Inserting a dielectric onto a capacitor and finding capacitance as a fxn of x

[casanova2528]

I'm not too fond of writing the math down here, so I need only those who know their circuits to reply.

Battery connection is maintained with one capacitor.
du goes up as you insert a dielectric
dc goes up
dq goes up

However, foe those who've done the thinking and solution...we find that dE of the system is negative!

The dielectric is basically sucked in if there is no external force opposing the attraction! That is very counter intuitive because we would assume that the increase in du as the dielectric is inserted would come only from the external force that would oppose the insertion!

How can you explain this truth about the sucking in of the dielectric? After all, charge flows from the battery onto the plates as the dielectric is inserted.

The mathematical answer lies in this part of the step to find the -dE answer.

dq (v) = dU battery. As opposed to using

.5 dq (v) = du battery.

I'm not sure how to explain this. Help me !

 

[casanova2528]

Here's another situation...do the opposite. Taking the dielectric out while battery is connected will decrease potential energy, charge on the plates, and Capacitance. However, you need to pull out the dielectric even though you decrease the potential energy. If you do the math...you see that dE of the system increases which implies that external work must be done to pull it out. How can I explain this better?

 

[casanova2528]

any ideas?

 

[berkeman]

The extra energy comes from the battery, minus some efficiency loss in the wiring. To pull in the dielectric, the battery supplies current to increase the Q to match the higher C

Q = CV

The dielectric is pulled in for the same reason that a magnetic slug is pulled into a solenoid coil that is energized with a current. The energy to pull in the slug or dielectric piece comes from the power source.

 

[casanova2528]

Why do we use Du = V dQ1 to find the -W of battery= V dQ1

INSTEAD OF using

dU = .5 dQ1 V to find the -W of battery = .5 dQ1 V ??

 

[casanova2528]

any thoughts?

 

[casanova2528]

what I'm trying to ask is what is the difference between (.5q squared)/c = (.5c vsquared) = (.5 qv)

And

(v dq) ?

 

[casanova2528]

when the battery puts charges on the plate, the battery is charging up the capacitor and changing the U by (.5)(c)(v squared). Why is it necessary to conclude that the charge from battery is going across a potential difference of v? Why can't I conclude the charge from battery will charge up the plate according to (q)/(c) = v?

 

[casanova2528]

I don't see where the q from battery goes thru a potential difference of v foe a capacitor. For a resistor, it's evident... But not for a dielectric in a capacitor.

Help!

 

[casanova2528]

Berkeman??

 

[MATLABdude]

Potential doesn't actually require a charge to move anywhere; you just make use of a fictitious differential charge to figure out what the energy (potential) between these two plates is. Where you to actually move charges from one plate to the other (say, via a resistor hooked in parallel, with the original battery used to charge up the capacitor removed), the voltage would drop.

As an analogy, think about a heavy block of metal on a shelf. The block has a certain amount of gravitational potential energy (relative to the ground), and were the shelf not there, it would acquire most (drag forces, etc.) of this and turn it into kinetic energy before it hits the ground.

Same deal with the capacitor. The electron can take a non-straight path (through the wire) from one plate to the other (but it'd lose the energy in doing so--just as if you were to take the block off the shelf and place it on the ground).