26-1 (SJ, phys1120P)
CAPACITORS: are any physical objects that can store charge. I guess they're
called capacitors because they have a capacity to store charges(!?) They're
useful - your flash camera has a big capacitor in it, to store up a bunch of
charges (which it quickly releases to flow through the flash bulb!)
Any time you have two conducting surfaces near
each other, you have a capacitor. (We call the
surfaces "plates", even if they're funny shapes)
+Q
-Q
When we say a capacitor has "charge Q", we mean
+Q on one plate, and -Q on the other, like in the picture here. (Of course, the
net charge is really zero, but we still say it’s “charged” up!)
These two plates have a voltage difference between them. (It takes WORK to
move a little test charge from one plate to the other). Let's think about this:
What if you double Q on both plates?
=> you'll double the number of E field lines
=> which means you've doubled E
=> since voltage difference between the objects comes from integrating E
(see last chapter!) that means: doubling Q doubles V. They're proportional!
Q ∝ V , or Q=CV .
(Of course, you can also turn it around and define C= Q/V)
• C is a constant, and called the "capacitance".
• C depend on the precise shapes, and the materials involved, but not on Q.
• A big C, ("big capacitance") means you can store a LOT of charge for a
SMALL voltage difference.
• In a sense, big C means it's "easy" to store up a bunch of charges.
A capacitor can store charge (and energy) for you - that's why they're useful.
Think of a closet with lots of bowling balls up on the shelf at high gravitational
potential. You can let them fall down whenever you like, to do lots of work (or
damage) later. Your closet shelf has only a limited capacity to hold bowling
balls. Similarly, metal plates of a given voltage difference have a certain
limited capacity to hold charge, and that's what "C" tells you.
The units of capacitance are [Coulombs/Volt] = C/V .
We call 1 C/V = 1 Farad = 1 F.
(The units are getting confusing now! 1 V = 1 J/C, so 1 F=1 C^2/J too.)
• 1 F is a really big capacitor! It would hold 1 C (a lot of charge!) with only
one volt difference between the plates. Most normal capacitors are more like
micro or nano Farads.
26-2 (SJ, phys1120P)
My favorite example is the "parallel plate capacitor".
Last chapter we showed the potential difference between d
the plates is ΔV = E d .
We often drop the " Δ " for convenience, and say: V=Ed.
(It's a bad notation, but standard.) V really means "voltage difference" here!
+Q
-Q
For large parallel plates, area A, the E field between them is given by a simple
formula (derived last chapter from Gauss' law):
E = σ/ε0 = (Q/A)/ε0
The formula says the more Q you have, the bigger E is. (That makes sense,
yes?) Also, the bigger the area of the plates, the more the charge is spread out,
and that should weaken the E field, so the A (area) dependence in that formula
also seems reasonable. The formula doesn't depend on "d" (how far apart the
plates are.) This is perhaps a bit surprising, but it’s correct: E is UNIFORM, it
doesn't change with distance from either plate!
We know V=Ed for a parallel plate capacitor.
r r
(Why? Let's review: it's because ΔV = − ∫ E ⋅ dr . That's a key formula from last chapter!
f
i
For a parallel plate capacitor, E is constant, this just gives ΔV= -E. d
If you integrate FROM the negative plate TO the positive one, i.e. moving "against" the
direction of E, that dot product will be negative, because E points opposite d.
€ to give positive ΔV = Ed. Which is what I claimed)
The two negatives cancel
Combining this with the formula for E (just above) gives V =
1Q
d.
ε0 A
Finally, C = Q/V (that defines C, remember)
So that last equation for V yields (can you check the algebra yourself?)
C=
€
ε0 A
d
(This formula is only for ideal, infinite,€parallel plate capacitors!)
• As advertised, C is a constant depending only on the size and shape.
• Since V=Ed, and E is uniform: the closer the plates, the smaller V is.
Smaller V (for a fixed charge) means C=Q/V is BIGGER.
If you can get plates close together, you get a better (bigger C) capacitor!
So that "1/d" dependence in the formula makes physical sense. (If the two
plates are really close, the +'s on one plate are tightly attracted to the -'s on the
other. If I may anthropomorphize, they are happy, and you can fit LOTS of
them on the plate! That's what big C means! )
Similarly, if the area of the plates is bigger, the formula says you have more
"capacity" to hold charge. That makes good intuitive sense to me; you have
more area to spread those charges out onto, so you can easily hold more
charge. (If they're allowed to spread apart from each other, they're happier :-)
26-3 (SJ, phys1120P)
+Q
-Q
If you connect a wire up to a capacitor
(like on the left) the charges on the
capacitor plates are free to travel. The +
and - attract one another, the capacitor
will discharge quickly, possibly with
some nice sparks. There's energy in
there!
0
0
It takes work to charge up a capacitor!
Consider e.g. a battery charging up a capacitor.
A battery "wants" the plates to be at a certain
voltage difference V. (E.g., a 12 V car battery
always tries to make things connected to it to be
12 V apart.) To do this, it forces charges to go
from one plate onto the other plate, fighting
against an E field all the while.
+Q
Battery
-Q
How much work does the battery do, charging the capacitor up from 0 to V?
The first charge moves easily, it takes almost no
work, because at first the plates have no charge.
(There's no E field, and charges are pretty much
free to move around wherever they want.)
Battery just
pushed the 1st
charge from one
plate to the other
On the other hand, if the capacitor is fully charged (with a voltage V across it)
you'd expect that moving Q charges across that voltage V would cost energy =
Q*V. (Because after all, V means "potential energy per charge). But the real
answer is the average of 0 and QV (on average, while charging , the voltage is
only V/2!).
(See next page for a rather more formal argument to get the same answer!)
Bottom line:
Energy to charge up = Energy stored in capacitor = U = QV/2 .
Remember, the symbol "U" is used for potential energy.
(Don’t confuse “potential energy” with “potential”!!)
Since Q=CV, we can rewrite this as U = CV^2/2, or U=Q^2/(2C).
Depending on what’s held constant, these forms are sometimes useful.
Please don't confuse the symbol C (Capacitance) with C (Coulombs!)
26-4 (SJ, phys1120P)
How much work does it take to charge a parallel plate capacitor from 0 to
Q? This will also be the total energy stored in that charged capacitor:
Each charge that you move from the bottom plate to the top plate has to move
from a low potential to a higher potential. That takes a little work, call it dW.
Each time you move one more charge, the E field gets bigger, and the potential
difference between the plates increases. So the next charge will need even
MORE dW!
The total work =
∑ dW = ∫ dW
=
∫
q=Q
q= 0
ΔV dq
(Think about that last one: I'm saying that the work dW to move a little bit of
charge dq from one plate to the other, with voltage difference ΔV, is dq ΔV.
That's from
€ the basic
€ definition: voltage difference = (potential energy difference)/ charge.
q
d dq
Aε0
(The first step came from remembering ΔV = Ed, and the second step from
So, total work =
∫
Q
E d dq =
0
∫
Q
0
recalling our formula for E, derived 2 pages ago.)
€
We can
do this integral: most of the stuff in there is constants:
d
U = total energy stored = total work =
Aε0
∫
Q
0
d q2
q dq =
Aε0 2
2
=
0
dQ2
2Aε0
We can rewrite this in several ways. First, remember that we showed
Capacitance C =
ε0 A
, so (convince yourself!) U = (1/2) Q^2/C.
d
€
€
Here's another way, which is very cool. Remember that E = (Q/A)/ε0
So it's
€ also true that U =
d
d
1
2
Q2 =
(ε0 EA) = ε0 (Ad) E 2
2Aε0
2Aε0
2
Think about that last line: since (A d) is the volume of the capacitor,
we've just seen that U = (volume) *
€
This leads me to think of
1
ε0 E 2 .
2
1
ε0 E 2 as the "energy per volume".
2
It implies that ELECTRIC FIELDS
STORE ENERGY.
€
It's a very important and different way of thinking about things. Instead of
thinking of the charges
as storing the energy, think of that E field as storing it!
€
This is no proof, but it's a very general result. Anywhere you have E field, you
have energy, and the energy/volume is
€
1
ε0 E 2 . Even in empty space!
2
26-5 (SJ, phys1120P)
Recap: Energy stored in any capacitor is U =
1
1 Q2 1
= CV 2 = QV
2
2
2 C
These can each be useful, depending on what you know/are interested in.
(E.g. If you have a battery involved, you'll probably have a fixed "V" between
the plates, and so U=
1
€
€
useful.
)
CV 2 might be most
2
€
But they are all very general, true for any capacitor geometry.
1
ε0 E 2 . E fields store energy!
2
€
εA
BUT: only for parallel plates can we argue C = 0 . (NOT a general formula!)
d
Also general: U = (volume)*
Example: A 12 V car€battery is hooked up to a capacitor with plates of area
0.3 m^2, a distance 1 mm apart. How much charge builds up on the plates?
€
Answer: The battery will charge the capacitor up until the voltage is 12 V.
Since Q=CV, we only need to find C. But we know that
C = ε 0 A / d . So we're all set:
Q = CV = ε 0
A
C2  0.3 m 2 
V = 8.85⋅10 −12

 12V
d
N m 2  10 −3 m 
C2
V = 32 nC
Nm
(I used 1 V=1 J/C: check for yourself that that nasty combination of units simplifies like I
claimed, to Coulombs) . Also note the capacitance involved here:
εA
C2
= 8.85 ⋅10−12
C= 0
(.3 m2 )/(.001m) = 2.6 nF . (Fairly small.)
2
d
Nm
= 32 ⋅10 −9
€
How much energy is stored in the capacitor now?
U=Q*V/2 = 32E-9 C * 12V / 2 = 0.2 micro Joules. Not so much!
You
€ must use high technology to get big capacitances: you need BIG areas and
VERY small distances between the plates. (1 mm is not really that close!)
C2
Coul
Coul
Coul
Farad
=
=
=
=
Units: [ ε0 ]=
2
Nm
J
Nm
m
m
m Volt m
Coul
Coul
εA
When I see a formula like C = 0 it makes sense to think of ε0 as being
€
d
"Farads/meter".
It also tells me that I need A/d = 10^11 meter to get a
€
capacitance on the order of 1 Farad!!!
€
€
Reminder: Where exactly is the energy stored, in a capacitor? It's stored in
the E field! Wherever you have electric fields, there is stored energy.
The energy is stored in the "space" between the plates, in the form of electric
field energy.
26-6 (SJ, phys1120P)
In diagrams, we will represent capacitors with a simple symbol like this, even
if the capacitor in reality isn’t physically “parallel plates”. (Sometimes the “C”
is left off the picture too)
Parallel and series capacitors:
We might combine capacitors: e.g, we could take two capacitors
C1
C2
and connect them side by side, called "parallel", like this:
(Solid lines represent "ideal wires": perfect conductors!)
Remember, in electrostatics, conductors are equipotentials
throughout, so every point along an ideal wire is at the exact SAME voltage!
In our diagrams, we often indicate a voltage drops as shown to the
left: here, the top plate is at V(top), the bottom plate has a voltage
V
V(bottom, and what we mean by "V" in the fig is instead really
ΔV = V(top)-V(bottom)
I claim that our two parallel capacitors are really
equivalent to ONE capacitor, with a "total" or
"effective" or "equivalent" capacitance given by
C(parallel) = C1+C2.
C1 C2
equiv to
C//=C1+C2
Informal proof: Suppose C1 has charge Q1 on it, and C2 has charge Q2 on it.
Let's call the voltage difference between the top and bottom of the picture "V".
Notice that this voltage difference must be the same for both capacitors (!!)
So V = Q1/C1 = Q2/C2, or if you prefer, Q1 = C1V, and Q2=C2*V.
Thus Q1+Q2 = (C1+C2)*V, or V = (Q1+Q2)/(C1+C2) = Q(tot)/(C1+C2)
I now just read off the denominator: C(effective) = C(parallel) = C// = C1+C2
For capacitors in series:
I claim 1/C(series) = 1/C1 + 1/C2.
C1
equiv to
C2
C(series)
Informal Proof: This time, since C1 and C2 are connected, I claim Q1 = Q2.
(Has to be! Think about it - the section in the middle starts off uncharged, and
then as charges shift around at the top and bottom, charge can SEPARATE to
go up or down, but whatever -'s go up, the same # of +'s must go down!)
Now suppose you have a total voltage drop V from the top to the bottom.
V = V1+ V2 (the total drop = the sum of the two drops added up!)
And since Q1 = C1*V1 and Q2=C2*V2: V = V1+V2 = Q1/C1 + Q2/C2
But calling Q1=Q2=Q tells us V = Q/C1 + Q/C2 = Q(1/C1 + 1/C2)
Finally, we set V = Q/C(series) (which really DEFINES C(series)!)
and I see 1/C(series) = 1/C1+1/C2.
26-7 (SJ, phys1120P)
DIELECTRICS:
Any insulating material (paper, plastic, etc) can be called a dielectric. Most real
capacitors have dielectric materials between the plates.
This helps to keep the plates apart! After all, the plates are +Q
oppositely charged, and so attract each other strongly. If
they ever touched, the capacitor would discharge, or "short
Dielectric
out", and be useless.
So, with a dielectric in there, you can make "d" quite small, -Q
and remember that helps make the capacitance bigger.
But the main reason for putting a dielectric in there is something different: it
actually decreases the E field in the capacitor! Why?
Real materials (like dielectrics) are "polarized" by a strong E field.
That means the E field in the capacitor effectively pulls
some "-" charge towards the top of the dielectric (nearer
+Q
the "+" plate), and some "+" towards the bottom (nearer
the "-" plate) In the region throughout the middle, it looks
like some of the "Q" around you has been weakened, or
shielded, or canceled out.
The net effect is that E is reduced throughout the dielectric.
-Q
And since V=Ed, the voltage between the plates is reduced.
Since C=Q/V, if Q is fixed and V is reduced, C gets bigger.
In this way, dielectrics make the capacitance bigger.
(Your capacitor can hold MORE charge for a given voltage, with a dielectric in
there, because the surface of the dielectric effectively shields out some of the E
field from the middle region)
Recall the formula C =
1 A
A
= ε0
.
4πk d
d
That's if there's no dielectric. If there IS a dielectric, we just argued C is bigger.
It turns out for most dielectric materials, C is bigger by some constant factor
which depends only on the material, i.e.
1 A
A
Cwith dielectric = K
= K ε0
4πk d
d
K will be some constant for any given dielectric material Bigger K means you
get a bigger C. (Paper, e.g. has K=3, roughly.)
26-8 (SJ, phys1120P)
Capacitors are everywhere: in circuits, radios, computers, TV's...
It's handy to have a "charge storage" device!
The final (brief) topic of this chapter is a real world application of capacitors:
CRT's, or "cathode ray tubes".
vertical capacitor
hot metal "-" cathode
screen
"+" anode (with holes)
horizontal capacitor
A "cathode ray" is an old-fashioned name for electrons. The "cathode" is a
heated piece of metal, set at a very low voltage. The "anode" is set at a high
potential (so the "cathode" and "anode" basically form a capacitor) Electrons
boil off the hot cathode, and then they are accelerated towards the high voltage
("+" charged) anode. The anode is a grid with lots of holes, so many electrons
can fly right on by and cruise towards the screen.
They pass through a pair of capacitors (vertically and horizontally oriented)
which have a voltage that "sweeps". As electrons pass through these
capacitors, they feel the force from the E field, which bends the path of the
electron. Since the voltage is swept, the electrons are also swept. They fly on
by, and hit the screen, which glows where the electrons hit. So you see them
sweeping by, and this makes the whole screen glow. By turning the anode on
and off, you can make the electrons go through or not, thus making bright or
dark spots, which allows you to make an image.
This device is used in old computer monitors and TV, oscilloscopes, EKG
traces, you still see them all over. We’ll play with an oscilloscope in lab, where
you can control the voltages by hand and mess around with manipulating
electron beams...