem as stated. After changing d and increasing V, is it no longer a 3300uF capacitor, it is a 1235.7uF capacitor.

Bob

 

[hevans1944]

...it is a 1235.7uF capacitor.

Good point, Bob. But actually it is a 1237.5 μF NOT 1235.7 μF: (3300) (9) / (24) = (3300) (0.375) = 1237.5.

 

[Harald Kapp]

That "solution" for a 24 V, 3300 μF capacitor is so impractical I didn't even mention it

Totally impractical, but theoretically correct.

is it no longer a 3300uF capacitor, it is a 1235.7uF capacitor.

Nobody said it should be.

Well if there is no source of energy available its practically impossible.

Not only practically, but theoretically, specially if the capacity is supposed to remain the same during the process (which is possibly to be assumed but not stated explicitly). The energy stored within a capacitor is 1/2 * C * V², thus by increasing the voltage by whatever method from 9 V to 24 V the energy stored increases by a factor of 7.1. As this energy has to come from somewhere some source of energy is required.

 

[BobK]

Good point, Bob. But actually it is a 1237.5 μF NOT 1235.7 μF: (3300) (9) / (24) = (3300) (0.375) = 1237.5.

Dyslexia strikes again!

Bob

 

[duke37]

A capacitor with a solid dielectric has most of the energy stored in the dielectric. If the dielectric is slid out from between the plates, how does the energy get free from the dielectric? Is the energy retained. How are electret microphones made?

 

[Arouse1973]

Electric field stored in the dielectric. Please explain more?

 

[Hopup]

You could use ionized gas as plate of the capacitor this way it could be relatively easy to do.

 

[davenn]

A capacitor with a solid dielectric has most of the energy stored in the dielectric.

not quite correct .... the energy is stored in the electric field BETWEEN the plates, REGARDLESS of what the (non conductive) dielectric is air, or solid
The electric field will just happen to be distributed within the dielectric ( cant be any other way)

Electric field stored in the dielectric. Please explain more?

yeah he didn't quite get that right

You could use ionized gas as plate of the capacitor this way it could be relatively easy to do.

I think that would be impractical, as it would give the capacitor an initial positive charge on that plate that would have to be overcome during use

 

[Bluejets]

With Harald's suggestion in mind, there migh be a connection between the original teachers request and the fact that there is a pd in the atmosphere.

 

[Harald Kapp]

there is a pd in the atmosphere.

What is a pd, please?