Capacitance and Dielectrics

©2011 by Bryan Pflueger

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Capacitors

A capacitor is any two conductors seperated by an insulator, such as air or another material.

Each conductor has equal magnitude, but opposite charge so the net charge across the two conductors is zero.

In a circuit a capacitor is represented by:

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CapacitorsTo charge a capacitor you can simply connect each conductor to the opposite end of each battery. Once the plates have achieved a magnitude of Q the battery can be disconnected.

There will be a fixed potential across the conductors equal to the potential difference of the battery.

In previous sections we discussed that both the electric field and the potential difference are both directly proportional to the charge Q. By doubling the charge we double both the electric field and potential difference. Capacitance is the ratio of charge to potential difference, therefore it is independent of the charge and potential.

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Calculating Capacitance

The simplest version of a capacitor is the parallel plate capacitor, which has two conducting plates of area A.

Using Gauss's law we can find that the electric field between a parallel plate capacitor is:

Since the electric field is uniform and the plates are separated by a distance d, the electric potential is equal to:

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Calculating Capacitance

Capacitance is the ratio of charge to potential difference across the plates, which can be written as:

By plugging in what we just solved for v, we will see that the capacitance is only dependent on the area of the plate and their separation.

(Capacitance of a parallel plate capacitor in a vacuum)

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Unit of Capacitance

Capacitance is the ratio of charge to the potential difference.

So the unit for capacitance is coulomb/volt better known as a Farad, which was named in honor of the English physicist Michael Faraday. A Farad is denoted by a capital F.

1 F = 1 Farad = 1 C/V = 1 coulomb/volt

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1 The plates of a parallel plate capacitor are 3x10-3m apart and they are each 2cm squares. What is the capacitance of the capacitor?

A 5.9x10-11 F

B 4x10-15 F

C 1.2pF

D 3.6µF

E 2.4pF

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2 A parallel plate capacitor has a charge of 6µF. The electric field present between the plates is 3x107 N/C and the plate's separation is 4cm. What is the capacitance of the capacitor?

A 5nF

B 12pF

C 5µF

D 5MF

E 5pF

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Spherical Capacitor

-

--

-- - - -

----

-----

++

+ + + + ++

+++

++

++

++++

++ +

Q

-Q

so, Electric Field for point charge

If the electric field is the same for a point charge then so is the potential difference.

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Cylindrical Capacitor

Lrarb

Before we found the electric potential for a cylinder to be equal to:

The equation for capacitance is:

The charge on the cylinder is represented as:

The capacitance per unit of length is:

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Capacitors in Series and Parallel

+-

C1

C2

x

y

z

The potential difference across the battery results in both of the two capacitors to begin charging. The first plate of C1 will acquire a positive charge equal to Q which will displace the charge on the second plate making it negative and the first plate of C2 positive, which in turn will make the second plate of C2 to become negatively charged. This is not an immediate change in the charge, it slowly builds up, but in the end the charge is equal for each capacitor.

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Capacitors in Series and ParallelWhen we discussed Circuits the last two years you could always replace any combination of resistors with one that has the equivalent resistance. For resistance in series we would just add their values to find the net resistance and just draw a new circuit only using the one resistor. We can do the same for Capacitors, but instead of just adding their values like the resistors, when they are in a series circuit we have to take their reciprocals.

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3 Three capacitors are connected in series, C1, C2, and C3. Their capacitance's are 4µF, 3µF, and 6µF respectively.

A 13µF

B 1.33µF

C 1µF

D 7µF

E 2.78µF

+-

C1

C2

C3

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4 Three capacitors are connected in series, C1, C2, and C3. Their capacitance's are 2µF, 7µF, and 13µF respectively.

A 22µF

B 7µF

C 20µF

D 1.39µF

E 2.6µF

+-

C1

C2

C3

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Capacitors in Series and Parallel

+- C1 C2

x

y

The voltage across each capacitor is equal to the potential difference of the battery because in a parallel circuit the voltage is the same, but in this case the charge is the sum of those on the capacitors.

The charges on each of the capacitors are:

therefore:

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Capacitors in Series and ParallelWhen we discussed Circuits the last two years you could always replace any combination of resistors with one that has the equivalent resistance. For resistance in parallel we would add their reciprocals to find the net resistance and just draw a new circuit only using the one resistor. We can do the same for Capacitors, but instead of adding their reciprocals like the resistors, when they are in a parallel circuit we just have to add their values.

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5 Three capacitors are connected in parallel, C1, C2, C3. Their capacitance's are 3μF, 4μF, and 2μF

A 2µF

B 1.1µF

C 9µF

D 3µF

E 2.5µF

+-

C1 C2 C3

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6 Three capacitors are connected in parallel, C1, C2, C3. Their capacitance's are 12μF, 5μF, and 7μF

A 23.8µF

B 24µF

C 12µF

D 8µF

E 32.4µF

+-

C1 C2 C3

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7 Five capacitors are placed into a circuit as shown below. The five capacitors are C1, C2, C3, C4, and C5. Their capacitance's are 2µF, 1µF, 3µF, 2µF, and 1µF respectively. What is the net capacitance of the circuit?

A 2µF

B 23/4 µF

C 3µF

D 6µF

E 1µF

+-

C1

C2 C3

C4

C5

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Energy Storage in a Capacitor and Electric-Field Energy

For a capacitor to be used in a practical manner we need to know how much electric potential energy it can store. This can be achieved by determining how much work is required to charge each of the plates. The small charges we will add up to equal the net charge will be denoted as dq and the potential difference can be given as Q/C.

The work done to charge the capacitor is also the same as the amount of work required to discharge it. The uncharged capacitor has a potential energy of zero so the work to charge the capacitor is equal to the stored potential energy.

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Energy Storage in a Capacitor and Electric-Field Energy

Since the work done to charge the capacitor is equal to the potential energy stored on the capacitor and since we know C=Q/V the potential energy can be represented as:

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Energy Storage in a Capacitor and Electric-Field Energy

Another way of looking at the amount of energy stored in a capacitor is to look at the electric field it produces because of the charged plates. We can say that the electric potential energy is spaced throughout the electric field therefore it has an energy density, which is denoted by a small u. The energy density is the ratio of electric potential energy to the volume between the plates.

Equation for Capacitance:

Potential Difference across the plates:

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Energy Storage in a Capacitor and Electric-Field Energy

The equation for the energy density is:

The equation is derived from the simple case of a parallel plate capacitor, however it works for every type of capacitor in a vacuum and for every electric field also in a vacuum.

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Dielectrics

Dielectrics are materials which are placed between the parallel plates or any other configuration for a different number of reasons.

It helps to maintain the shape of the capacitor, preventing the walls from coming in contact with one another.

It allows the capacitor to reach a higher potential difference then it could normally before dielectric breakdown, which is the ionization of the air around the capacitor which would result in charge leaving the capacitor.

It also enables the capacitor to increase its capacitance.

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Dielectrics

When a dielectric is placed between the plates of a parallel plate capacitor the voltage drops to a smaller value then its original, but the charge on the capacitor remains the same.

Original Capacitance Capacitance with Dielectric

and Q is the same,therefore C > Co

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Dielectric ConstantThe Dielectric Constant is denoted by a capital Kappa and it is the ratio of the final capacitance to the original capacitance.

The new equation for Capacitance is now represented as:

By adding a dielectric between the plates of a capacitor with a constant charge decreases the potential difference, but also the electric field.

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9 A capacitor is completely charged and afterwards the battery is disconnected. What will cause the potential difference across the capacitor to decrease?

A Increase the Plate's surface area

B Add a dielectric with a greater # C Decrease the distance

D All of the above

E Both A and C

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10 A Cylindrical capacitor is filled half way with a dieclectric of #. What is the net capacitance of this configuration?

A

B

C

D

E

L

r

R

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r

R

11 A spherical capacitor is filled halfway with a dielectric of #. What is the net capacitance?

A

B

C

D

E

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Dielectric's effect on Electric Energy Density

By adding a dielectric you alter the way the capacitor normally behaves. You allow it to carry a greater potential difference and also have a greater electric field passing through it without dielectric breakdown occuring.

The normal electric field between the plates is which can be determined through Coulomb's law for the infinite charged disk.

However when the dielectric is added the net charge density along the surface of the capacitor plate is the difference of that on the plate and the one induced on the dielectric because of the effects of polarization.

The net electric field is now represented as:

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ra

rb

12 The outer shell has a charge of +Q and the inner shell has a charge of -Q. Half of one side of a cylindrical capacitor of length l is filled with a dielectric of #. What is the magnitude of the electric field at a radius r, if ra < r < rb, for both parts of the capacitor?

A

B

C

D

E

Dielectric Vacuum

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