calculating the measure of electric field in capacitor with Gauss's law?

[skaddi]

Hiiiiiiiiiii...........
I have a problem with calculating the measure of electric field in capacitor with Gauss's law.
you know, in this calculating we use the electric charge of just one surface but why we ignore the other one and its electric charge?:-?
thanks for your attention.....

 

[Ibkay]

Re: electric field of capacitor

Gauss's law is used to relate the total or net charge enclosed within a surface to the electric field emanating or exiting from that surface (normal to the surface, and all pointing outwards from the surface). The key to using Gauss's law is to clearly and unambiguously define the surface in question as well as properly taking into consideration the direction of the electric field.

Now for a parallel plate capacitor. If we define a surface that encloses BOTH plates, then the net or total charge enclosed is zero (+q and -q from each plate) and by Gauss's law, this means that OUTSIDE of the enclosed capacitor, we expect zero net field, which corresponds to our observation in practice (this also includes fringing effects of the fields because the fringing fields entering the surface are equal to those exiting the surface. Therefore, the net field exiting the surface is zero).

On the other hand, if we are interested in the electric field within the capacitor (i.e. within the dielectric of the capacitor), then the surface we will draw can ONLY cover ONE of the plates. If you take some time to imagine this, you would agree that you cannot draw a surface that cuts through the dielectric and at the same time covers the two plates. For your Gaussian surface to cut through the dielectric, it can only envelope one plate.

This is the reason why applying Gauss's law to the parallel plate problem always involves only one of the two plates, and the net charge enclosed is therefore +q or -q (either one works but you must correct for the direction of the electric field which by convention should be pointing outside of the Gaussian surface).

Hope this helps!

 

[skaddi]

Re: electric field of capacitor

Thanks for your attention sir..!
but you know, you define my question by some facts but I want to know more about the reason of this by some intuition ..
Again Thanks...

 

[Ibkay]

Re: electric field of capacitor

Okay, let me try to re-phrase.

The electric field of a typical parallel plate capacitor exists WITHIN THE DIELECTRIC. It originates on the positive charge on the first plate (+q), travels through the dielectric and terminates on the negative charge on the opposite plate (-q).

Now, the objective of our analysis is to apply Gauss's law to evaluate the magnitude of this electric field WITHIN THE DIELECTRIC of the capacitor.

If you take a close look at the equation form of Gauss's law, you would observe that it equates enclosed charge on one side to the electric field surface integral on the other side (sorry it's not easy writing integrals and equations here!).

The key point to note from Gauss's law is that if you know the net charge enclosed within a volume, then you can determine the electric field emanating from the surface of that enclosing volume, which is fairly intuitive since electric fields exist as a consequence of charged particles. In most cases, we use the symmetry of the problem to simplify the surface integral and extract the electric field itself.

Now, to take advantage of Gauss's law for the parallel plate capacitor, we have to define our enclosing surface in such a way that the surface will PASS THROUGH THE DIELECTRIC, while at the same time enclosing a KNOWN quantity of charge. By doing this, we already know one side of the equation which is the enclosed charge (+q or -q depending on the plate we decide to enclose). The other side of the equation which is the surface integral of the electric field will therefore give us the electric field WITHIN THE DIELECTRIC because we deliberately defined our surface to pass through the dielectric, and the electric field emanating from this surface must consequently be equal to the electric field within the dielectric and thus the capacitor.

As described in the earlier post, you can only take one plate if you also choose your enclosing surface to pass through the dielectric. If you enclose the two plates of the capacitor in your Gaussian surface, the surface will not pass through the dielectric and we will therefore not solve our problem which is to evaluate the electric field within the dielectric i.e. of the capacitor.

Hope this helps, it's easier described with images, but most texts illustrate it well...

 

[skaddi]

Re: electric field of capacitor

now I can understand your purpose somehow...
thanks a lot...:-D

 

[John Steave]

Re: electric field of capacitor

"I would use sensors connected to a computer to measure the voltage and i could easily measure the plate separation distance.

Would the original electric field strength be measured using the maximum voltage reached when the capacitor is attached to the battery? Or would i have to remove the battery before measuring this voltage?

When the dielectric is inserted I think that i have to remove the battery to measure the voltage difference, otherwise will the voltage will just be the same? If this is true will i get accurate readings? I mean the voltage won't decay unless i am actually discharging the capacitor right?

I'm just not sure if this will work especially since i was thinking of making the capacitor myself
I’m not recommending but perhaps you could check out this for your better understands about capacitors.
**broken link removed**
https://www.physicsforums.com/showthread.php?t=96099
https://www.allaboutcircuits.com/vol_1/chpt_13/1.html
Thanks!!!!!