Potential difference and conductors

[henry3369]

Homework Statement

I just need help understanding what establishes a positive charge on the top conductor in C1:
http://imgur.com/fppHtMN

My book says that first there is a positive potential difference, V_ab

Homework Equations

The Attempt at a Solution

So since V_ab is positive, a is at a higher potential. Does that mean the electrons from the top conductor of C1, move to point a which leaves a positive net charge? And my book also says that the negative charges from the bottom conductor of C2, acquires its negative charges from b. Are these electrons coming from the source that is providing the potential difference, such as a battery?

 

[STEMucator]

Suppose we short the source $V_{ab}$ for a moment with a wire. Initially, when the plates of each capacitor are uncharged, the potential difference between each plate is zero, i.e $V_1 = V_2 = 0$.

Now hook up the source $V_{ab}$ to the circuit, this will immediately start the flow of charge carriers due to the applied potential difference. The plates of each capacitor will become oppositely charged, and the potential difference between the plates of each capacitor increases until the sum of both potential differences $V_1$ and $V_2$ equals the potential difference between the terminals of the battery $V_{ab}$.

At that point, the top plate of the upper capacitor and the positive terminal of the battery are at the same potential, and there is no electric field in the wire between them. Hence the top plate will be positively charged.

Similarly, the bottom plate of the lower capacitor and the negative terminal of the battery reach the same potential, and there is no electric field in the wire between them. Hence the bottom plate will be negatively charged.

Thus with the electric field being zero for each capacitor, there is no further drive of electrons. The current in the circuit stops because both capacitors are fully charged (with a net charge of zero on each of them). The potential difference and charge are then related by $q = CV$ for each capacitor.

 

[henry3369]

Suppose we short the source $V_{ab}$ for a moment with a wire. Initially, when the plates of each capacitor are uncharged, the potential difference between each plate is zero, i.e $V_1 = V_2 = 0$.

Now hook up the source $V_{ab}$ to the circuit, this will immediately start the flow of charge carriers due to the applied potential difference. The plates of each capacitor will become oppositely charged, and the potential difference between the plates of each capacitor increases until the sum of both potential differences $V_1$ and $V_2$ equals the potential difference between the terminals of the battery $V_{ab}$.

At that point, the top plate of the upper capacitor and the positive terminal of the battery are at the same potential and there is no electric field in the wire between them. Hence the top plate will be positively charged.

Similarly, the bottom plate of the lower capacitor and the negative terminal of the battery reach the same potential, and there is no electric field in the wire between them. Hence the bottom plate will be negatively charged.

Thus with the electric field being zero for each capacitor, there is no further drive of electrons. The current in the circuit stops because both capacitors are fully charged (with a net charge of zero on each of them). The potential difference and charge are then related by $q = CV$ for each capacitor.

I'm still sightly confused about how it conductor acquires a positive charge. When you say that when the potential difference V_ab is applied, the charge carriers begin to flow. Are these coming from the conductors?

 

[STEMucator]

I'm still sightly confused about how it conductor acquires a positive charge. When you say that when the potential difference V_ab is applied, the charge carriers begin to flow. Are these coming from the conductors?

Think of the source $V_{ab}$ as a battery. Recall some familiar concepts about voltage and energy.

When you place a battery in a conducting path, you pump charge carriers through the conducting path because of the energy you supply to it. These are the charge carriers that accumulate on the plates of each capacitor.

 

[henry3369]

Think of the source $V_{ab}$ as a battery. Recall some familiar concepts about voltage and energy.

When you place a battery in a conducting path, you pump charge carriers through the conducting path because of the energy you supply to it. These are the charge carriers that accumulate on the plates of each capacitor.

I just started this unit so and usually when I calculated potential difference is between two points in a vacuum. Isn't the only thing that can move in the wire the electrons from the battery? My book explains that once the battery is connected, positive charge accumulates on the top conductor. This confuses me because the only way that can occur is if the conductor loses electrons. It seems like from your explanation, connecting a battery causes charges to flow through the wire. Isn't the only thing that can flow out of the battery electrons? Which have to come from the negative terminal at point b?

 

[STEMucator]

Isn't the only thing that can move in the wire the electrons from the battery?

The only things moving in the wire are the charge carriers themselves, which are being pushed by the energy of the battery (in the direction of conventional current flow). This movement stops when $V_1 + V_2 = V_{ab}$ because the battery can't push the carriers anymore against an equal amount of energy.

My book explains that once the battery is connected, positive charge accumulates on the top conductor.

Yes, and the top plate of the top conductor reaches the same potential as the positive terminal of the battery.

This confuses me because the only way that can occur is if the conductor loses electrons. Isn't the only thing that can flow out of the battery electrons?

Usually, a current arrow is drawn in the direction positive charge carriers would move, even if the actual charge carriers are negative and move in the opposite direction. So the capacitor plates can accumulate positive charge carriers.

 

[cnh1995]

Charging of capacitors is just a rearrangement of charge..Try this train of thoughts (it's not at all "technically" proper but might give you some insight.)
Electrons start moving towards +ve terminal all at once but can't cross the dielectric. So, electrons from -ve terminal of battery get stuck on outer plate of C2 making it -vely charged..The electrons which have left the inner plate of C2 leave it +vely charged and get stuck on the inner plate of C1, making it -vely charged. Electrons from outer plate of C1 are attracted towards +ve battery terminal, so,that plate gets +ve charge. Charge lost by one plate is deposited on the other..For detailed conduction mechanism in circuits, visit youtube and watch "surface charges and circuits"..