Capacitance

A capacitor is a device for storing separated charge.

No single electronic component plays a more important role today than the capacitor.  This device is used to store information in computer memories, to regulate voltages in power supplies, to establish electrical fields, to store electrical energy, to detect and produce electromagnetic waves, and to measure time.  Any two conductors separated by an insulating medium form a capacitor.

A parallel plate capacitor consists of two plates separated by a thin insulating material known as a dielectric.   In a parallel plate capacitor electrons are transferred from one parallel plate to another. 

We have already shown that the electric field between the plates is constant with magnitude E = σ/ε₀ and points from the positive towards the negative plate.

The potential energy difference between the negative and positive plate therefore is given by

∆U = U_pos - U_neg = -q Σ_neg^pos E∙∆r = q E d.
When summing, ∆r points from the negative to the positive plate in the opposite direction from E. 
Therefore E∙∆r  = -E∆r, and the minus signs cancel. 
The positive plate is at a higher potential ∆V = ∆U/q than the negative plate.

Field lines and equipotential lines for a constant field between two charged plates are shown on the right.  One plate of the capacitor holds a positive charge Q, while the other holds a negative charge -Q.  The charge Q on the plates is proportional to the potential difference V across the two plates.  The capacitance C is the proportional constant,

Q = CV,  C = Q/V.

C depends on the capacitor's geometry and on the type of dielectric material used. The capacitance of a parallel plate capacitor with two plates of area A separated by a distance d and no dielectric material between the plates is

C = ε₀A/d.

(The electric field is E = σ/ε₀. The voltage is V = Ed = σd/ε₀.  The charge is Q = σA. Therefore Q/V = σAε₀/σd = Aε₀/d.)
The SI unit of capacitance is Coulomb/Volt = Farad (F). 
Typical capacitors have capacitances in the picoFarad to microFarad range.

The capacitance tells us how much charge the device stores for a given voltage.  A dielectric between the conductors increases the capacitance of a capacitor.  The molecules of the dielectric material are polarized in the field between the two conductors.  The entire negative and positive charge of the dielectric is displaced by a small amount with respect to each other.  This results in an effective positive surface charge on one side of the dielectric and a negative surface charge on the other side of the dielectric.  These effective surface charges on the dielectric produce an electric field, which opposes the field produced by the surface charges on the conductors, and thus reduces the voltage between the conductors.  To keep the voltage up, more charge must be put onto the conductors.  The capacitor thus stores more charge for a given voltage.  The dielectric constant κ is the ratio of the voltage V₀ between the conductors without the dielectric to the voltage V with the dielectric, κ = V₀/V, for a given amount of charge Q on the conductors.

In the diagram above, the same amount of charge Q on the conductors results in a smaller field between the plates of the capacitor with the dielectric.  The higher the dielectric constant κ, the more charge a capacitor can store for a given voltage. For a parallel-plate capacitor with a dielectric between the plates, the capacitance is

C = Q/V = κQ/V₀ = κε₀A/d = εA/d,
where ε = κε₀.  The static dielectric constant of any material is always greater than 1.

Typical dielectric constants

If a dielectric with dielectric constant κ is inserted between the plates of a parallel-plate of a capacitor, and the voltage is held constant by a battery, the charge Q on the plates increases by a factor of κ.  The battery moves more electrons from the positive to the negative plate.  The magnitude of the electric field between the plates, E = V/d stays the same.

If a dielectric is inserted between the plates of a parallel-plate of a capacitor, and the charge on the plates stays the same because the capacitor is disconnected from the battery, then the voltage V decreases by a factor of κ, and the electric field between the plate, E = V/d, decreases by a factor of κ.

Embedded Question 2:

(a)  A parallel-plate capacitor initially has a voltage of 12 V and stays connected to the battery. If the plate spacing is now doubled, what happens?
(b)  A parallel-plate capacitor initially is connected to a battery and the plates hold charge ±Q. The battery is then disconnected. If the plate spacing is now doubled, what happens?

Hint: 
If the capacitor is connected to the battery, then the voltage stays constant. It stays equal to the battery voltage. The battery is a charge pump. It can pump charge from one plate to the other to maintain a constant potential difference.

If the battery is disconnected from the capacitor, the charge on the plates stays constant.
No battery <--> no charge pump.  Charge cannot move from one plate to the other. Therefore the voltage changes when the plate separation changes.

Discuss this with your fellow students in the discussion forum!

External link:  PhET Capacitor Lab (Basic)

Energy stored in a capacitor

The energy U stored in a capacitor is equal to the work W done in separating the charges on the conductors.  The more charge is already stored on the plates, the more work must be done to separate additional charges, because of the strong repulsion between like charges.  At a given voltage, it takes an infinitesimal amount of work ∆W = V∆Q to separate an additional infinitesimal amount of charge ∆Q. 
(The voltage V is the amount of work per unit charge.) 
Since V = Q/C, V increases linear with Q.  The total work done in charging the capacitor is

W = U = Σ V ∆Q = V_averageQ = ½VQ.
Using Q = CV we can also write U = ½(Q²/C)  or  U = ½CV².

Problem:

Each memory cell in a computer contains a capacitor to store charge.  Charge being stored or not being stored corresponds to the binary digits 1 and 0.  To pack the cells more densely, trench capacitors are often used in which the plates of a capacitor are mounted vertically along the walls of a trench etched into a silicon chip.  If we have a capacitance of 50 femtoFarad = 50*10^-15 F and each plate has an area of 20*10^-12 m² (micron-sized trenches), what is the plate separation?

Solution:

• Reasoning:
  The capacitance of a parallel plate capacitor with two plates of area A separated by a distance d and no dielectric material between the plates is C = ε₀A/d.
• Details of the calculation:
  C = ε₀A/d, d = ε₀A/C = (8.85*10^-12*20*10^-12/(50*10^-15)) m = 3.54*10^-9 m.
  Typical atomic dimensions are on the order of 0.1 nm, so the trench is on the order of 30 atoms wide.

For any insulator, there is a maximum electric field that can be maintained without ionizing the molecules.  For a capacitor this means that there is a maximum allowable voltage that that can be placed across the conductors.  This maximum voltage depends the dielectric in the capacitor.  The corresponding maximum field E_b is called the dielectric strength of the material.  For stronger fields, the capacitor 'breaks down' (similar to a corona discharge) and is normally destroyed.  Most capacitors used in electrical circuits carry both a capacitance and a voltage rating.  This breakdown voltage V_b is related to the dielectric strength E_b.  For a parallel plate capacitor we have V_b = E_bd.