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## MCAT

### Unit 8: Lesson 13

Electrostatics# Electric field

We can think of the forces between charges as something that comes from a property of space. That property is called the electric field. Created by Sal Khan.

## Want to join the conversation?

- At0:42, how do two protons or positive charges repel each other? They do repel each other, but how?(32 votes)
- For answering this you must know Feynman's Diagram which shows possible ways of how and why does positive and positive charges repel(5 votes)

- Is it possible to have vector addition of gravitational and electric field to get a total field ?(14 votes)
- No, it isnot that simple. Physicists have been trying to unify the forces into a single description for over 100 years.(33 votes)

- In a way, does Coulomb's Law mean that every particle, no matter how far away they are from each other, has some effect/force on the other?? Just that its too small to really make a difference...(20 votes)
- Why positive electric fields of lines are radially outwards?(10 votes)
- The test charge (a unit charge that is used to test the effect of the source charge) is assumed to be positive. So, the source charge, which is the charge from which the electric fields are generated, pushes the test charge outwards. And so, the electric field lines are radially outwards.

Totally sure now. Hope that helped. :)(17 votes)

- I did not get what exactly an electric field is? Can someone give a simple definition?(12 votes)
- It has been awhile. But I just saw this video.

Basically Electric field is the surrounding of the charges particles. You can imagine the charge particle is center of circle, the electric field will be everything in the circle(3 votes)

- The Lightning Which We See In The Sky Is Said To Be A Result Of Static Electricity And Static Electricity Does Not Move Or Flow Then How Does Electricity From Clouds Reach The Ground(6 votes)
- As i understand it, due to static electricity elektric charges will be build up. Now you have elektric voltage, which, when it's high enough, will unload into a spark, or lightning. When the charge in the clouds is high enough it can even result in a spark between cloud and ground, because of the polarity of the ground.

Elektricity is the result of moving electrons. So what you see as lightning are actually moving electrons towards for example the ground.

Btw we dont fully understand how lightnings are created.(6 votes)

- If the electric field lines generated by a positively charged particle are radially outwards and those of an electron are radially inwards,then this is visualised as though the positively charged particle's electric field force is pushing the negative charge further away from itself while being simultaneously pulled towards the negative charge whose electric field lines point inwardly towards itself.Right?What I'm mainly confused about is that if opposite charges attract each other,then the respective electric fields' force vectors of the positive and negative charges should ppint inwardly towards each charge;say A and B are two oppositely charged particles,hence both particles are pulling upon each other,shouldn't we represent the force vector of the force of A on B as pointing inwardly towards A and the force vector of B on A as pointing inwardly towards B.If that's true,then why did Sal at11:11schematize the electric field lines issued from A (which is +ve) point outwardly on B(which is -ve) as though it is repelling it.

Thanks in advance(4 votes)- No problemo.

Most important thing in your work is to understand what the arrows indicate.

The direction of the arrow indicates the directon of the force acting on 'a positive charge'.

Thats why they are 'outwards' for a proton and 'inwards' for an electron.

and thats why a proton sitting in an electrons electric field would feel a force attracting it towards the centre of the electron. because the arrows point in and the proton is positive...

and for an electron sitting in a protons electric field, if the arrows from the proton point outwards, then the force is inwards... this is because the arrow is the direction of force that would act on a positive charge.... electron is negative, so the force is in opposite direction...ie inwards.

ok??(5 votes)

- When u put distance instead of radius.. we actually measure the distance between the centres of the charges or the distance between their respective outer surfaces ?(3 votes)
- A charge is actually represented by a single point. Therefore distance is measured from thecentres of respective charges(5 votes)

- To explain the charge interaction between a charged and neutral object we use the concept of polarisation. So if I talk about a positively charged balloon and a wall, then the electrons are attracted towards the balloon and get crowded on the wall near the balloon. Why don't they enter the balloon?(3 votes)
- Because the balloon is made of rubber, and charge can't easily move onto or through insulating materials like that. Also, the charges are still attracted by the positive charges in the wall.(5 votes)

- I have a question in this video. About 6min 45s, you talked about the direction of the force. And I agree that the q particle will be move farther from the Q particle. But my question is, would the Q particle is also going to be move farther from the q particle?? and if it does, which particle is going to be exerted more force??(3 votes)
- Yes Q particle is also going to move away as q does.

I have tried to know how much force both charges exert on each other.

Lets say Q particle has 2 Coulomb charge and q has 1 Coulomb charge.You can calculate the electric field created by charges Q and q as E(Q)=F/q= k.Q/d2 and E(q)=F/Q= k.q/d2 respectively.In this way you get E(Q)=1.8*10^10 N/C

E(q)=9*10^9 N/C

We can calculate force exerted on charge q by charge Q as F=E(Q)*q= 1.8 x 10^10 N

And force exerted on charge Q by charge q as

F=E(q) * Q = 9*10^9 x 2= 1.8 x 10^10 N

So we got to know both charges experiences same Force due to their somehow different electric fields.

Let me know if it helps.My first ever answer on Khan Academy since I joined yesterday.(5 votes)

## Video transcript

Let's imagine that instead of
having two charges, we just have one charge by itself,
sitting in a vacuum, sitting in space. So that's this charge here, and
let's say its charge is Q. That's some number,
whatever it is. That's it's charge. And I want to know, if I were to
place another charge close to this Q, within its sphere of
influence, what's going to happen to that other charge? What's going to be the
net impact on it? And we know if this has some
charge, if we put another charge here, if this is 1
coulomb and we put another charge here that's 1 coulomb,
that they're both positive, they're going to repel each
other, so there will be some force that pushes the
next charge away. If it's a negative charge and I
put it here, it'll be even a stronger force that pulls it
in because it'll be closer. So in general, there's this
notion of what we can call an electric field around
this charge. And what's an electric field? We can debate whether it really
exists, but what it allows us to do is imagine that
somehow this charge is affecting the space around it
in some way that whenever I put-- it's creating a field that
whenever I put another charge in that field, I can
predict how the field will affect that charge. So let's put it in a little more
quantitative term so I stop confusing you. So Coulomb's Law told us that
the force between two charges is going to be equal to
Coulomb's constant times-- and in this case, the first
charge is big Q. And let's say that the second
notional charge that I eventually put in this field
is small q, and then you divide by the distance
between them. Sometimes it's called r because
you can kind of view the distance as the
radial distance between the two charges. So sometimes it says r squared,
but it's the distance between them. So what we want to do if we want
to calculate the field, we want to figure out how much
force is there placed per charge at any point around this
Q, so, say, at a given distance out here. At this distance, we want to
know, for a given Q, what is the force going to be? So what we can do is we could
take this equation up here and divide both sides by this small
1, and say, OK, the force-- and I will arbitrarily
switch colors. The force per charge at this
point-- let's call that d1-- is equal to Coulomb's constant
times the charge of the particle that's creating the
field divided by-- well, in this case, it's d1--
d1 squared, right? Or we could say, in general--
and this is the definition of the electric field, right? Well, this is the electric field
at the point d1, and if we wanted a more general
definition of the electric field, we'll just make this a
general variable, so instead of having a particular distance,
we'll define the field for all distances
away from the point Q. So the electric field could be
defined as Coulomb's constant times the charge creating the
field divided by the distance squared, the distance we are
away from the charge. So essentially, we've defined--
if you give me a force and a point around this
charge anywhere, I can now tell you the exact force. For example, if I told you that
I have a minus 1 coulomb charge and the distance is equal
to-- oh, I don't know. The distance is equal to let's
say-- let's make it easy. Let's say 2 meters. So first of all, we can say,
in general, what is the electric field 2 meters
away from? So what is the electric
field out here? This is 2, right? And it's going to be
2 meters away. It's radial so it's actually
along this whole circle. What is the electric
field there? Well, the electric field at
that point is going to be equal to what? And it's also a vector
quantity, right? Because we're dividing a vector
quantity by a scalar quantity charge. So the electric field at that
point is going to be k times whatever charge it is divided
by 2 meters, so divided by 2 meters squared, so that's 4,
right, distance squared. And so if I know the electric
field at any given point and then I say, well, what happens
if I put a negative 1 coulomb charge there, all I have to do
is say, well, the force is going to be equal to the charge
that I place there times the electric field
at that point, right? So in this case, we said the
electric field at this point is equal to-- and the units for
electric field are newtons per coulomb, and that
makes sense, right? Because it's force divided
by charge, so newtons per coulomb. So if we know that the electric
charge-- well, let me put some real numbers here. Let's say that this
is-- I don't know. It's going to be a really large
number, but let's say this-- let me pick
a smaller number. Let's say this is 1
times 10 to the minus 6 coulombs, right? If that's 1 times 10 to the
minus 6 coulombs, what is the electric field at that point? Let me switch colors again. What's the electric field
at that point? Well, the electric field at
that point is going to be equal to Coulomb's constant,
which is 9 times 10 to the ninth-- times the charge
generating the field-- times 1 times 10 to the minus
6 coulombs. And then we are 2 meters
away, so 2 squared. So that equals 9 times 10 to
the third divided by 4. So I don't know, what is that? 2.5 times 10 to the third or
2,500 newtons per coulomb. So we know that this is
generating a field that when we're 2 meters away, at a radius
of 2 meters, so roughly that circle around it, this is
generating a field that if I were to put-- let's say I were
to place a 1 coulomb charge here, the force exerted on that
1 coulomb charge is going to be equal to 1 coulomb times
the electric fields, times 2,500 newtons per coulomb. So the coulombs cancel out, and
you'll have 2,500 newtons, which is a lot, and that's
because 1 coulomb is a very, very large charge. And then a question you should
ask yourself: If this is 1 times 10 to the negative 6
coulombs and this is 1 coulomb, in which direction
will the force be? Well, they're both positive,
so the force is going to be outwards, right? So let's take this notion and
see if we can somehow draw an electric field around a
particle, just to get an intuition of what happens when
we later put a charge anywhere near the particle. So there's a couple of ways to
visualize an electric field. One way to visualize it is if I
have a-- let's say I have a point charge here Q. What would be the path of a
positive charge if I placed it someplace on this Q? Well, if I put a positive charge
here and this Q is positive, that positive charge
is just going to accelerate outward, right? It's just going to go straight
out, but it's going to accelerate at an ever-slowing
rate, right? Because here, when you're really
close, the outward force is very strong, and then
as you get further and further away, the electrostatic force
from this charge becomes weaker and weaker, or you could
say the field becomes weaker and weaker. But that's the path of a--
it'll just be radially outward-- of a positive
test charge. And then if I put it here, well,
it would be radially outward that way. It wouldn't curve the
way I drew it. It would be a straight line. I should actually use
the line tool. If I did it here, it would be
like that, but then I can't draw the arrows. If I was here, it would
out like that. I think you get the picture. At any point, a positive test
charge would just go straight out away from our charge Q. And to some degree, one measure
of-- and these are called electric field lines. And one measure of how strong
the field is, is if you actually took a unit area
and you saw how dense the field lines are. So here, they're relatively
sparse, while if I did that same area up here-- I know
it's not that obvious. I'm getting more
field lines in. But actually, that's not a good
way to view it because I'm covering so much area. Let me undo both of them. You can imagine if I had a lot
more lines, if I did this area, for example, in that area,
I'm capturing two of these field lines. Well, if I did that exact same
area out here, I'm only capturing one of the field
lines, although you could have a bunch more in between here. And that makes sense, right? Because as you get closer and
closer to the source of the electric field, the charge
gets stronger. Another way that you could have
done this, and this would have actually more clearly shown
the magnitude of the field at any point, is you could
have-- you could say, OK, if that's my charge Q, you
could say, well, really close, the field is strong. So at this point, the vector,
the newtons per coulomb, is that strong, that strong, that
strong, that strong. We're just taking
sample points. You can't possibly draw them
at every single point. So at that point, that's
the vector. That's the electric
field vector. But then if we go a little bit
further out, the vector is going to be-- it falls off. This one should be shorter, then
this one should be even shorter, right? You could pick any point and you
could actually calculate the electric field vector, and
the further you go out, the shorter and shorter the electric
field vectors get. And so, in general, there's all
sorts of things you can draw the electric fields for. Let's say that this is a
positive charge and that this is a negative charge. Let me switch colors so I don't
have to erase things. If I have to draw the path of
a positive test charge, it would go out radially from
this charge, right? But then as it goes out, it'll
start being attracted to this one the closer it gets to the
negative, and then it'll curve in to the negative charge and
these arrows go like this. And if I went from here, the
positive one will be repelled really strong, really strong,
it'll accelerate fast and it's rate of acceleration will slow
down, but then as it gets closer to the negative one,
it'll speed up again, and then that would be its path. Similarly, if there was a
positive test charge here, its path would be like
that, right? If it was here, its path
would be like that. If it was here, it's path
would be like that. If it was there, maybe its path
is like that, and at some point, its path might never get
to that-- this out here might just go straight
out that way. That one would just go straight
out, and here, the field lines would just
come in, right? A positive test charge would
just be naturally attracted to that negative charge. So that's, in general, what
electric field lines show, and we could use our little area
method and see that over here, if we picked a given area, the
electric field is much weaker than if we picked that
same area right here. We're getting more field lines
in than we do right there. So that hopefully gives you
a little sense for what an electric field is. It's really just a way of
visualizing what the impact would be on a test charge
if you bring it close to another charge. And hopefully, you
know a little bit about Coulomb's constant. And let's just do a very
simple-- I'm getting this out of the AP Physics book, but they
say-- let's do a little simple problem: Calculate the
static electric force between a 6 times 10 to the negative
sixth coulomb charge. So 6 times-- oh, no, that's
not on an electric field. Oh, here it says: What is the
force acting on an electron placed in an external electric
field where the electric field is-- they're saying it is 100
newtons per coulomb at that point, wherever the
electron is. So the force on that, the force
in general, is just going to be the charge times the
electric field, and they say it's an electron,
so what's the charge of an electron? Well, we know it's negative, and
then in the first video, we learned that its charge is
1.6 times 10 to the negative nineteenth coulombs times
100 newtons per coulomb. The coulombs cancel out. And this is 10 squared, right? This is 10 to the positive 2, so
it'll be 10 to the minus 19 times 10 to the positive 2. The force will be minus
1.6 times 10 to the minus 17 newtons. So the problems are
pretty simple. I think the more important thing
with electric fields is to really understand intuitively
what's going on, and kind of how it's stronger
near the point charges, and how it gets weaker as it goes
away, and what the field lines depict, and how they can be used
to at least approximate the strength of the field. I will see you in
the next video.