Conducter in a Parallel Capacitor

[Trenthan]

So just to check i understand what i just read. from the thread "dielectric constant"


By putting a conductor e.g copper in a electric field we polarize it, and thus induce an electric field opposing the orignal field of the capacitor without it

........Copper in mid
Positve Plate...+.l...l.-...+.l...l.-..Negative plate
......+.l...l.-...+.l...l.-
......+.l...l.-...+.l...l.-
......+.l...l.-...+.l...l.-
......+.l...l.-...+.l...l.-


Electric field...-------> ---------> --------> (field goes through conductor in middle)
Induced field.....<--------

If the size of the induced field matched the size of the field going through the copper in the middle they would cancel and effecitvely the overall field would just be 0 in the conductor placed in the middle. So the resultant field would just be --------> ---------> effectively?



Cheers Trent
Sorry if I've broken any rules first post

**edit** had to add dots due to it colapsing the spaces and it was just a bunch of lines and letters sry its made it a lot harder to understand

 

[Defennder]

Yes you're right. The E-field in a conductor at electrostatic equilibrium is zero. Although I sometimes wonder what would happen if we applied an E-field whose magnitude exceeds all that of the electrons of the conductor combined. Clearly the E-field in the conductor would be non-zero, no?

 

[Trenthan]

you raise a very interesting point, ill have to think about that one hmmm...

 

[Trenthan]

while on the topic of a conductor in the field wouldn't it be correct to say the capacitance would be

If we use C=(e₀*A)/w

e₀=8.85*10^-12,

"w" is width from positive plate to negative plate, therefore since the field cancels in the middle couldn't we just substitute (w/2) into C=(e₀*A)/w, **assuming that the width of the copper inside the field is half "w/2", therefore "w/4" lies on both sides of the copper inbetween the plates (lets assume it air for simplicity).

The copper inside the field doesn't contribute towards capactiance, since there is no electric field therefore no potential difference inside the copper.

So in this case capactiance would in fact increase by adding in the copper, since "w" has decreased thus capacitance is higher. However, this is where I am confussed, doesn't it only increase if it is a insulator in the capacitor, and also since there is nolonger a potential in the copper wouldn't it infact decrease? goes with the idea a conductor lowers capacitance

 

[Trenthan]

Sorry about posting a post after a post but it wouldn't let me edit anything in for some reason

Just curious a conductor in between two parallel plates let's say the plate width was 6mm, and a 2mm conductor is inserted**

Therefore the capacitance C=(e₀*A)/d

Lets say A is 1 to make it easier
Therefore C = (e₀*1)/.004, in every case and it doesn't matter if the conducter is in the middle, touching the left plate, or the right plate does it?. Capacitance is the same in all conditions

Got this from this website hopefully it is correct if some1 could just double check that its realiable it seems reliable from what I've read from it http://dev.physicslab.org/Document....ctrostatics_DielectricsBeyondFundamentals.xml

Cheers TRent