Uncharged capacitor connected to a charged capacitor

In summary, the conversation discusses whether or not charges will redistribute when an uncharged capacitor is connected to a charged capacitor. The textbook states that the positive charge on one face of the capacitor will hold the negative charge in place, and therefore no charge redistribution will occur. However, if the two free ends of the setup are connected, a potential difference will be established across the connecting wire, causing charges to flow and potentially leading to charge redistribution.
  • #1
Titan97
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Homework Statement


An uncharged capacitor is connected to a charged capacitor as shown:
Capture.PNG

The ends are left as it is.
Will the charges redistibute?

Homework Equations


None

The Attempt at a Solution


I think it won't.

Let the first capacitor be charged with charge Q. Before closing the switch, the charges are like:
Capture.PNG

If ##x## charge flows from the negative plate of first capacitor to the second capacitor, then a same ##x## has to flow from the first plate as well since opposite faces of a capacitor should have charges of same magnitude. So no charge transfer is possible.

This, however, is the reason given in my textbook:
" The positive charge on one face of capacitor will hold the negative charge in place. "

Is it correct? Because if I were to connect the two free ends to each other, then also no charge should redistribute. (I am also thinking about the problem 3.2 from Prucell and Morin).
 
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  • #2

See the case 1.
 
  • #3
@gracy that's a totally different case. The wire does not form a loop.
 
  • #4
Titan97 said:
This, however, is the reason given in my textbook:
" The positive charge on one face of capacitor will hold the negative charge in place. "

Is it correct? Because if I were to connect the two free ends to each other, then also no charge should redistribute. (I am also thinking about the problem 3.2 from Prucell and Morin).

The textbook is correct. Before the switch is closed the charges are held on the inside surfaces of the capacitor plates of the left-hand side capacitor by their mutual attraction. The leads of the charged capacitor will be at some potentials with respect to the zero potential reference at infinity. When the switch is closed there is no external potential difference added that can overcome this attraction. What does happen, however, is that the entire right hand capacitor assumes the potential of the first capacitor's right hand plate thanks to sharing a conductive path. Some very negligible charge displacement may occur in the conductors as a result, about the same as you would expect if the second capacitor were just a short length of wire instead of a capacitor.
upload_2016-4-14_9-44-4.png


If you close the loop by connecting the outer leads of the setup then a potential difference is established across the connecting wire and current will flow in that wire (very briefly if the wire's resistance is very small. Instantaneously if we're talking about an ideal conductor).
upload_2016-4-14_9-55-59.png

Current will flow until this potential difference across the wire is eliminated, that is, when the potentials across the capacitors reach an equilibrium.
 
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  • #5
@gneill is my reasoning correct?
 
  • #6
Titan97 said:
@gneill is my reasoning correct?
Which reasoning are you referring to? The following bit?
Titan97 said:
Because if I were to connect the two free ends to each other, then also no charge should redistribute.
If so then no, because the potential difference established across the connecting wire will force charges to flow in that wire. When the charge moves, it can be pulled from the left hand plate of the left capacitor so long as an equal charge is pushed onto its right hand plate. The closed loop facilitates that.

upload_2016-4-14_11-48-11.png
 
  • #7
I meant this:
Titan97 said:
If xxx charge flows from the negative plate of first capacitor to the second capacitor, then a same xxx has to flow from the first plate as well since opposite faces of a capacitor should have charges of same magnitude. So no charge transfer is possible.
 
  • #8
Look at the direction of the conventional current flow that I drew in on the figure in post #6. The parcel of charge ΔQ moves from the positive plate of capacitor C1, through the wire, and onto the right plate of C2. In turn, this pushes the same amount of charge ΔQ off of the left plate of C2 and onto the right plate of C1 where it cancels the same amount of negative charge there. This all takes place in step, so that the charges on the plates of C1 remain equal and opposite. The same goes for the plates of C2.
 
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FAQ: Uncharged capacitor connected to a charged capacitor

What is an uncharged capacitor?

An uncharged capacitor is a passive electronic component that has the ability to store electrical energy in the form of an electric charge. It consists of two conductive plates separated by a dielectric material, and can be found in various electronic devices such as radios, TVs, and computers.

What is a charged capacitor?

A charged capacitor is an electronic component that has an electric charge stored on its conductive plates. It can be charged by connecting it to a power source, such as a battery or another charged capacitor, and can store and release energy in the form of an electric field.

What happens when an uncharged capacitor is connected to a charged capacitor?

When an uncharged capacitor is connected to a charged capacitor, the uncharged capacitor will begin to charge as it equalizes with the charged capacitor. This is because the charged capacitor has a higher potential energy than the uncharged capacitor, causing an electric current to flow between the two to balance out the energy levels.

Will the total charge on the capacitors change when they are connected?

No, the total charge on the capacitors will remain the same when they are connected. This is due to the conservation of charge, which states that charge cannot be created or destroyed, only transferred from one object to another.

What is the equation for calculating the voltage of an uncharged capacitor connected to a charged capacitor?

The equation for calculating the voltage of an uncharged capacitor connected to a charged capacitor is V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. This equation is known as the capacitance formula and can be used to determine the voltage of any capacitor in a circuit.

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