hi,

I know that capacitance is measured by the following equation:

Q = C V

So, the capacitor that can store one coloumb of electrical charge by applying one volt between its terminals, is said to have a capacitance of 1 Farad.

And the capacitor which can store one coloumb of electrical charge by applying 2 volts between its terminals, is said to have a capacitacne of 0.5 farad.

So, does this mean, we can get voltage difference between the capacitor terminals equals to 2 volts with the same amount of charges? 1 coloumb also?

 

Do not understand the question?

 

Ok , sorry, and exuce me for my bad english,I am egyptian.

if we have two capacitors, A and B, capacitor A is 1 farad, capacitor B is 0.5 farad, so if we connected A with a voltage source = 1 volt dc, and leave it until it is fully charged, so capacitor A now contains 1 coloumb and the voltage difference between its terminals is 1 volt, right? yes

well, if we connect capacitor B with the same voltage source = 1 volt dc, and leave it until it is fully charged , so capacitor B contains now 0.5 coloumb, and the voltage difference is also 1 volt.

So my question is, How could we get the same emf by less amount of charges ( 0.5 coloumb )?!

I know the equation but I need the physical explianation.

 

I think you have confused me.

Are you not taking current into this consideration?

 

If you fill a 1 litre water jug and a 0.5 litre water jug, they are both full.

If the base area of the larger jug is twice the base area of the smaller the water will stand at the same hieght in both.

A capacitor is fully charged (full) when its voltage equals the applied voltage, as you have described. Just like the jugs.

 

Voltage isn't charge, that's pretty much the explanation.

Voltage depends on the applied source, amount of charge depends on the geometry as well.
Your geometry determines the relationship between the two quantities.

Look at some of the other equations.
The simplest one for constant electric field is:

V = E * d

Electric field multiplied by distance is voltage.

For a point charge,
E = K * q / r^2

Electric field, not voltage, is fundamentally proportional to charge.

By having a different geometry and material in your capacitor you have different relationship between charge and electric field (changing K) and a different relationship between E and V (changing d in that simple case).
This lets you have a different amount of charge for the same voltage.

 

I am confused.

I know that a battery has two terminals - and + , the - terminal contains extra electrons, the + terminal contains extra protons, and this establishes an emf between the two terminals. If we increase the amount of extra electrons and extra protons, we get higher voltage difference, and vice versa.

So, how can two batteries have different amount of charges inside them, and both of them are said to have 12 v for example. The battery that contains more coloumbs should push electrons by higher amount of electric force (volt).

In other words;

If we use a voltmeter to measer the voltage difference between a 0.5 farad capacitor contains 0.5 coloumb, the voltmeter will give us 1 volt.
If we use a voltmeter to measer the voltage difference between a 1 farad capacitor contains 1 coloumb, the voltmeter will give us 1 volt.

How could we get the same value of electrical pushing force (voltage difference), by different number of coulombs?

 

I am confused.

So, how can two batteries have different amount of charges inside them, and both of them are said to have 12 v for example. The battery that contains more coloumbs should push electrons by higher amount of electric force (volt).

How could we get the same value of electrical pushing force (voltage difference), by different number of coulombs?

in your example, ? Volts is ? volts -- just if you discharge the cap, the one that is larger will hold a charge longer simply because it has a larger capacity. The voltage would be the same.

 

Because the number of electrons you can pull out of the 1F cap is X2 the number of electrons you can get out of the ½F cap. It has X2 the charge.

As with wattage, voltage is only ½ the formula.

 

Before you proceed further I seriously suggest you review your understanding of voltage and current.

Here is a quick primer.

http://www.kpsec.freeuk.com/voltage.htm