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### Course: MCAT>Unit 9

Lesson 12: Capacitors

# Capacitors article

## What are capacitors?

Capacitors store energy by holding apart pairs of opposite charges. Since a positive charge and a negative charge attract each other and naturally want to come together, when they are held a fixed distance apart (for example, by a gap of insulating material such as air), their mutual attraction stores potential energy that is released if they are re-united. The simplest design for a capacitor is a parallel-plate, which consists of two metal plates with a gap between them: electrons are placed onto one plate (the negative plate), while an equal amount of electrons are removed from the other plate (the positive plate).
As you may recall, charges create electric field lines that point away from positive charges and towards negative charges. In a parallel-plate capacitor, the electric field lines point straight across the gap between the two plates. We know that electric fields and voltage differences go hand-in-hand, and so it also turns out that the two plates are at different voltages. The size of this voltage difference ($V$) is related to the charges on the two plates (Q):
$Q=C\cdot V$
The constant $C$ is called the capacitance. It determines how much of a charge difference the capacitor holds when a certain voltage is applied. If a capacitor has very high capacitance, then a small difference in plate voltage will lead to a huge difference in the number of electrons (total charge $Q$) on the two plates.
Once opposite charges have been placed on either side of a parallel-plate capacitor, the charges can be used to do work by allowing them to move towards each other through a circuit. This usually requires them to travel through a circuit (as current) and perform some task, like illuminating a light bulb, along the way. The total energy that can be extracted from a fully charged capacitor is also related to the capacitance and voltage,
$E=\frac{1}{2}C\cdot {V}^{2}$
If you attach a capacitor (with capacitance $C$) to a battery (at voltage $V$), it will slowly develop a charge on each plate ($Q$) as electrons build up on one plate and then exit the other. Once you remove the battery, this difference in charge between the two plates remains indefinitely, until the capacitor is connected to a circuit (such as a light bulb) through which it can discharge. Once this occurs, charges will slowly pass out of one plate of the capacitor, move through the circuit, and onto the other plate. Capacitors function a lot like rechargeable batteries. The main difference is a capacitor’s ability to store energy doesn’t come from chemical reactions, but rather from the way that its physical design allows it to hold negative and positive charges apart. This makes capacitors very fast at charging and discharging, much faster than batteries. They are essential for applications where rapid bursts of current are needed, such as camera flashes.

## Do capacitors store charge?

Capacitors do not store charge. Capacitors actually store an imbalance of charge. If one plate of a capacitor has $1$ coulomb of charge stored on it, the other plate will have $-1$ coulomb, making the total charge (added up across both plates) zero. If you short circuit the capacitor by connecting the two plates with a wire of negligible resistance, you’ll see a sudden rush of current (depending on the size of the capacitor, this can result in sparks) as the electrons on the $-1$ coulomb plate rush onto the $+1$ coulomb plate. This sudden rush of current releases all the energy that’s stored in the capacitor.
To help us understand parallel plate capacitors, consider this situation. Imagine you start with two metal plates with no difference of charge ($Q=0$). You attach a battery, which at first adds a single electron to one side of the capacitor. The electron has an electric field that repels other electrons, and this field reaches through space and pushes on the electrons in the other plate, causing that plate to acquire an induced positive charge. Now your first plate has a charge of $-1e$, and the far plate has a charge of $+1e$, where e is the way we normally write the elementary charge of a single electron.
Now imagine repeating this process over and over, until a considerable amount of negative charge has built up on one plate and induced an equal positive charge on the other plate. At some point, the existing negative charge on the first plate will be so repulsive that it prevents you from adding any more negative charges to that plate. In this case the capacitor is fully charged. This maximum charge $Q$ corresponds to the final voltage of the charged capacitor in the relation $Q=C\cdot V$.
But how does the shape of the capacitor affect this process?
• Distance: The further apart the two plates are, the less the free electrons on the far plate feel the push of the electrons that you’re adding to the negative plate, making it harder to add more negative charges to the negative plate. If the plates were infinitely far apart, you would just be adding negative charges to an already negative metal surface, which would be pretty hard. If the plates were very close to each other or even touching, you essentially would be making current flow through a short circuit, which would be easy. This means that the capacitance of a parallel plate must be inversely related to the plate separation.
• Area: It’s a lot easier to add charge to a capacitor if the parallel plates have a huge area. Two wide metal plates would give two repelling like charges a greater range to spread out across the plate, making it easier to add a lot more negative charge to one plate. Likewise, a very small plate area would cause the electrons to get cramped together earlier, making it harder to get a large difference in charge for a given voltage. From this we can guess that the capacitance of a parallel plate must be related to the plate areas.
These two principles can be expressed as the parallel-plate capacitor formula:
$C=\text{ɛ}\frac{A}{d}$
A is the area of the plates, and d is the distance between the plates. $\text{ɛ}$ is a constant called the permittivity, which determines how easily the air between the plates allows an electric field to form. If a different insulating material is used inside the gap, this constant will have a different value, and so materials with a higher value of this constant generally make better capacitors.

## Consider the following… cardiac defibrillators

Sometimes the regular rhythm of your heart pumping blood around your body stops beating regularly. It turns out that the most effective way to make the heart start ticking regularly is simply to shock it with a giant capacitor. When a patient’s heart beats too fast or doesn’t beat in the correct sequence (called fibrillation), paramedics attach two electrodes to the patient’s chest. These electrodes are connected to a defibrillator, which consists of a battery and a giant capacitor. When paramedics use defibrillators, batteries slowly charge the capacitor by adding electrons to one plate and pulling an equal number of electrons off of the other plate. Once the capacitor is charged to the set voltage, paramedics rapidly discharge the capacitor through the electrodes on the patient’s chest in hopes of resetting the heart beat.

## Want to join the conversation?

• How light is produced when electrons flow through a wire?
(4 votes)
• I'll assume you mean in a lightbulb? In an incandescent bulb, the general structure is that the wires run into the bulb, then there is a resistor in the center of the bulb and the rest of the wire which leaves to complete the circuit. The flow of electrons passing through the wire hits the resistor, which slows them down, causing friction and heating up the resistor in the bulb, called the filament. When the filament heats up enough (due to enough electricity passing through) it begins to glow and produce light. One reason that people are moving away from incandescent bulbs is because the vast majority of the energy put into the bulb is used to create heat rather than light since the bulb has to be very hot before it produces light. This is also why if you have ever tried to unscrew a lightbulb after it was on for awhile, it was very hot to the touch. This means a lot of the energy is wasted and there has been a large movement to change the bulb technology from incandescent to fluorescent lamps which utilize a more complex process involving expelling the electrons into a gas to eventually produce photons through a series of radiative processes.
(31 votes)
• "C = ɛ A/d

...and L is the distance between the plates."

I'm assuming L and d are supposed to both represent distance?
(20 votes)
• Yes, in this case L and d both represent the same value. I'm not sure why this article isn't consistent in it's usage of variable names.
(20 votes)
• I don't think this is necessarily a mistake, but definitely confusing. The article specifically says "capacitors DO NOT store charge," while the very first video on capacitors specifically says that they DO store charge.
(9 votes)
• It is meant that they do not store any net charge.
(19 votes)
• What will happen if one plate is comparatively smaller or larger than other ?
(5 votes)
• I believe it will store charge in relation to the smaller charge since both plates store equal and opposite charges. So the smallest plate has the smallest area and hence is the limiting factor
(5 votes)
• In C = ɛ A/d, how do you find the permittivity of an object other than air?
Thank you in advance :)
(3 votes)
• In the parallel-plate capacitor formula, we can substitute permittivity of an object other than air as:
k=e/e' , where e' is the absolute permittivity of free space , e is the permittivity of the medium or object and, k is the dielectric constant of the medium or the object.
So, after solving, the capacitance in a different medium than air becomes,
C'=Ck where,
C' is the new capacitance
C is original capacitance
Hope this helps!;)
(3 votes)
• can any one elaborate how distance effect capacitance?
(2 votes)
• We know that V=Ed and that C=Q/V.
Substituting for V, we get C=Q/Ed. So, we see an inverse relation between capacitance and the distance of separation between them. So if the distance is increased, say by a factor of 2, the capacitance would decrease by a factor of 2.
Hopefully this made sense...
(1 vote)
• when charging a capacitor how electrons move from +ve terminal to -ve terminal & wthout interruption to the flow. please explain it with an example [chemical rxns in battery]. thank u!
(2 votes)
• could anyone explain this statement "The further apart the two plates are, the less the free electrons on the far plate feel the push of the electrons that you’re adding to the negative plate, making it harder to add more negative charges to the negative plate."
(2 votes)
• In previous video Sal said Q/V = 1/4K(pi).A/d where 1/4K(pi) is epsilon and 1/4K(pi).A/d is Capacitance C.

In capacitors article it was said that capacitance
C = (epsilon) A/d.

Can anybody pls distinguish and explain how it can be?
I'm little confused but also not confused.
(2 votes)
• What is the role of Battery in charging the battery? How does it help?
(1 vote)