Why Does the Negative Plate of a Capacitor Become Negatively Charged?

[beg]

I don't understand why the plate linked to the negative battery becomes negatively charged.

To begin with, each plate is neutral so each contain equal amounts of positive and negative charge. When connected to the battery, the plate linked to the postitive battery terminal becomes positively charged because electrons in a conductor can move, and are attracted from the plate to the positive battery terminal, leaving the plate with only protons. But what causes the electrons on the negative battery terminal to move up to the other plate?

Am I right in also thinking that the capacitor will have the same Potential difference across it after its charged? so it will have a more electrons than protons?

 

[Freeze3018]

Think this way:
The electrons of the plate connected to +ve terminal of Cell are attracted towards the +ve terminal of cell(Leaving the plate Electrons Deprived and +vely charged) ...Since the incoming electrons to the cell must be somehow equal to outgoing electrons, The electrons from the -ve terminal of the cell are pushed to the Plate connected to -ve terminal (Leaving it electron Rich and -vely charged)...
Hope u'll find this helpful...!

 

[Freeze3018]

Am I right in also thinking that the capacitor will have the same Potential difference across it after its charged? so it will have a more electrons than protons?

What do u mean by same potential difference ? If u mean that of battery then yes if u leave the connection of the plates not disconnected for sufficient time YES..

The net charge of the Capacitor(2 plate system) is Zero...But one plate will have the same no of electrons deficient as the other plate is sufficient of...!

 

[DaveW]

The same thing that pushes electrons onto the (-) plate is the same reason why a charged sphere has electrons on the outside only... electrons repel. They'll try to fill all space (metal in this case) available to them to spread out more. Hence they "spread out" onto the negative side of the cap. Letting them "spread out" means a certain loss of potential energy in the battery, ie, it takes WORK to charge the cap.
That potential energy (some energy is lost in charging process) is now stored in the cap.

Hope that helps!
DaveW

 

[beg]

Freeze - So you're saying after the capacitor is charged, it has a net total of ZERO? ...but for it to have a voltage, wouldn't it have to have a higher number of electrons on one of the plates? So if the battery is 9v, then it would have 9 times more electrons on the -plate, than there are protons on the +plate?

DaveW - So then why don't the electrons from the cell repel the electrons on the plate, and just stay in the battery? When the cell is first connected the plate will be neutral, and become gradually more and more negatively charged, so what would attract the electrons to keep traveling down a path that is negative?

 

[NascentOxygen]

why don't the electrons from the cell repel the electrons on the plate, and just stay in the battery? When the cell is first connected the plate will be neutral, and become gradually more and more negatively charged, so what would attract the electrons to keep traveling down a path that is negative?

There is only one thing that causes electrons to flow around a circuit, and that is a potential difference (voltage). While ever the voltage from the battery is greater than that on the capacitor, electrons will continue to flow to the capacitor's -ve plate (and to flow from its +ve plate). When so much charge has accumulated on the plates that the capacitor voltage exactly equals that of the battery, then no more electrons will be pushed from the battery and the capacitor can be considered fully charged.

 

[DaveW]

I don't understand why the plate linked to the negative battery becomes negatively charged.

To begin with, each plate is neutral so each contain equal amounts of positive and negative charge. When connected to the battery, the plate linked to the postitive battery terminal becomes positively charged because electrons in a conductor can move, and are attracted from the plate to the positive battery terminal, leaving the plate with only protons. But what causes the electrons on the negative battery terminal to move up to the other plate?

Am I right in also thinking that the capacitor will have the same Potential difference across it after its charged? so it will have a more electrons than protons?

I thought of another way to explain what is going on.

The same work that is done in PULLING electrons AWAY from the protons in one plate, is the same work that is done in PUSHING more electrons TOWARD the pre-existing electrons in the other plate. The battery-capacitor circuit is just finding the equilibrium situation. Energy "spreads out" from the battery to the capacitor.

As far as having the same potential difference after charging, that depends on the energy in the battery. Let's say instead of a battery, you have 2 identical caps, one fully charged, one fully discharged.
What happens when you connect them?

The answer to that is rather surprising.

 

[beg]

Is the following correct?

When the plates have no charge and are first linked up.

You can approximate the Electric Pontential (Voltage) at the (-)Plate and the (-)Cell Terminal, by using the Electric potential due to a point charge at each. If there is a Electric Potential Difference, then the Electrons will flow from the (-)Cell Terminal to the (-)Plate.
You can approximate the Electric Pontential (Voltage) at the (+)Plate and the (+)Cell Terminal, by using the Electric potential due to a point charge at each. If there is a Electric Potential Difference, then the Electrons will flow to the (+)Cell Terminal from the (+)Plate.