Capacitance Question

[ljcox]

I thought the question was theoretical. Energy cannot be destroyed.
0.5 C1 V1^2 = E =0.5 C2 V2^2
If C magically changes by moving the plates then V^2 must change.

LJCOX brings up an interesting point. Theory may not equal reality. It takes energy to move the plates.

What I said does not contravene the conservation of energy theorem.

When the plates are pulled apart, work (ie. energy input) must be done to separate them.

When the plates are allowed to move towards each other, energy is taken from the electric field and converted into some other type of energy, eg. heat. There is no gain or loss of total energy in either case.

 

[bananasiong]

The question is asked here:
https://www.csun.edu/~gsl05670/labs/cap_plate_sep_volt.htm
The answer is homework.

I have never had this kind of lab work

All the while I thought that the charge will remain the same as well as the energy when the distance of the parallel plate capacitor changes. ljcox has proven that I was wrong!!

 

[ronsimpson]

Len,

We agree that the voltage on a capacitor changes when the spacing changes. Examples are some kinds of microphones and a type if capacitive keyboard.

There is an example in textbooks where the plate spacing is changes (without adding or subtracting energy). Maybe this can only happen in theory. A 2:1 change in spacing causes the voltage to change by 0.707 or 1.414. That is based on E=E and Charge<>Charge.

You state charge = charge and E=E. The only way for that to happen is that it must take energy to move the plates. For Q=Q and E=E+E then the voltage change = 2:1.

I am glad you wrote it takes energy to move the plates. That explains why different people have different answers to the textbook question.

Thanks
Ron

 

[abbarue]

A strange phenomenon I witnessed

A while back I had assembled a power supply that put out about 500 volts.
I built a bridge rectifier out of four 1000 volt 6 amp diodes.
The amperage output of the transformer was only about 200mA, which is
well below the 6 amp rating of the the diodes.
Also the 500 volt output of the transformer is only half the 1000 volt rating, of the diodes.
I also used a 40 watt lightbulb in series between the transformer, and the rectifiers, just in case I shorted the leads by accident.
While I used two 470mfd capacitors connected in series across the output, all worked well.
I used this setup for hours on end without the diodes even getting warm.

I made 2 homemade capacitors out of Al foil and plastic tarp,
they measured about 300mfd each on the capacitance meter.
I then tried connecting one homemade capacitor in series with a 100 watt light bulb, across the bridge output.
As soon as I powered up the circuit, the capacitor began to shrink because
of the static charge buildup. and then all 4 diodes zapped out.
I say zapped out, because they were still cold to the touch when I checked them.
So this indicated the voltage of the diodes was exceeded and not the amperage.

I made another bridge out of 4 new diodes and tried using the other homemade cap.
Same thing happened again. When I checked the caps on the meter they
still measured about 300mfd so I knew they hadn't shorted out internally.

My understanding is when four 1000 volt diodes are connected as a
bridge rectifier, the bridge should handle 2000 volts because two diodes are
always connected in series at any given time in the circuit.

So how did the homemade capacitor build up a charge greater then 2000 volts, from a 500 volt input.
Plus the fact that the 100 watt light bulb in series with it would divide the voltage up some more.
This strange phenomenon is what lead me to believe that the voltage increases when the charged plates of a capacitor
are brought closer together. This didn't just happen once, but twice.
I have no other explanation for what happened to the diodes. Good thing I got them for 17 cents each.

I was so upset with those capacitors I ripped them apart and threw them in the garbage.
But I haven't been able to forget about this mystery since, that is why I started this thread.
I had also wondered if this had something to do with the casimir effect.

 

[ljcox]

Len,

We agree that the voltage on a capacitor changes when the spacing changes. Examples are some kinds of microphones and a type if capacitive keyboard.

There is an example in textbooks where the plate spacing is changes (without adding or subtracting energy). Maybe this can only happen in theory. In theory, we have to satisfy the conservation of energy theorm (or as Einstein pointed out conservation of Mass - Energy, but that issue is not relevant to this discussion) so perhaps the change kept the E constant but changed the Q to compensate. I don't see how you can keep both constant. A 2:1 change in spacing causes the voltage to change by 0.707 or 1.414. That is based on E=E and Charge<>Charge. What do you mean by "Charge<>Charge"?

You state charge = charge and E=E. The only way for that to happen is that it must take energy to move the plates. For Q=Q and E=E+E then the voltage change = 2:1. A voltage change of 2:1 would imply an E changre of 4:1 due to the square.

I am glad you wrote it takes energy to move the plates. That explains why different people have different answers to the textbook question.

Thanks
Ron

You're welcome Ron, it is an interesting question.

 

[ljcox]

A strange phenomenon I witnessed

A while back I had assembled a power supply that put out about 500 volts. I assume you mean 500 Volt RMS.
I built a bridge rectifier out of four 1000 volt 6 amp diodes.
The amperage output of the transformer was only about 200mA, which is
well below the 6 amp rating of the the diodes.
Also the 500 volt output of the transformer is only half the 1000 volt rating, of the diodes. I also used a 40 watt lightbulb in series between the transformer, and the rectifiers, just in case I shorted the leads by accident.
While I used two 470mfd capacitors connected in series across the output, all worked well.
I used this setup for hours on end without the diodes even getting warm.

I made 2 homemade capacitors out of Al foil and plastic tarp,
they measured about 300mfd each on the capacitance meter.
I then tried connecting one homemade capacitor in series with a 100 watt light bulb, across the bridge output.
As soon as I powered up the circuit, the capacitor began to shrink because
of the static charge buildup. and then all 4 diodes zapped out.
I say zapped out, because they were still cold to the touch when I checked them.
So this indicated the voltage of the diodes was exceeded and not the amperage.

I made another bridge out of 4 new diodes and tried using the other homemade cap.
Same thing happened again. When I checked the caps on the meter they
still measured about 300mfd so I knew they hadn't shorted out internally.

My understanding is when four 1000 volt diodes are connected as a
bridge rectifier, the bridge should handle 2000 volts because two diodes are
always connected in series at any given time in the circuit. Wrong!

So how did the homemade capacitor build up a charge greater then 2000 volts, from a 500 volt input.
Plus the fact that the 100 watt light bulb in series with it would divide the voltage up some more.
This strange phenomenon is what lead me to believe that the voltage increases when the charged plates of a capacitor
are brought closer together. This didn't just happen once, but twice.
I have no other explanation for what happened to the diodes. Good thing I got them for 17 cents each.

I was so upset with those capacitors I ripped them apart and threw them in the garbage.
But I haven't been able to forget about this mystery since, that is why I started this thread.
I had also wondered if this had something to do with the casimir effect.

I don't know why the home made caps caused the failure and not the original caps.

I don't know what the casimir effect is.

However, as in red above there are some errors in your reasoning.

If you draw out the rectifier bridge and consider the instant when the sine wave input is at its maximum. the voltage across 2 of the diodes is 500 * sqrt 2 = 707 VOLT.

The diodes are never in series as you claim.

EDITED to correct an error

 

[abbarue]

Whenever I trace the current through a bridge rectifier it always travels
through 2 diodes. One on the way into the bridge then into the load.
and another as the current leaves the load into the bridge and then out,
to the transformer.
The other 2 diodes are acting as gates blocking the current from going the
other way.
So I always see the current traveling through 2 diodes at a time, when I
follow the circuit.
Where can I find an explanation of what you are saying.
This is totally new to me, and goes against my logic.

Here is a diagram showing voltage flow through a bridge rectifier.
**broken link removed**
It clearly shows the voltage moving through 2 diodes at a time in series.
In dia. A it moves through CR2 , LOAD, CR4 in series
in dia. B it moves through CR1 , LOAD, CR3 in series

So you can see why you have me confused.

 

[ljcox]

Yes, the current flows through 2 diodes. What I thought you meant is that 2 diodes were in series across the supply voltage.

The diagram shows current flow, not voltage flow. There is no such thing a voltage flow.

Also, the diagram uses electron flow (- to +), I use conventional current flow (+ to -).

Supporters of electron flow argue that electrons flow from - to + which is true, but this assumes that electrons are the only charge carriers.

For example, protons and positrons carry a positive charge as do holes in semiconductors & positive ions.

So there is nothing special about electrons except that they are the most common charge carrier.

Diodes and transistors are drawn with their arrows pointing in the conventional flow direction.

I find it better to use conventional flow as it is easier to write the nodal equations when analysing circuits.

 

[Nigel Goodwin]

Your diodes were too low a voltage, 1000V is too low for a 500V supply - the 1000V rating is PIV (Peak Inverse Volts), which is what a 500V supply will place across the rectifiers.

 

[abbarue]

I know that conventional electronics uses current flow, I would have drawn
the circuit with current flow from pos. to neg. this was just the first diagram
I found on google to use as an example.

So do I understand this right?
500 volts AC goes from -500V to +500V.
which is a difference of 1000V total.
But the rms voltage would be even higher.
Depending on whether I'm using 500V rms or not.
I used a variac on the input and adjusted it until
my digital meter read 500V DC on the output of the bridge.

 

[ljcox]

I know that conventional electronics uses current flow, I would have drawn the circuit with current flow from pos. to neg. this was just the first diagram I found on google to use as an example.

So do I understand this right?
500 volts AC goes from -500V to +500V. No, 500 V AC is the RMS. The peak is 500 * Sqrt 2 = 707 Volt. So it goes from 707 to - 707.
which is a difference of 1000V total.
But the rms voltage would be even higher. Wrong, see above.
Depending on whether I'm using 500V rms or not.
I used a variac on the input and adjusted it until
my digital meter read 500V DC on the output of the bridge.

Does your digital meter read true RMS? If not, it probably measures the average. The output of the bridge is a rectified sine wave, so the average would be (and this is from memory) 0.637 of the peak. 500/0.637 = 785 Volt, ie. the peak is 785 V.

The diodes should have been able to withstand this voltage given that they are rated at 1000 PIV.

So I suspect that the insulator of your home made cap may broken down once the voltage reached a particular level thus causing a large current through the diodes hence destroying them.

You seem to be confused by a few issues so I've drawn the attachment to try to explain.