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### Course: MCAT > Unit 9

Lesson 10: Electrostatics# Electric potential energy

Explore the concept of electrical potential energy and its similarities to gravitational potential energy. Understand how work is required to move an object within a gravitational or electric field, and how this work translates into potential energy. Discover how potential energy changes can impact kinetic energy and velocity. Created by Sal Khan.

## Want to join the conversation?

- At11:11Sal said that if we let go of the particle all of 30J would be converted to kinetic energy.

I get that if left, the particle would move away from the positively charged plate, but its still in the electric field so it should have some potential energy remaining, right?(96 votes)- From the author:Yes, the 30J is really the
**increase**in potential from it's starting point to the ending point 3 meters closer to the plate. If we let go at the ending point, the particle would start accelerating in the direction of the field (upward), and, in this case, when it passes its starting, all of that**increase**in potential energy will be converted to kinetic energy.

The particle, however, still has the field acting on it and will still be accelerated upwards (so you could say that it still has a positive potential relative to a point further away from the plate).(268 votes)

- around9:10sal says that to get the charge moving downwards, we have to exert a force of 10N. But if we exert that force in the downward direction, seeing that the metal plate is ALSO exerting a force by the same amount, won't the charge just stay stationary over there (like suspended in the electric field)???

PLEASE HELP ME ON THIS ONE :(

VERY IMPORTANT(64 votes)- for one moment a force little greater than 10N is applied, so that the body gains a velocity. after that the force can be equal to 10N.(In that case net force will be zero and the charged body will continue to go with the velocity it has gained at first.(15 votes)

- I got a question about electric potential energy, though may not be related to Sal's video. When a positive charge is brought near a positive point charge. The work done will be changed to the electric potential energy and stored in the charge. However, when a negative charge is brought away from the positive charge, the negative charge gains electric potential energy.When r keeps increasing, the electric potential energy stored in the negative charge will be extremely large?? Hope you guys can understand my question....(10 votes)
- I understand now! for opposites charges, work is done to pull them away from each other. it changes into PE and stored inside the charge. But their separation is getting larger as well, the attractive force becomes smaller, then the work needed is smaller too. So the pulling force decreases. It means that the electric PE increases at a decreasing rate!(16 votes)

- At7:25Sal says that a field of 5N/C is quite strong. According to Google a Newton is about 1/5 of a pound. So the field is something like 1 pound/coloumb.

Why is this so strong? I was under the impression that a coloumb was a fairly large amount of charge. 1 pound for a high amount of charge does not seem so strong.(6 votes)- It's quite strong for an electrical field.(2 votes)

- why does the temperature of a solid conductor increase when the conductor is carrying current?(5 votes)
- Electricity gives off heat. The more electricity the more heat.(4 votes)

- im always confused at this fact .

if an object weighing mg newtons needs to be pulled up,it requires an opposite force EQUAL to mg newton.(should'nt the force be more than the downward force)(because if force to pull up would be equal it would be suspended in air {if applied from rest}).(3 votes)- good question

OK...so

IF it is pulled up at a constant speed then it does not need extra force

but if it is pulled up accelrating then it does need extra force

(this is by Newtons first law...)(6 votes)

- The potential is constant throughout a given region of space . Is the electrical field zero or non zero(3 votes)
- two ways you might want to think about it...

1) if the potential is constant, what is the potential differnce like? and therefore what must the field be like?

2) what does the graph of potential look like? How is the graph of field related to the graph of potential?

Let us know how you get on(5 votes)

- at3:40,sal says that the work done is equal to force of gravity into h,but shouldn't the force be a little bit more if we want to lift it upward against the force of gravity?(6 votes)
- Yes. In later in the video he goes on to say that exact same thought. His equation shows 10n/m but the faster you wanted to go the more you would apply. (10n - SLIGHT SPEEED).(0 votes)

- i have 2 questions

1.what is the intuition behind a constant like G(gravitational constant)and K(coulombs constant),where do we get the values of them?

2.when sal said that we will apply a force of 10N DOWNWARDS wouldn't that force be balanced by the upward force since the field is also applying a force of 10N on the charge in the opposite direction that is UPWARDS?(2 votes)- Constants like G and K simply arise from the fact that we need to relate on type of unit to another. For example, F = G(m1)(m2)/d^2. The G makes newtons out of mass^2/distance^2. K does something similar.(7 votes)

- Are electric fields and magnetic fields related? Do they co-exist and if so, do they do so always?(3 votes)
- They are just two different aspects of the same thing. That thing is called the electromagnetic interaction/force and is one of the 4 fundamental interactions of nature.

The reason, why the interaction is talked about as the combination of two different things is mostly historical as people didn't recognize the two things as one until Maxwell. (Could be wrong on the history :P)

The electric field is the irrotational (curl-free/gradient-field) aspect of the phenomenon. It's source is the scalar potential Φ (given by the distribution of electric charge). The magnetic field on the other hand is the solenoidal (source-free) part of the field. It is caused by some vector potential**V**.

Concerning the second part of your question: In case of a static potential Φ, the**B**aspect is zero. So you could say, they don't always coexist. On the other hand, "they" are in some sense one and the same thing.

-> http://en.wikipedia.org/wiki/Helmholtz_decomposition(3 votes)

## Video transcript

Let's review a little bit of
what we had learned many, many videos ago about gravitational
potential energy and then see if we can draw the analogy,
which is actually very strong, to electrical potential
energy. So what do we know about gravitational potential energy? If we said this was the surface
of the Earth-- we don't have to be on Earth, but
it makes visualization easy. We could be anywhere that has
gravity, and the potential energy would be due to the
gravitational field of that particular mass, but let's
say this is the surface of the Earth. We learned that if we have some
mass m up here and that the gravitational field at this
area-- or at least the gravitational acceleration-- is
g, or 9.8 meters per second squared, and it is h-- we could
say, I guess, meters, but we could use any units. Let's say it is h meters above
the ground, that the gravitational potential energy
of this object at that point is equal to the mass times the
acceleration of gravity times the height, or you could view
it as the force of gravity, the magnitude of the
force of gravity. You know, it's a vector, but
we can say the magnitude of the vector times height. And so what is potential
energy? Well, we know that if something
has potential energy and if nothing is stopping it
and we just let go, that energy, at least with
gravitational potential energy, the object will start
accelerating downwards, and a lot of that potential energy,
and eventually all of it, will be converted to kinetic
energy. So potential energy is energy
that is being stored by an object's situation or kind of
this notional energy that an object has by virtue
of where it is. So in order for something to
have this notional energy, some energy must have
been put into it. And as we learned with
gravitational potential energy, you could view
gravitational potential energy as the work necessary to move
an object to that position. Now, if we're talking about work
to move something into that position, or whatever, we
always have to think about, well, move it from where? Well, when we talk about
gravitational potential energy, we're talking about
moving it from the surface of the Earth, right? And so how much work is required
to move that same mass-- let's say it was here at
first-- to move it from a height of zero to
a height of h? Well, the whole time, the
Earth, or the force of gravity, is going to
be F sub g, right? So essentially, if I'm pulling
it or pushing it upwards, I'm going to have to have-- and
let's say at a constant velocity-- I'm going to have to
have an equal and opposite force to its weight
to pull it up. Otherwise, it would accelerate
downwards. I'd have to do a little bit more
just to get it moving, to accelerate it however much, but
then once I get it just accelerating, essentially I
would have to apply an upward force, which is equivalent to
the downward force of gravity, and I would do it for a
distance of h, right? What is work? Work is just force
times distance. Force times distance, and it
has to be force in the direction of the distance. So what's the work necessary
to get this mass up here? Well, the work is equal to the
force of gravity times height, so it's equal to the gravitational potential energy. Now this is an interesting
thing. Notice we picked the reference
point as the surface of the Earth, but we could
have picked any arbitrary reference point. We could have said, well, from
10 meters below the surface of the Earth, which could have been
down here, or we could have actually said, you know,
from a platform that's 5 meters above the Earth. So it actually turns out, when
you think of it that way, that potential energy of any form,
but especially gravitational potential energy-- and we'll
see electrical potential energy-- it's always in
reference to some other point, so it's really a change in
potential energy that matters. And I know when we studied
potential energy, it seemed like there was kind of an
absolute potential energy, but that's because we always assume
that the potential energy of something is zero the
surface of the Earth and that we want to know the
potential energy relative to the surface of the Earth, so it
would be kind of, you know, how much work does it take to
take something from the surface of the Earth
to that height? But really, we should be saying,
well, the potential energy of gravity-- like this
statement shouldn't be, you know, this is just the absolute potential energy of gravity. We should say this is the
potential energy of gravity relative to the surface of the
Earth is equal to the work necessary to move something, to
move that same mass, from the surface of the Earth to
its current position. We could have defined some other
term that is not really used, but we could have said
potential energy of gravity relative to minus 5 meters
below the surface of the Earth, and that would be the
work necessary to move something from minus 5 meters
to its current height. And, of course, that
might matter. What if we cut up a hole and
we want to see what is the kinetic energy here? Well, then that potential
energy would matter. Anyway, so I just wanted to do
this review of potential energy because now it'll make
the jump to electrical potential energy all that
easier, because you'll actually see it's pretty
much the same thing. It's just the source of the
field and the source of the potential is something
different. So electrical potential energy,
just actually we know that gravitational fields are
not constant, we can assume they're constant maybe near the
surface of the Earth and all that, but we also know that
electrical fields aren't constant, and actually they
have very similar formulas. But just for the simplicity of
explaining it, let's assume a constant electric field. And if you don't believe me that
one can be constructed, you should watch my videos that
involve a reasonable bit of calculus that show that a
uniform electric field can be generated by an infinite
uniformly charged plane. Let's say this is the side view
of an infinite uniformly charged plane and let's
say that this is positively charged. Of course, you can never get
a proper side view of an infinite plane, because you
can never kind of cut it, because it's infinite in every
direction, but let's say that this one is and this
is the side view. So first of all, let's think
about its electric field. It's electric field is going to
point upward, and how do we know it points upward? Because the electric field is
essentially what is-- and this is just a convention. What would a positive charge
do in the field? Well, if this plate is positive,
a positive charge, we're going to want to
get away from it. So we know the electric field
points upward and we know that it's constant, that if these
were field vectors, that they're going to be the same
size, no matter how far away we get from the source
of the field. And I'm just going to pick
a number for the strength of the field. We actually proved in those
fancy videos that I made on the uniform electric field of an
infinite, uniformly charged plane that we actually proved
how you could calculate it. But let's just say that this
electric field is equal to 5 newtons per coulomb. That's actually quite strong,
but it makes the math easy. So my question to you is how
much work does it take to take a positive point charge-- let
me pick a different color. Let's say this is the
starting position. It's a positive 2 coulombs. Once again, that's a massive
point charge, but we want easy numbers. How much work does it take it to
move that 2-coulomb charge 3 meters within this field? How much work? So we're going to start here
and we're going to move it down towards the plate 3
meters, and it's ending position is going to be
right here, right? That's when it's done. How much work does that take? Well, what is the force of
the field right here? What is the force exerted on
this 2-coulomb charge? Well, electric field is just
force per charge, right? So if you want to know the force
of the field at that point-- let me draw that
in a different color. The force of the field acting on
it, so let's say the field force, or the force of the
field, actually, is going to be equal to 5 newtons per
coulomb times 2 coulombs, which is equal to 10 newtons. We know it's going to be upward,
because this is a positive charge, and this is a
positively charged infinite plate, so we know this is an
upward force of 10 newtons. So in order to get this charge,
to pull it down or to push it down here, we
essentially have to exert a force of 10 newtons
downwards, right? Exert a force of 10 newtons in
the direction of the movement. And, of course, just like we did
with gravity, we have to maybe do a little bit more than
that just to accelerate it a little bit just so you have
some net downward force, but once you do, you just have
to completely balance the upward force. So just for our purposes, you
have a 10-newton force downward and you apply that
force for a distance of 3 meters, the work that you put to
take this 2-coulomb charge from here to here, the work
is going to be equal to 10 newtons-- that's the force--
times 3 meters. So the work is going to equal
30 newton-meters, which is equal to 30 joules. A joule is just a
newton-meter. And so we can now say since it
took us 30 joules of energy to move this charge from here to
here, that within this uniform electric field, the potential
energy of the charge here is relative to the charge here. You always have to pick a point
relative to where the potential is, so the electrical
potential energy here relative to here and this
is electrical potential energy, and you could say P2
relative to P1-- I'm using my made-up notation, but that gives
you a sense of what it is-- is equal to 30 joules. And how could that help us? Well, if we also knew the mass--
let's say that this charge had some mass. We would know that if we let go
of this object, by the time it got here, that 30 joules
would be-- essentially assuming that none of it got
transmitted to heat or resistance or whatever-- we know
that all of it would be kinetic energy at this point. So actually, we could
work it out. Let's say that this does have
a mass of 1 kilogram and we were to just let go
of it, right? We used some force to bring it
down here, and then we let go. So we know that the electric
field is going to accelerate it upwards, right? It's going to exert an upward
force of 5 newtons per coulomb, and the thing's going
to keep [COUGHS]-- excuse me-- keep accelerating
until it gets to this point, right? What's its velocity going
to be at that point? Well, all of this electrical
potential energy is going to be converted to kinetic
energy. So essentially, we have 30
joules is going to be equal to 1/2 mv squared, right? We know the mass, I said, is 1,
so we get 60 is equal to v squared, so the velocity is the
square root of 60, so it's 7 point something, something,
something meters per second. So if I just pull that charge
down, and it has a mass of 1 kilogram, and I let go, it's
just going to accelerate and be going pretty fast once
it gets to this point. Anyway, I'm 12 minutes into this
video, so I will continue in the next, but hopefully, that
gives you a sense of what electrical potential energy
is, and really, it's no different than gravitational
potential energy. It's just the source of the
field is different. See you soon.