Let's say we take a ceramic capacitor and charge it to 5V 50pF. I leave the cap 5V biased for the remainder of the experiment.

I hook one pole of the capacitor to an oscilloscope and begin measuring the potential/signal at that point.

Then I drop a heavy, flat headed rod from height directly impacting and smashing the capacitor.

Question is, what shows on the oscilloscope tracing? What timeframe and what potential levels? What waveform and spectrum?

Assume as the ultimate result of the collision the plates will short and discharge after some relatively short period.

 

There are too many variables at work here, what kind of answer are you expecting?

 

@cbecket13 --- As I understand your question, you are curious as to the behaviour of pre-existing' charge with increasing capacitance???

 

He may also be thinking that ceramic might show a piezo electric effect...

 

Also...need to explain your Oscope connection with the cap better A cap and an Oscope hooked to one leg is vague and as presented would not be a connection which would give any indication of a signal. What is your ground reference?

 

The time constant for discharge of a 50pF cap into the ~1meg input impedance of a scope is only 50uS. You'll need to be quick with your experiment .

 

As Hypatia's comment suggests, if the amount of energy is fixed and then you increase the capacitance the voltage will change.

In other words (symbols):

W = 1/2 x C x V^2

At the beginning of the event, W is constant and C increases. What happens to V?

 

sounds like someone found out about the Kasimir effect. although a 50 pf cap is a bit small.

 

So what's the point of hitting a capacitor with a hammer? Does it make your radio work better?

 

I was looking for the effect of the rapidly approaching plates might have on the charge or voltage, if any. In other words, the effects of accelerating charged plates toward each other. Ceramic, , piezo etc. was not what I was thinking of. I thought of the oscilloscope was sampling voltage at a very high rate.

 

As Hypatia's comment suggests, if the amount of energy is fixed and then you increase the capacitance the voltage will change.

In other words (symbols):

W = 1/2 x C x V^2

At the beginning of the event, W is constant and C increases. What happens to V?

Voltage drops. Linearly?

 

So what's the point of hitting a capacitor with a hammer? Does it make your radio work better?

You bring to mind the arc-water explosion effect. So what, you say. Big bang, who cares. But it gave us electrohydraulic forming.

 

Also...need to explain your Oscope connection with the cap better A cap and an Oscope hooked to one leg is vague and as presented would not be a connection which would give any indication of a signal. What is your ground reference?

Actually, my "oscilloscope" is a computer sound card line input and Audacity software. I can't really get ns sampling, unfortunately.

 

There are too many variables at work here, what kind of answer are you expecting?

Ok let's work on that. Put two circular conducting plates aligned in cylindrical fashion in a vacuum. Put a C of +/- charge on each of them. Set up to measure charge on the plates, electric field, magnetic field, electromagnetic emissions inside and outside the system. Put them a kilometer apart. Accelerate one plate @ 9.8 m/s^2 toward the other. (Ok, use the arc-water explosion effect to accelerate the plate. (That was a joke.)j What change if any do we see on our measurement equipment.

 

I don't know about arc-water explosions either, but I do have a feel for area/distance = charge/volts.
Reduce distance and voltage decreases.
No exponents in there, so the answer seems linear to me.

 

none till you get real close together, disance works on square law for most of your measurements.

 

Start off small and slow. Find what ever you can for plates. One square inch or 10 square inches.

Line them up and put a charge on them. Connect scope, vary distance.

Vary distance at various angles.

Put a charge on one plate only.

Vary neutral plate and watch scope.

Now try a large tuning capacitor.

Also, put a charge on a blown up balloon. Place plate near balloon. Slowly release air from balloon and watch plate.

Get a feel for charge, distance and area.

To see how speed effects these things, you will need better and faster equipment.

The potential between two stationary charges is dependent on the amount of charge and the distance between them.

But when the distance between the two charges is changing................the RATE of this distance change(relative velocity).......Adds to the potential.

This velocity factor can be dominant, although it’s hard to notice in our scale.

In our scale, we are not use to those speeds. But on the fundamental scale, light-speed and a little under is most common.

If you use a manufactured capacitor, the collision will cause an UN-symmetrical plate distance change.

And I suggest buying or building an electroscope. It’s a great tool.

 

Let's say we take a ceramic capacitor and charge it to 5V 50pF. I leave the cap 5V biased for the remainder of the experiment.

I hook one pole of the capacitor to an oscilloscope and begin measuring the potential/signal at that point.

Then I drop a heavy, flat headed rod from height directly impacting and smashing the capacitor.

Question is, what shows on the oscilloscope tracing? What timeframe and what potential levels? What waveform and spectrum?

Assume as the ultimate result of the collision the plates will short and discharge after some relatively short period.

I THINK NO TRACE WIL BE SHOWN UP AND NO LEVELS AT ALL.

 

Let's say we take a ceramic capacitor and charge it to 5V 50pF. I leave the cap 5V biased for the remainder of the experiment.

I hook one pole of the capacitor to an oscilloscope and begin measuring the potential/signal at that point.

Then I drop a heavy, flat headed rod from height directly impacting and smashing the capacitor.

Question is, what shows on the oscilloscope tracing? What timeframe and what potential levels? What waveform and spectrum?

Assume as the ultimate result of the collision the plates will short and discharge after some relatively short period.

A ceramic capacitor is brittle and will just shatter, so only a few fragments will remain attached to the lead wires - and only a fraction of the total charge will be evident.

Vaguely remember reading something about the US military developing an EMP weapon based on imploding an inductor with a shaped charge while simultaneously hitting it with a capacitor discharge.

 

I THINK NO TRACE WIL BE SHOWN UP AND NO LEVELS AT ALL.

@JUNELER: Please don't use all capitals when writing a post -- it's seen by the online community as shouting. As a result, it reduces the effectiveness of your post. Instead, use all caps for isolated words or phrases that you want to specifically emphasize.