Voltage vs Charge: Capacity & Applied Voltage

[david845]

If voltage is the difference in charge between 2 points, then why for a capacitor of a larger area or thinner dielectric cross section, do they say that it can store more charge given the same apppied voltage? Isnt voltage the difference in charge between 2 points? So if you can store more charge wouldn't that change the voltage between the 2 plates? I thought you can store up to the applied voltage value anyway??

 

[Nugatory]

If voltage is the difference in charge between 2 points,...

It's not, and that is the basis of your confusion.

There is a relationship in a capacitor: charge = capacitance times voltage, so increasing the voltage increases the amount of stored charge, but changing the design of the capacitor (for example, by increasing the surface area) can increase the capacitance and you can get more charge that way as well.

If you google for "parallel plate capacitor" you will find the relevant equations and how area of plates, distance between plates, charge, electric field strength, voltage, and capacitance are all related. Give that a try, come back with a more focused question if you get stuck and we can help you over the hard spot.

 

[A.T.]

Isnt voltage the difference in charge between 2 points?

No, its work per charge:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html

 

[sophiecentaur]

If voltage is the difference in charge

Where could you possibly have read that?

 

[Drakkith]

If voltage is the difference in charge between 2 points, then why for a capacitor of a larger area or thinner dielectric cross section, do they say that it can store more charge given the same apppied voltage? Isnt voltage the difference in charge between 2 points?

Voltage is the difference in electric potential between two points.

Electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point in an electric field. We typically use infinity as the reference point.

Imagine we have two points in an electric field and we find their respective electric potentials. If the two potentials are not the same, then it requires that net work be performed to move a charge between the two points. This work can be positive if you have to work against the electric field, or negative if the electric field is doing the work for you.

When we set up a voltage in an electric circuit we are setting up a situation where there is a difference in the electric potential between the different points in the circuit. This voltage drives a current through the circuit and allows us to create complex circuits to do things like run motors or electronic devices.

Note that all of these terms, electric potential, voltage, and etc, are just ways of describing what the electric field is doing. When we say that the voltage causes a current to flow we mean that the electric field is in a particular configuration that makes it generate a current in a circuit. It is the electric field that does the work, not the voltage itself.

So if you can store more charge wouldn't that change the voltage between the 2 plates? I thought you can store up to the applied voltage value anyway??

Let's look at a couple of specific situations:
1. Keeping plates static and changing the number of charges on the plate.
2. Moving the plates while keeping the number of charges on the plates constant.
3. Changing the size of the plates while keeping their separation and number of charges constant.

In example 1, changing the number of charges on each plate changes the voltage. Increasing the number of charges increases the voltage and vice versa. Easy example. No surprises here.

In example 2, moving the plates apart increases the voltage of the capacitor. This is because the voltage for a capacitor is given by $V=\int_0^d E \, ds = E\int_0^d \, ds = Ed$, where $E$ is the electric field (taken to be constant here because the field lines between the parallel plates are straight lines and we are ignoring the field lines near the fringe) and $d$ is the distance between the plates. You can imagine that a unit positive test charge placed near the positive plate will accelerate for a greater time and distance when the plates are far apart, giving it a higher kinetic energy when it finally reaches the negative plate. Conversely, moving the plates closer together while keeping the number of charges static will decrease the voltage.

Interestingly, moving the plates apart decreases the capacitance of the capacitor. Since we just found that moving the plates apart increases the voltage if we hold the number of charges on each plate constant, that means that it requires a larger and larger applied voltage to put the same number of charges on each plate as we move the plates apart. Hence the capacitance decreases per the equation $C=\frac{Q}{V}$ since $Q$ is constant and $V$ increases.

In example 3, changing the area of the plates while keeping the number of charges and the separation constant will also alter the voltage. Imagine that we take our charged capacitor and we quickly weld on a neutral conductor to each plate such that the area of each plate doubles. The charges on each plate will quickly spread out and the end result is that the density of charges on each plate decreases. The causes a corresponding drop in the density of the field lines between the plates and, since the density of the field lines represents the magnitude of the electric field, the magnitude of the E-field between the plates decreases. This makes sense, since a test charge placed near one of the plates will find that many of the charges have now moved further away from it after welding on the neutral conductors. The contribution of those charges on our test charge is reduced, reducing the magnitude of the force felt by that test charge.

The end result is that the voltage between the plates decreases when we increase the area of the plates while keeping the number of charges and separation constant. Put another way, if we increase the area of the plates, we can put more charges on each plate for the same applied voltage. This also increases the capacitance of the capacitor per the capacitance equation I provided in example 2.

So we have more than one way of altering the voltage on a capacitor, only one of which involves altering the number of charges on each plate.