Reversing the polarity of capacitor?

[al_201314]

Hi everyone,

I don't know if this could be understood here but I don't have any way of putting the diagram here so I'll try my best to illustrate it as clear as possible I hope it's understandable and please bear with me!

If I connect to capacitors, one 1uF and the other 2uF in parallel (no emf drawn in, just a closed box with the 1uF one on the left side and the 2uF one on the right side of the box) with a P.d of 6V across them. In this case, the positive plates of the 2 capacitors are on the same side (top) on the circuit.

I can find the total charge in them to be 18uC.

But why is it that, when I reverse the plates on the 2uF capacitor, the total charge becomes 6uC? (12uC - 6uC)?

I can't understand the concept even after some thinking through. Appreciate any help.

Many thanks in advance.

 

[Tide]

HINT: What would have happened if you had done exactly the same thing with two identical capacitors?

 

[al_201314]

The total charges would remain the same?

 

[al_201314]

On further thoughts I thought that there would be a charge redistribution but I would think that the total charges should still remain the same?

 

[berkeman]

Don't think of the 18uC as a net charge. The net charge of a capacitor is zero. There is a *separation* of charge in a capacitor. So in your original question, there is a separation of how much charge on the 1uF cap and how much separation of charge on the 2uF capacitor? When you reverse one, then yes, there is a redistribution of the electrons, and some of the separation of charge is canceled out. Think about Tide's hint -- what would be the separation of charge if both caps were 1uF? Then think about what is different when one cap is 2uF instead of 1uF.

 

[al_201314]

Here's what I've been able to come up with.

Refering to the values in my question, I worked out the separation of 6uQ in the 1uF capacitor and 12uQ in the 2uF capacitor.

So, when I reverse the plates of the 2 uF capacitor after both of them have been charged to the above values by 6V, there is a redistribution of charges. But what I don't understand is that how this distribution goes about. The positive plate of the 1uF will be reduced in terms of number of positive charges by the negative plate of the 2uF capacitor? Or does it happen 2 ways, ie, positive and negative charges move?

Apologise for being slow here couldn't really figure more here.

Thanks again.

 

[berkeman]

No apologies needed. Think of the problem in this way:

Initial configuration:

 +6     +12
----   -----
----   -----
 -6     -12

The 1uF cap has a separation of 6C in one direction. That means that 6C of electrons were pumped from the + to the - side of the capacitor, and that separation is static. Same deal with the 2uF cap, 12C of electrons were pumped from the + plate to the - plate and left there. So the - plates have an excess of electrons, and the + plates have a lack of electrons. The net charge of each capacitor is still zero, though. Since the capacitors each have the same voltage across them, no charges flow if they are connected in parallel with the same polarity shown above.

But now disconnect the parallel caps, and reverse the left cap. Just before you re-connect them in parallel, they will look like this:

 -6     +12
----   -----
----   -----
 +6     -12

Now when you connect the two caps back together, the -6C of electrons on the left cap will redistribute between the two capacitors, and that -6C of electrons replaces 6C of the electrons that were pumped off the left cap's + plate before. The net charge left on the top plates is +6C, and a similar thing happens on the bottom plates, leaving -6C total on the bottom two plates. That leaves the overall 3uF of capacitance with a separation of charge of 6C, with the + side still the top and the - side on the bottom. Quiz question -- what is the final voltage?

 

[al_201314]

Thanks a lot berkeman. I think I'm almost there. Just a bit more on the concept.

About the 6V across the capacitors, this voltage must come from a battery or from a charging device, right? In the second part of the question when we reverse the plates of one of the capacitor, I assume that the battery is no longer there. Am I right?

Quiz question -- what is the final voltage?

6=3(V) where V= 2V. I'm pretty sure I got it I hope I have!

Thank you!

 

[berkeman]

Thanks a lot berkeman. I think I'm almost there. Just a bit more on the concept.

About the 6V across the capacitors, this voltage must come from a battery or from a charging device, right? In the second part of the question when we reverse the plates of one of the capacitor, I assume that the battery is no longer there. Am I right?

That's how I interpreted the question also.

6=3(V) where V= 2V. I'm pretty sure I got it I hope I have!

I don't understand your answer. Before the first capacitor is reversed, the caps are both at 6V, right? What is the voltage after you reverse the 1uF cap?


EDIT -- Oh, I understand what you wrote now. The multilple uses of the "V" confused me. Yes, the final voltage is 2V. Good job.