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## Class 12 Physics (India)

### Course: Class 12 Physics (India) > Unit 2

Lesson 9: Capacitance & parallel plate capacitors# Capacitance

Introduction to the capacitance of a two-plate capacitor. . Created by Sal Khan.

## Want to join the conversation?

- Why do we need a capacitor?(155 votes)
- Or, in the case of musical equipment, they store charge that the circuit requires to operate other components. Once the ON switch is hit, they release the charge stored to fire up parts of the circuit. Like a quick burst of power (i.e. camera flash). Also, they're used in Equalizers, crossovers and tone control as they're important to create filters. Especially important in vacuum tube amplifiers.(135 votes)

- Why do the field lines at3:10bulge out when they get to the edges?(31 votes)
- just imagine that there were a positive charge kept near the edge of the positively charged plate then the charge would be pushed outward because the horizontal components due to the charges on the left are not cancelled.but when is more near to the negatively charged plate there would be a net attractive force on it and it would curve inwards hence giving the field lines that kind of shape.

the field lines at the edges of a capacitor plate have the same shape as of the field lines due to an electric dipole.you might want to see earlier videos to get that(29 votes)

- Whats the difference between a capacitor and a battery?(12 votes)
- Batteries use an ongoing chemical reaction to generate a potential difference between two terminals. Capacitors can store charge like a battery, but have no internal chemistry to maintain a potential; however they can discharge very rapidly whereas batteries have a limit on how much current they can output. Once the charge in a capacitor has been used up, it cannot be replenished until the capacitor is charged by an external current. Capacitors have high power density, but low energy density. Batteries have low power density, but high energy density.

If someone can invent a device which has both high power and energy density, they will be very rich.(46 votes)

- why a capacitor does not allows dc current and why it allows ac(6 votes)
- You can imagine a capacitor like a "balloon", and the DC as a man blowing on the balloon.

The balloon at the start allows the man to keep blowing into it but at some point the balloon can no longer stretch and no more flow of air can get in.

For AC you can imagine it as man who inhales and exhales in the balloon (Don't try this at home XD) The balloon expands as the man exhales and contracts while the man inhales the air he blew in. You can see that air is constantly flowing in and out of the balloon.

The analogy is an oversimplification but it gives you a rough idea of how a capacitor works for electricity in AC and DC. DC stands for Direct Current and so the electric current only flows in directly in one direction (man constantly blowing), while AC stands for Alternating Current where it alternates electricity from one direction to the opposite direction back and forth (man inhaling and exhaling). The flow of the air in my analogy is the flow of electricity and so there's an initial flow during DC when the capacitor is empty/discharged but soon doesn't allow a flow when its fully charged whereas the capacitor constantly charges and discharges during AC allowing a constant flow of electricity.(54 votes)

- At8:40he goes from work= charge*Electric Field*distance to work=Electric Field*change in joules. How did he arrive at the second definition? Would that not give you (N/C)*J, which is not the units that work is in.(8 votes)
- Since the charge is 1 in his example, he drops the "charge" from the work equation. It then becomes E*d, with units J. While it is easy to confuse d as distance, when placed next to the unit J, with the notation for an infintesimal part of the unit, this is not what was intended in the circumstance here! In other words, it is d=distance, not dJ="change in Joules."(9 votes)

- why we always talk about INFINITE charged plates?why infinite??(5 votes)
- Makes the problem easier. An infinite plate has a constant electric field everywhere. The field from a finite plate is actually much more complicated to solve for.(9 votes)

- How to know when is the Capacitor fully charged? Is there a Formula?(4 votes)
- A theoretical capacitor is never fully charged! It just keeps following an asymptote closer and closer to maximum charge. Graph the formula provided by Marco and you will see that Vc never stops getting closer and closer to V.

Real capacitors of course eventually stop accepting any measurable amount of current.(5 votes)

- Does current flow or not when capacitors are connected in a circuit?(2 votes)
- Raghav's answer is partially right.

DC current can flow through a capacitor when it is not fully charged. While he is correct that no current flows across the capacitor gap, a current will still flow through the circuit because as electrons accumulate on one side of the plate, they will repel the electrons on the other side of the plate through electrostatic repulsion causing a DC current to flow. Once the capacitor is fully charged, then no more electrons can accumulate and the DC current will be zero.(7 votes)

- In8:38, where did he get Joules from? It was 1C(E)d =E(d)J ? I that later he divides that J/1C = V, but where did the J come from?(3 votes)
- Work has to be in Joules, right? Work is force*distance, and force is qE and distance is d, so the units of qE*d must somehow be joules. Sal knows that he used SI units all along, so he already knows that the formula for work is going to give him Joules, so he writes Joules.

To see that indeed it does come out to be joules, note that the units of E can be expressed EITHER as N/C, or V/m (those are the same thing). If we take E in V/m, then you have (just using units )

work = C*(V/m)*m, which simplifies to C*V, and a volt is a joule per coulomb, so you end up with joules.

If you want to use the other route , you have C*N/C*m, which is n*m, which is, again, a joule.(5 votes)

- when do we get a uniform electric field and wen do we get a varying electric field???

what are the sources and how will we come to know??(3 votes)- There are certain ways by which you come to know that you've got a uniform electric field. Coming by the "lines of force approach", a uniform electric field will be obtained by the help of equidistantly placed,parallel straight lines. (Think of it, you know that the tangent to the lines of force,give you the direction,a straight line would be the best option). Or think of closely packed and distantly spaced field lines in a field. In the former, the electric field is obviously stronger,whereas in the latter,the electric field is weaker,thereby making the field non-uniform.(2 votes)

## Video transcript

We learned several videos ago
that if I had an infinite uniformly charged plane-- let me
draw one right here, and I won't draw it infinite and
I'll tell you why in a second-- that if we had an
infinite uniformly charged plane, and let's say this
one's positive, that the electric field generated
by it is constant. Those are the field lines. They should all be
the same size. And the strength of the field,
or the magnitude of the field, is equal to 2 times Coulomb's
constant times pi times the charge density of the plate. So if this is infinite-- so
what was charge density? We defined it when we proved
that this truly is a uniform electric field, but what
is charge density? Charge density is just the total
amount of charge divided by the area, or charge
per area. Well, if we have an infinite
plane, the area's going to be infinite, and so if this is a
constant number, this is also going to be infinite, so it's
kind of hard to work with. But what we also know is that
when we have a non-infinite plane that has some finite area,
that near the center of it and fairly close to it, it
approximates an infinite uniformly charged plane. So with that said, let's see if
we can figure out some of the properties of the voltage
and how the voltage relates to the charge. If we were to have two
parallel-- let me draw it before I say it, because I
think saying it'll just confuse it. So let's say I have two plates,
that plate-- and then I'll do this one in a different
color-- and I have that plate, and let's say
they're the same size and they both have area A. Let's say that I place plus
Q worth of charges here. So this is plus Q so this is
positively charged, right? I could draw a bunch
of charges here. Let's say this is minus Q, so
this is negatively charged. So what's the electric field
going to look like between these two? Well, it's essentially going to
be the combination of the electric field generated by
this plate on top of the electric field generated
by this plate. And they're both going to be
constant close to the center, assuming that they're
reasonably-- and let's say that they're d apart. Assuming that d isn't too big,
near the center we're going to have a constant electric
field. For example, this green one is
going to be generating-- its field lines are going to look
something like this. Near the center,
it's constant. These are meant to
look constant. Near the center, it'll look like
that, and it'll start to bulge out when you
get to the edges. Once again, near the center,
it's constant. That one should've
been at an angle. They start to bulge out,
and it'll look something like that. And similarly, this purple plate
will generate a constant electric field, and since it's
negatively charged, the field lines will be going towards it,
not away from it, so its field lines are going to look
something like this. Near the center, they'll be
constant, and its field lines are going to look something
like that. As you can see, they're going
to be of the same magnitude and in the same direction,
and they also will bulge out there. So the big picture is that you
just kind of have twice the electric field as you would have
if you just had one of these plates. So let's say we're operating
near the center of these, where we have roughly a constant
electric field and see if we can figure out the
relationship between the voltage across these two plates
and the area, and maybe the distance between
the two plates. So we know that the electric
field generated by any one of these charge plates-- I'll do it
in the blue of this color. So for the bottom plate right
here, what is the electric field generated? It's 2K pi times sigma. Sigma is just the total charge
divided by the area. So Q/A, right? And we know that the total
electric field generated by this one is going to be
essentially the same thing. I mean, we could say it's a
minus, a negative, because it's going towards it, but it's essentially the same thing. Because we see that they overlap
just drawing the field line, so the electric field
from that one, and we know that they go in the
same direction. If this was somehow-- well, this
is negative, so the field lines go towards it. So plus 2K pi and this is Q/A. We could have said minus and
then had a minus Q/A, but we know that they go into same
direction, so we know that they're going to be additive. And so we know that the total
electric field is going to be 4K pi Q/A. So now we know the
exact strength of the electric field. Let's see if we can figure out
the voltage difference between this point and this point. What was voltage difference,
just as a review? Well, voltage difference is
the electrical potential energy per charge if the charge
was here versus here. So how much more potential
energy per coulomb is there for a charge to be here
relative to here? So another way to view it is a
charge here, a positive charge here, because by default we're
always assuming a positive charge when we talk about
positive numbers and the direction of the field lines
or what the positive charge would do. So by default, a positive charge
here really wants to go up to this negative plate,
although we later learned that most of the movement in
electronics and electricity, it's actually the negative
charge that's moving. It's the electrons moving. But let's say we did
have a positive ion or a positive charge. The voltage is a measure of if
any charge is here, how badly does it want to move to this
point if it has a way to move? If we have air here or if we
have a vacuum here, it might be difficult or impossible
for it to move up here. But maybe if we were to connect
a wire where the charges could freely conduct,
then it will move. And the voltage is just
kind of how badly does it want to move? You could almost view it
as electrical pressure. And maybe I'll do a whole video
on trying to get an intuitive understanding of
voltage, because that really is probably the most important
thing to get an intuitive understanding of, if you ever
want to study electrical engineering or whatever. But anyway, back
to the problem. We know that the combined
electric field is this, right? It goes upwards in
that direction. So what is the electric
potential at this point relative to this point or
the potential difference from here to here? Well, that's the amount of
energy per charge it would take to move a positive charge
from here to there, right? Remember, electric potential
energy is the amount of work necessary to move a charge from
there to there, and then the voltage is how much
to do it per charge. Let me write that down. So the work necessary to move a
charge from there to there-- let's say a 1-coulomb charge, it
will be 1 coulomb times the electric field, because we're
always going to have to be going against the
electric field. So we have to apply an equal
and opposite force. So the force that is going to
be the electric field-- so far, this just generates this
force. coulomb times electric field, charge times electric
field, tells us the force on the charge, right? That's force, and
then we have to multiply that times distance. Force times distance. So we see the work necessary
is going to be the electric field times d joules-- the J is
joules-- and so what is the voltage difference or the
electric potential difference between this point
and this point? Let's call that point a. Let's call that point b. So Va minus Vb, which is the
voltage difference, that's essentially the electric
potential energy difference divided by the charge. Or, per charge. Well, here, the charge was just
1, so we can just divide by 1, and we see that it is
equal to the electric field times the distance. And the units are going to be
joules because we divided both sides by charge joules per
coulomb, or volts, right? That's just the units. So what does that equal? So the voltage difference-- so
we can say change in voltage. The voltage difference is equal
to the electric field, which we know is constant 4K
pi Q over A times distance. Or we could rewrite this. Let's see if we could write
Q as a function of V. So if we just do a little bit of
algebraic manipulation, we can get Q is equal to what? We would essentially divide
both sides by 4 pi Kd and multiply both sides by
A, so we would get A over 4K pi d voltage. And why is this interesting? Why did I go through all of
this work to get this relationship? Well, what it shows you, if you
look at this, if we assume that the area of the plates
aren't changing-- that's a constant; this is definitely a
constant-- and if we assume that the distance between the
plates don't change, what we see is that there's a
proportional difference between the voltage and the
amount of the combined charge in the plates. And that's interesting because,
before doing this, maybe voltage is somehow
proportional to the square of the charges or to the square
root, but now we know that it's directly proportionally. And actually, this term right
here has a name, and it is called capacitance. And so another way of rewriting
this, if we divide both sides by voltage, we get
Q/V is equal to 1 over 4K pi area over distance. And so what it essentially says
is that the amount of energy that-- well, actually,
I don't want to go into that yet. But for a given configuration,
and the configuration is defined by the area of the
plates and the distance-- for a given configuration, if I know
the amount of charge that I put onto the plates, if I did
a minus Q here and a plus Q here, I know the voltage
across the plates or vice versa. If I know the voltage across
the plates and I know its configuration, I know how much
charge there is, and this is called capacitance, and the
unit for capacitance is called the Farad. And if you become an electrical
engineer or even take a couple of electrical
engineering courses, you'll become very familiar
with this. And one other thing to point
out; this term right here, just so you know a little
bit of terminology. This term right here. This 1 over 4K pi, this is often
called epsilon nought, or just epsilon, and that's
called the permittivity of free space or permittivity
of the vacuum. And maybe in a future lecture
or a future video, I'll talk more about why it's
called that. But anyway, I'm already well
over the time limit. So I just wanted to give you a
sense of, one, that you can calculate the voltage across
what we call, in this case, a capacitor. It has capacitance. That voltage, you can
kind of view it as the electric pressure. How bad does the charge here
want to move here? And if you put a wire here,
you'll learn in a second-- not in a second, in several
videos-- that that charge will flow. Or actually the negative charges
will flow this way and generate current. And we'll do that when we start
learning a little bit more about electricity. For any given configuration,
it has a corresponding capacitance, and then given
that capacitance, if I put some amount of charge, I can
figure out the voltage, or if I know there's some voltage, I
can figure out the charge. Anyway, I will see you
in the next video.