Why do Capacitors discharge?

Question

I understand how capacitors charge and i know they discharge but i am so confused why they discharge. How do they suddenly know when they are full to discharge. I am doing a school report and really need to be able to explain why rather than just saying they do.

Answer 1

They discharge because there is something else (for example, a resistor or network of resistors) connected between their terminals, and the potential difference across the capacitor causes a current to flow through that something else.

If you're asking about self-discharge (when nothing is connected to the capacitor), it's because the dielectric between the capacitor plates is not perfectly non-conductive, so it acts like a (often very high-valued) resistor connected between the capacitor terminals, and again the potential difference across it causes a current to flow through it.

Answer 2

I assume you are already familiar with Coulomb force/potential. As you should know the force is attractive between charges with different sign and repellent between charges with the same sign.
When you apply a voltage to a capacitor, i.e. creating an electrical field $E$ (which is equal to the the negative gradient of the electric potential) between both plates, the charges experience a force along $E$. This Force $$F = q*E$$ divides the charges, hence there will be a positive and a negative plate. The reason why I am still explaing this to you, is that discharging is very similar.
As you split up the charges (we are still charging), you again create an inhomogenous charge distribution, hence a new electric field will be created, call it $\tilde E$, which acts in the opposite direction as $E$ does. Now the source of this field are the charges on the plates of the capacitor, not an external voltage source (as it was for $E$), hence $\tilde E$ will grow with more charges on the plates. The capacitor will be charged, until $$E + \tilde E = E_{total} \approx 0$$

So now we will let the capacitor discharge by removing the external voltage source (else we would just have a close to static system, i.e. a nearly fully charged capacitor, nothing would happen because $E_{total} = 0$).
Make yourself clear, that $E_{total}$ is the reason for everything happening during the charging and discharging process, you can explain the whole process just by analyzing the total electric field.
Now that $E=0$, $\tilde E =E_{total}$ (for t = 0), hence the charges will now experience a force in the direction opposite to the force during the charging process. Again it is important, that $\tilde E$ depends on the charges left on the plates and is not constant. It will decrease over time during the discharging process.
You can think of it in a more simple way: All the electrons on one of the plates are pushing each other away because of the Coulomb force (as mentioned at beginning), the same applies to the "positive" charges on the other plate (there are no actual positive charges on that plate, it's just a model). Hence they are pushing each other from the plate and get attracted to differently charged plate. So electrons move away from the negative plate towards the positiv and vice versa. But this will only happen till the charges get distributed homogenously (there will be some oscillation), then the system reaches equilibrium and that is, generally speaking, a state nature strives for.

Answer 3

i am so confused why they discharge.

Keeping always in mind that a capacitor stores electrical energy (and not electric charge), a capacitor in a circuit discharges when the attached circuit 'draws' on the stored energy in the capacitor. That's really all there is to it.

A simple example is to take a capacitor that has been charged (how it has been charged doesn't matter) to voltage $V_0$ and to connect a resistor across it.

The capacitor, being charged, has a voltage across and so, when the resistor is connected the resistor has the same voltage across as the capacitor.

But a resistor with a voltage across necessarily has a current through (Ohm's law). This current must be sourced by the capacitor and as a result, the stored energy in the capacitor decreases since power the power is out of the capacitor and in to the resistor (recall the resistor power law).

As the stored energy decreases, the voltage across decreases which (again by Ohm's law), means the current through decreases and so the rate at which the energy decreases is also decreasing.

This leads to the exponential decrease in voltage across the capacitor.

Note that there was never a time that the capacitor was full. This same description would apply if the capacitor had a fraction of the initial voltage $V_0$ or a multiple of it. The reason the capacitor discharged was that the external circuit, a resistor in this case, 'draws' on the stored energy in the capacitor (a resistor never supplies energy but can only dissipate it).

Answer 4

Capacitors discharge to send their charges to ground, or to get zero electric potential. Because all charges whether they are positive or negative have the tendency to flow the ground so as you provide them a path to flow to ground or zero electric potential they will flow to the zero potential to minimize their electric potential energy.

Same as water flows from a water tank at height (high potential energy) to ground to lose potential energy.