Voltage & Capacitors: How Does It Create Charge?

[gkangelexa]

For a given capacitor that is connected to a battery, the amount of charge acquired by each plate is proportional to the magnitude of the potential difference V between them.
Q = CV

Is it correct to say that the voltage (potential difference V) applied across the capacitor plates is what causes the charge?
How does the voltage from the battery create the charge?


thanks!

 

[cabraham]

No. Charge is a fundamental quantity, along with time, mass, & length. Voltage is a ratio of work to charge. Voltage from a to b is defined as the work per unit charge expended transporting a charge from a to b.

Charge is basic. Voltage is defined in terms of work & charge. Nothing "causes charge" that we know of. Charge just basically is & that's all we can say.

Claude

 

[gkangelexa]

Claude,
You reminded me of another point of confusion...
voltage is work/ charge... V = -W/q = (U2-U1)/q... since the difference in potential energy is equal to the negative of the work done.

How do you relate those equations to this one concerning Electric energy storage in a capacitor? U = 1/2 QV

 

[srivi]

yes, we can say Potential differencce is cause for the charge but it does not mean that charge is created.

when a battery is connected to a circuit , the electrons are produced due to the chemical reactions in the battery .These electrons flow into the capacitor and the voltage
across the capacitor rises until the supply voltage and the electrons flow stops .

once the capacitor is charged the voltage of the capacitor is same as that of the battery.

 

[cabraham]

Claude,
You reminded me of another point of confusion...
voltage is work/ charge... V = -W/q = (U2-U1)/q... since the difference in potential energy is equal to the negative of the work done.

How do you relate those equations to this one concerning Electric energy storage in a capacitor? U = 1/2 QV

By definition C = Q/V, i.e. 1 farad = 1 coulomb / volt. Then Q = CV. U = (1/2)QV = (1/2)(CV)V = (1/2) C V^2.

Claude