Chapter 24: Capacitance and Dielectrics · Chapter 24: Capacitance and Dielectrics Capacitors and Capacitance A capacitor (“cap”, for short) is a device for storing charge. A

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Chapter 24: Capacitance and Dielectrics

Capacitors and Capacitance A capacitor (“cap”, for short) is a device for storing charge. A capacitor consists of two conductors, called the plates, separated by an insulator. The insulator is called the dielectric. Common materials used for the diectric include air, mica, polystyrene, polyester, and ceramic materials. Capacitors are used in camera flashes, DC power supplies, amplifiers, filters, and in tuning applications, to name just a few.

Ch. 24: Capacitance and Dielectrics

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The Defining Equation for the Capacitance We “charge up” a capacitor by connecting it to a voltage source (battery, e.g.). When this is done, the charges on the plates are the same, but opposite in sign. We call the magnitude of charge on either plate “Q ”. It turns out that Q is just proportional to the magnitude of the voltage between the plates. This happens because the magnitude of the electric field E between the plates is proportional to Q and the voltage VΔ is proportional to E . The constant of proportionality between Q and VΔ is called the capacitance, C . So: Q C V= Δ It is customary to call the absolute value of the potential difference between the plates just “V ”, so from now on, I will write the above equation: Q CV= (1) This is the “defining equation” for the capacitance, C . (That is, Eq. (1) defines what C means.)


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Notes: • Unit for C : C

V≡ “Farad”, F (after Michael Faraday)

• C just depends on geometry (area of plates, distance of separation) and on the material used for the dielectric.


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Calculating Capacitances Three special cases for which it’s “easy” to write down C :

1. Parallel-plate capacitor 2. Cylindrical capacitor 3. Spherical capacitor

General Procedure: • Write down E between the plates from Gauss’s law.

• Get VΔ between the plates from f

i

r

r

V E drΔ = − ⋅∫

• Get C from Q QCV V

= =Δ


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Case 1: Parallel-plate capacitor In Chapter 22, found E due to a single “infinite” sheet:

02

E σε

=

Let E+ be the electric field due to the positive plate and E− be the field due to the negative plate. Then consider three regions: • Region I: above the positive plate • Region II: between the plates • Region III: below the negative plate

Everywhere in Region I:

0

ˆ2

E jσε+ =

and:

0

ˆ2

E jσε− = − ,

so the net field in Region I is:


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0E E E+ −= + = Similarly, in Region III:

0

ˆ2

E jσε+ = −

and:

0

ˆ2

E jσε− = ,

so 0E = . In Region II, however, we have:

0

ˆ2

E jσε+ = −

and:

0

ˆ2

E jσε− = − ,

so:


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0

ˆE jσε

= −

Now to get VΔ between the plates, imagine moving along a path from a point a somewhere on the bottom plate (at iy ) to a point b directly above a on the top plate (at fy ). Then:

( ) ( )0 0 0 0 0

ˆ ˆf f f

i i i

r y y

f ir y y

d QdV E dr j dy j dy y yA

σ σ σ σε ε ε ε ε

⎛ ⎞Δ = − ⋅ = − − ⋅ = = − = =⎜ ⎟

⎝ ⎠∫ ∫ ∫

Finally, the capacitance is:

( )0

Q QCV Qd Aε

= =Δ

0ACd

ε= (2)

Notice that I didn’t say anything about what the dielectric was made of. This means that I assumed that the dielectric was vacuum (literally nothing at all). If there is a dielectric between the plates, then Eq. (2) must be modified slightly. I’ll discuss this in more detail later.


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Note: • The capacitance does just depend on geometrical factors, as we

said. Specifically, C is large if: o A is large (big plates) o d is small (plates close together)


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Case 2: Cylindrical capacitor From Gauss’s law, the electric field between the plates is:

0

ˆ2

E rr

λπε

=

Now consider moving along a radial path from a point a on the inner cylinder (at ar r= ) to a point b on the outer cylinder (at br r= ). Then:

( )0 0 0

1ˆ ˆ ln2 2 2

f b b

i a a

r r rb

ar r r

rV E dr r drr drr r r

λ λ λπε πε πε

⎛ ⎞ ⎛ ⎞Δ = − ⋅ = − ⋅ = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫ ∫

The V in Q CV= is the absolute value of this potential difference:

0

ln2

b

a

rVr

λπε

⎛ ⎞= ⎜ ⎟

⎝ ⎠

And the capacitance is:

02

ln b

a

QQCV r

r

πε

λ= =

⎛ ⎞⎜ ⎟⎝ ⎠

The Q and the λ are related, however:


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Q Q LL

λ λ= ⇒ =

So:

02

ln b

a

LCrr

πε λ

λ=

⎛ ⎞⎜ ⎟⎝ ⎠

02

ln b

a

LCrr

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

(3)

In using Gauss’s law to get E , we assumed that the capacitor was very long (i.e., essentially infinitely long) compared with the spacing between the plates. If L were literally infinite, the capacitance would be infinite, from (3). No real cylindrical cap has an infinite capacitance, however, so it’s common to speak of the capacitance per unit length instead:

02

ln b

a

CL r

r

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

(4)


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Case 3: Spherical Capacitor From Gauss’s law, the electric field between the plates is:

20

ˆ4

QE rrπε

=

Now consider moving along a radial path from a point b on the outer sphere (at br r= ) to a point a on the inner sphere (at ar r= ). Then:

( )2 20 0 0

1 1 1ˆ ˆ4 4 4

f a a

i b b

r r r

a br r r

Q Q QV E dr r drr drr r r rπε πε πε

⎛ ⎞ ⎛ ⎞⎛ ⎞Δ = − ⋅ = − ⋅ = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

∫ ∫ ∫

Or:

04

b a

a b

r rQVr rπε

⎛ ⎞−Δ = ⎜ ⎟

⎝ ⎠

So the capacitance is:

04 a b

b a

r rCr r

πε⎛ ⎞

= ⎜ ⎟−⎝ ⎠ (5)


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Capacitors in Series and Parallel Key Terms • A component of a circuit is a part of the circuit (capacitor, battery,

resistor, etc.)

• A terminal of a component is a conducting lead to which other components can be connected.

• A node is the junction of two or more terminals of components.

• Two components are said to be in series if they share exactly one node, with nothing else connected to that node.

• Two components are said to be in parallel if each terminal of one component is connected to a unique terminal of the other component.

Note: • It is possible for components to be connected so that they are neither

in series nor in parallel with one another.


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Caps in Parallel Consider two capacitors, 1C and 2C , connected in parallel across a battery that supplies voltage V . The two caps have the same voltage across them, namely, the battery voltage, V . However, they store different charges, 1Q and 2Q , because the capacitances are different. The equivalent capacitance, eqC , of any network of capacitors is the single capacitor that could replace the entire network and store the same total charge. What is eqC for the parallel combination of 1C and 2C ?


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To answer this question, consider the “defining equation” for the capacitance, Q CV= , applied to 1C and 2C . For 1C : 1 1 1 1Q C V C V= = and for 2C : 2 2 2 2Q C V C V= = Now add these two: ( )1 2 1 2Q Q C C V+ = + ( )1 2totQ C C V= + This looks like tot eqQ C V= , with: 1 2eqC C C= + This means that if the parallel combination of 1C and 2C were replaced by a single capacitor eqC connected across the same battery voltage V , this single cap would store the same total charge as the parallel combination of 1C and 2C if eqC were chosen to be simply the sum of 1C and 2C .


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In general, for N caps in parallel across a battery supplying voltage V , we would have:

1 1

2 2

N N

Q C VQ C V

Q C V

==

=

Adding these, as before, I get: ( )1 2 1 2N NQ Q Q C C C V+ + + = + + + ( )1 2tot NQ C C C V= + + + This looks like tot eqQ C V= ,

with: 1 2eq NC C C C= + + + (6) So, caps in parallel just add. For N equal capacitors, C , in parallel: eqC NC= (7)


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Caps in Series Consider two caps, 1C and 2C , initially uncharged, connected in series. Now imagine connecting 1C and 2C in series with a battery (supplying voltage V ) and a switch, S , that is open (not conducting) until time 0t = , then closed (conducting) at 0t = . The bottom plate of 1C , the top plate of 2C , and the wire connecting them constitute one “hunk” of conducting stuff that is electrically neutral. This conducting “stuff” must stay electrically neutral since there is no way for charge to cross the insulating gap between the plates of 1C and

2C (well, unless you do something rather extreme to the caps, that is).


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When the switch S is closed, positive charge Q+ flows onto the top plate of 1C . This positive charge draws free electrons to the bottom plate of 1C . This migration of free electrons to the bottom plate of 1C stops when the charge on the bottom plate of 1C is Q− . However, this charge Q− has to come at the expense of charge someplace else. This means that there must be Q+ on the top plate of 2C . And this draws Q− to the bottom plate of 2C . The upshot: caps in series store the same charge.


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What equivalent capacitance eqC could replace the series combination of

1C and 2C and store charge Q? To answer this question, apply Q CV= to 1C and 2C :

1 1 11

QQ C V VC

= ⇒ =

2 2 22

QQ C V VC

= ⇒ =

Now add these:

1 21 2

1 1V V QC C

⎛ ⎞+ = +⎜ ⎟

⎝ ⎠

But 1 2V V V+ = , the battery voltage, so:

1 2

1 1V QC C

⎛ ⎞= +⎜ ⎟

⎝ ⎠

or:

1 2

1 1

1

C C

Q V⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠


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This looks like eqQ C V= , with the equivalent capacitance being:

1 2

1 1

1eq

C C

C =+

For this special case of two caps in series, we can rewrite eqC in an especially simple form:

1 2

1 2eq

C CCC C

=+

(8)


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For the general case of N caps in series:

1 1 11

2 2 22 1 2

1 2

1 1 1N

N

N N NN

QQ C V VCQQ C V VC V V V V Q

C C C

QQ C V VC

⎫= ⇒ = ⎪⎪⎪= ⇒ = ⎛ ⎞⎪⇒ = + + + = + + +⎬ ⎜ ⎟

⎝ ⎠⎪⎪⎪= ⇒ = ⎪⎭

1 2

1 1 1

1

NC C C

Q V⎛ ⎞

= ⎜ ⎟⎜ ⎟+ + +⎝ ⎠

This looks like eqQ C V= for a single cap, eqC , given by:

1 2

1 1 1

1

N

eqC C C

C =+ + +

This is usually written in the somewhat simpler form:

1 2

1 1 1 1

eq NC C C C= + + + (9)


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Finally, for N equal caps, C , in series, (9) gives:

terms

1 1 1 1 1

eqN

NC C C C C

⎛ ⎞= + + + = ⎜ ⎟⎝ ⎠

eqCCN

= (10)

This says that if we put N equal caps in series, the equivalent capacitance is one-N th of any one of them.


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Energy Stored in a Capacitor Because there’s some electric potential energy elecU associated with any configuration of charges, caps store energy as well as charge. But how do we write down an expression for this energy in terms of Q , C , and V ? To do this, imagine a capacitor C that’s initially uncharged. Now imagine charging up the cap by “grabbing” an infinitesimal bit of charge dq and lifting it from the lower plate to the upper plate. For the first little bit of charge moved, no work is required. (There is not yet any electric field that we have to “fight against” to lift dq .) But for each subseqent dq , we will have to do some work. If we imagine lifting dq quasistatically, then the infinitesimal amount of work, dW , required is equal to the infinitesimal amount of potential energy, elecdU , gained by dq . But

elecU Vq= , so: elecdU V dq= Let q be the magnitude of the charge on either plate of the cap at any time when the voltage between the plates is V . These are related by the defining equation for the capacitance:


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qq CV VC

= ⇒ =

So:

elecqdU dqC

=

To get the total energy elecU stored in the cap after a total charge Q is moved from the lower plate to the upper plate, we need to integrate:

0

Q

elec elecqU dU dqC

= =∫ ∫

2

2elecQUC

= (11)

Eq. (11) can be written in two alternative forms using Q CV= :

212elecU CV= (12)

and:

12elecU QV= (13)


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Energy Density Stored in Electric Field The energy density, u , stored in a capacitor is the energy per unit volume:

Volume

elecUu ≡ (14)

The volume bounded by the capacitor plates is: Volume Ad= , in which A is the area of the plates and d is the distance of separation. So:

elecUuAd

= (∗)


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It is possible to express the energy density u only in terms of the magnitude E of the electric field between the plates. To do this, recall that, for a parallel-plate cap, we found:

0 0

QEA

σε ε

= =

Rearranging this for Q gives: 0Q AEε= And the capacitance of a parallel-plate cap was found to be:

0ACd

ε=

Plugging these into (11) gives:

( )( ) ( )

20 2

00

12 / 2

AEU Ad E

A dε

εε

= = ,

and plugging this into (∗) above gives the energy density in terms of E :

( ) ( ) 201/ 2 Ad E

uAdε

=

20

12

u Eε=


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Notice that this result doesn’t depend on any other properties of the cap except the electric field between the plates. This suggests that the energy stored by a capacitor can be thought of as energy stored in the electric field between the capacitor plates. In fact, it turns out (won’t show this here) that the expression for the energy density just derived is correct no matter how the electric field gets there. Whenever there is an electric field in any region of space, there is some energy per unit volume stored in the field, and this energy density is given by the formula above. As a reminder that this energy density is stored in the electric field E , I will call this energy density “ Eu ”:

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12Eu Eε= (15)


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Capacitors With Dielectrics In deriving all of the results so far for capacitances of the parallel-plate, cylindrical, and spherical caps, as well as the energy stored in a capacitor, I haven’t said anything about the material used for the dielectric. This means that I assumed the dielectric was just a vacuum – literally nothing at all. But what happens if there is a dielectric present? To answer this question, imagine a parallel-plate cap – initially with no dielectric between the plates – that has been charged up and then disconnected from any other components (batteries, etc.). Let 0Q be the magnitude of the charge on either plate of the cap and 0E be the magnitude of the electric field between the plates. Now suppose you insert a dielectric between the plates. When you do this, atoms and molecules within the dielectric become polarized. This polarization gives rise to an additional electric field within the dielectric – and induced electric field, indE , that opposes the field 0E due to the charges on the plates, partially cancelling 0E .


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The net electric field E within the dielectric is, therefore, some fraction of 0E . We express this fact mathematically as follows:

0EEK

= (16)

The factor K is called the dielectric constant of the material used for the dielectric. Note: • K is a property of the material; it has to do with the extent to which

the material becomes polarized when placed in the field 0E . That is, K depends on how tightly bound the electrons in the material are to their nuclei.

• 1K ≥


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Effect of Dielectric on Capacitance How does the presence of a dielectric between the plates affect our earlier formulas for the capacitance C of the parallel-plate, cylindrical, and spherical capacitors? Consider the parallel-plate cap just discussed. We’ve seen that inserting a dielectric reduces the field between the plates by the factor K . But E is related to the magnitude of the voltage between the plates by: V Ed= So if E is reduced by the factor K , then V is reduced by the same factor:

0VVK

= , (17)

in which 0V means the voltage between the plates before the dielectric was inserted.


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From the defining equation for the capacitance, the capacitance after the dielectric is inserted is:

QCV

= ,

in which Q means the magnitude of the charge on either plate with the dielectric inserted. Because the cap was assumed to be disconnected from all other components before the dielectric was inserted, there is no way – if the dielectric does not conduct – for charge to get off of one plate and onto the other plate. Therefore: 0Q Q= , and the capacitance with the dielectric inserted is:

( )

0 0 0

0 0

Q Q QC KV V K V

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

0C KC= (18) This says that inserting the dielectric boosts the capacitance by the factor K .


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For the parallel-plate cap, recall that we found the capcitance without a dielectric to be:

0 0ACd

ε=

With a dielectric present, then, the capacitance is:

0AC Kd

ε= (19)

The product 0Kε is called the permittivity, ε , of the material used for the dielectric: 0Kε ε≡ (20) With this, we can write C as:

ACd

ε= (21)

For any capacitor, then, we can just use this same trick: to modify the formula to reflect the presence of a dielectric, just replace 0ε with ε .


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Effect of Dielectric on Energy Density Recall that the energy density stored in the electric field in a region of space where there is no dielectric present was:

20

12Eu Eε=

By a procedure similar to what we just did for the capacitance, we can derive the expression for the energy density stored in the electric field in a region of space where there is a dielectric. I won’t do this here, but the result turns out to be that we just replace 0ε with ε :

212Eu Eε= (22)


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Induced Surface Charge Density We’ve seen that when a dielectric is inserted into a capacitor, the atoms and molecules that make up the dielectric are polarized. This gives rise to some induced charge per unit area on the surfaces of the dielectric that are facing the plates. Let the induced surface charge densities on the surfaces of the dielectric be called indσ and indσ− . How are these related to σ and σ− , the surface charge densities on the capacitor plates? Consider a Gaussian “pillbox” that has two sides parallel to the plates – one buried in the conducting, positive plate of the capacitor, the opposite one buried in the dielectric. The only non-zero contribution to the flux through this pillbox is from the side that is buried in the dielectric. The flux through this side is EA. Therefore, Gauss’s law says:

( )0 0

indencl AQEAσ σ

ε ε−

= =

0

indE σ σε−

=


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But we also know that

0

0

EEK K

σε

= =

So:

0 0

ind

Kσ σ σ

ε ε−

=

which gives (after a little algebra):

11ind Kσ σ ⎛ ⎞= −⎜ ⎟

⎝ ⎠ (23)

Notes: • 1K ≥ , so indσ is always some fraction of σ .

• For materials that are easily polarizable, K is large. For materials with larger K , indσ is closer to σ than for materials with smaller K .

Documents

Chapter 24: Capacitance and Dielectrics · Chapter 24: Capacitance and Dielectrics Capacitors and Capacitance A capacitor (“cap”, for short) is a device for storing charge. A