Charged capacitors connected in series

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I considered N=2 .

Two similar charged capacitors are joined in series i.e positive plate of one is joined to negative of the other .

If I consider that there is no movement of charge since both the capacitors are similar and having equal charge I get option 2) . Is that correct ?

I don't have the answer key .

But I am just wondering why both the capacitors don't get discharged .Since positive plate of one is connected to negative plate of the second , shouldn't the capacitors get completely discharged ?

 

[gneill]

If I consider that there is no movement of charge since both the capacitors are similar and having equal charge I get option 2) . Is that correct ?

You should be able to determine whether or not this is the case this mathematically. What are the relevant equations?

But I am just wondering why both the capacitors don't get discharged .Since positive plate of one is connected to negative plate of the second , shouldn't the capacitors get completely discharged ?

Without a completed circuit charge won't move; the charges on the plates will be held there by the close proximity of the opposite charge on the other plate. Remember, there's essentially no electric field outside the region between the plates, and the individual capacitors are overall neutrally charged, so there's no potential difference to draw the charges away from the plates.

 

[Jahnavi]

You should be able to determine whether or not this is the case this mathematically. What are the relevant equations?

Yes . I also want to derive it mathematically . But I don't understand how maths is used in this problem . Q=CV is the only relevant equation I can think .

For N=2 , two similar capacitors are equally charged and then connected with each other such that positive of one is connected with negative of the other . A complete circuit is formed .

Without a completed circuit charge won't move;

I think a complete circuit is formed .

Isn't there a potential difference between the positive plate of first capacitor and negative plate of second capacitor ? This will drive the charge from positive plate to negative plate . This way both get discharged and become neutral .

 

[gneill]

Yes . I also want to derive it mathematically . But I don't understand how maths is used in this problem . Q=CV is the only relevant equation I can think .

Don't you have an equation for the energy stored in a capacitor?

For N=2 , two similar capacitors are equally charged and then connected with each other such that positive of one is connected with negative of the other . A complete circuit is formed .

No! A circuit, as the word suggests, is a closed path (like a circle). You should be able to trace a path from some starting location and without retracing any portion, arrive back at the starting point.

Isn't there a potential difference between the positive plate of first capacitor and negative plate of second capacitor ? This will drive the charge from positive plate to negative plate . This way both get discharged and become neutral .

Nope. No field outside of the regions between plates and the capacitors are overall neutral since they contain equal amounts of positive and negative charge.

 

[Jahnavi]

No! A circuit, as the word suggests, is a closed path (like a circle). You should be able to trace a path from some starting location and without retracing any portion, arrive back at the starting point.

I think we are interpreting the " Now if they are separated and joined in series " differently .

You are interpreting this as negative plate of first capacitor joined to positive plate of second . Negative plate of second capacitor is free .Positive plate of first is free .I am interpreting this as negative plate of first capacitor joined to positive plate of second AND negative plate of second joined back to positive plate of the first .

Is that wrong ?

 

[gneill]

I am interpreting this as negative plate of first capacitor joined to positive plate of second AND negative plate of second joined back to positive plate of the first .

Is that wrong ?

I think that, with the given list of answers to select from, the circuit cannot be closed. The two endmost capacitors must each have a free lead. If there was a closed circuit there would be no potential difference (where would you measure it?) and all the charges would be able to recombine and the net energy remaining would be zero.

 

[Jahnavi]

I think that, with the given list of answers to select from, the circuit cannot be closed. The two endmost capacitors must each have a free lead.

OK .

Don't you have an equation for the energy stored in a capacitor?

Total energy stored in the capacitors = N(1/2)CV²

After rearrangement , total energy stored = (1/2)(C/N)(NV)²

Both are equal . So total energy remains same but potential difference across the combination becomes NV .

This gives Option 2) .

Right ?

 

[gneill]

Sounds good to me

 

[Jahnavi]

the charges on the plates will be held there by the close proximity of the opposite charge on the other plate. Remember, there's essentially no electric field outside the region between the plates, and the individual capacitors are overall neutrally charged, so there's no potential difference to draw the charges away from the plates.

Why does the above logic fails in the below case (with the way I was interpreting ) ?

If there was a closed circuit there would be no potential difference (where would you measure it?) and all the charges would be able to recombine and the net energy remaining would be zero.

What I would like to understand is that if the capacitors are joined in series and a closed circuit is formed , why does charges recombine ? Why are charges on a plate not able to keep opposite charge on the other plate of the same capacitor attracted ? There is still no electric field outside the capacitor .

 

[gneill]

Suppose you have a string of charged capacitors connected in series and in a circular fashion with the final connection being made by a (currently open) switch. The potential difference across the open switch contacts, being the sum of the potential differences of all the capacitors, would exceed the potential difference between the plates of the two end capacitors. Hence there is a stronger field across the switch than there is between the plates of those capacitors.

Closing the switch, the charges on the plates will "see" a conductive path to follow across a significant (if short lived) potential difference. As the charges leave the the connected plates and start to recombine, this loosens the grip on the charges held on the opposite plates in the capacitors. In order to keep the facing plates at equal and opposite charges (and the capacitors overall neutral), the excess charges will migrate out of the capacitor leads and be driven towards the plate of the next capacitor in line where they can combine with the charges there. The process ripples down the line, driving a current through the circuit.

 

[Jahnavi]

Thanks