Understanding Electric Potential, Potential Energy, and Voltage Explained

in this video we're going to talk about electric potential and electric potential energy

let's begin our discussion with this what is electric potential electric potential

is defined as potential energy divided by charge in some physics textbooks you'll see

capital u represented as potential energy

now there are different types of potential energy that you need to be familiar with

there is elastic potential energy that is the energy stored in springs

there's gravitational potential energy which occurs any time an object is above ground which it has the ability to fall

and then there's electric potential energy which is what we're going to talk about

so if i use p e or u in this particular video i'm using it with reference to electric

potential energy so v v is electric potential

p e electric potential energy make sure you notice the difference between them they

sound similar but they're different electric potential is measured in volts one volt

is equal to one joule of potential energy per one coulomb of charge

so an electric potential of five volts means that a 1 coulomb charge has 5 joules of potential energy

an electric potential of 12 volts means that a 1 coulomb charge have 12 joules of potential energy

with respect to some reference point of course or you could say that

two coulombs of charge has 24 joules of potential energy so electric potential is simply the

ratio of electric potential energy per unit charge

so now that we've defined electric potential as potential energy per unit charge

let's relate it to something more familiar that is a voltage

voltage is electric potential difference

voltage is the change in electric potential by the way you may want to take down

these notes if you have a piece of paper and a pen with you or a sheet of paper rather

so voltage would be the difference between the potential at b and the potential at a

so you can write it this way so if you were to see vba that means vb minus va if you were to see vab

that means the difference in the potential at a minus the potential at b so let's say we have a resistor

a resistor is an electronic component that is used to basically control the flow of electric

current in a metal or in a circuit now let's say we have two points of

interest across this resistor we'll call this point a and point b actually let's

make this b sometimes my computer acts up let's do it that way let's say

b is at 80 volts a is at 20 volts

the change in v is 80 minus 20 that's 60.

the 80 volts represents the electric potential at b 20 volts is

the electric potential at a so this is vb and this is va

this is the voltage voltage is electric potential difference it is the difference in the electric

potential between two points hopefully this helps you to see the connection between voltage and electric

potential the electric potential at b is 80 volts the electric potential at a is 20 volts

but the voltage across the resistor is 60 volts so we could say vba

which is the electric potential at b minus a 80 minus 20 that's 60 or we could say vab

20 minus 80 that's going to be negative 60. so let's use another example to help you

see the difference between voltage and electric potential so let's call this b and let's call this

a so let's say for b this is at 90 volts

and a it's at negative 30 volts the electric potential at a is negative 30 volts

the electric potential at b is 90 volts so what is the voltage

across the resistor go ahead and calculate vba and vab so to calculate vba

it's going to be the final value vb minus the initial value a

so we're going from when you think of this we're going from a to b

initial to final so that's 90 minus

negative 30. so that's positive 120 so vba is positive 120.

now if we want to calculate vab in analyzing the direction we're going from

b to a so b becomes the initial value a becomes the final value

so it's final v a minus initial so a

va is negative 30 minus vb which is positive 90. so we get negative 120.

so let's think about what's happening here in terms of the direction in which we're moving

so during the first part we're going from a to b is the potential increasing or

decreasing as we go from a to b they change from negative 30 to 90.

so the potential is increasing therefore we got a positive value now for the second part we went from b

to a from a high potential to a low potential so the potential is decreasing thus we

can see why the change is negative so i wanted to help you to understand the reasoning behind these two formulas

so that if you understand what they mean you'll remember how to write them correctly

now let's talk about how we can calculate the work done when a charge moves to a certain voltage

let's say that the direction of motion is from point a to point b so the voltage

which in this case is going to be vba that is

the potential at b minus the potential at a and if you recall electric potential

is equal to the electric potential energy divided by the charge

so vb is going to be the potential energy at b divided by the charge

va is going to be the potential energy at a divided by the charge

now i'm going to factor out 1 over q doing so i'll get the potential energy of b

minus the potential energy at a or we could say the change in potential energy

so we have 1 over q times the change in potential energy going from a to b

now some other formulas that you need to be familiar with are these

work is equal to the negative i mean the positive change of kinetic energy but the negative change

of potential energy in a system where energy is conserved as kinetic energy increases

the potential energy decreases think of a ball that's fallen down as it falls from a certain height its

potential energy decreases but it's speeding up as it approaches the ground so the kinetic energy goes up

this is true when the only forces acting on the system are conservative forces

that's when mechanical energy will be conserved if there are non-conservative forces acting on the object

then this may not apply so just

something to keep in mind but what we need to take from this is that work

is equal to the negative change of potential energy so therefore the change in potential

energy is equal to negative w so we have one over q

times negative w which we can write as negative w over q and that equals delta v

so if delta v is equal to negative w over q we can multiply both sides by negative q

to get an expression for w so the work done

by a charge or on a charge as it moves to a certain voltage

is this equation it's negative q delta v now let's work on this example problem

how much work is required to move a negative 500 micro coulomb charge across a potential difference of 300

volts for this problem all we need to do is use this formula

the work is going to be negative q times delta v so we have the charge

q is negative 500 micro coulombs or 500 times 10 to negative 6 coulombs the voltage or the change in v

that's the potential difference across two points so that's positive 300.

now recall one volt is one joule per one column

let me write that better so it's one joule over one column so we can write this as joules per

coulomb so that's the unit coulombs will cancel and we're going to get the work in

joules so we have two negative signs which will make this positive

so it's going to be 500 times 10 to the negative 6 times 300

and the work done is going to be point 15 joules so that's how much work is required

to move this particular negative charge across a potential difference of 300 volts

number two a hundred joules of work was done by an electric field

on a five column charge i forgot to put the word charge

to accelerate it from point a to point b what is the voltage across points a and b

so let's draw a picture so let's call this point a and we're going to call this point b

so we're going to move a charge from point a to point b and this is a positive charge

so if it takes 100 joules of work to move this five coulomb charge from a to b

what's the voltage crossing so we could use this formula to get the voltage

to solve for v it's going to be the work divided by the charge is equal to delta v

now the work is 100 joules the charge is 5 coulombs i wouldn't really worry about the

negative sign too much i would have focused more on direction but we'll talk more about that

if you were to plug in the negative sign i mean you'd just get a negative answer so 100 divided by negative 5 is negative

20. you could simply say the voltage across points a and b

is 20 volts now here's a question for you if the electric potential at point a is

50 volts what is the electric potential at point b

now we know the difference is 20. the question is is it 20 volts higher or

20 volts lower would you say if we add 20 to 50 that would be 70

if we subtracted 50 by 20 it will be 30. we know the difference is 20 but which of these two values will be correct

now looking at this this is saying that the voltage is negative 20. that's indication that b should be 20

volts less than a which will indicate that 30 volts is the answer

but let's confirm it though let's say we have two

parallel metal plates say the first one has a positive charge

and the second one has a negative charge now the electric field is going to go from the positively

charged plate to the negatively charged plate now let's say that at the negatively

charged plate the electric potential is zero the electric potential at the positive

plate is going to be higher it could be 100 volts now what's going to happen if we put a

positive charge here will it move to the negative plate or the positive plate

a positive charge will fill a force that will accelerate it in the same direction as the electric

field so in this case it's going to be attracted to the

negatively charged plate so notice the direction in which the positive charge moves

the electric field causes the positive charge to move from a high potential to a low potential

now if there was a negative charge the electric force i mean the electric field rather would cause the negative

charge to move from a low potential to a high potential now it's important that you understand

that so i'm just going to summarize this for notes

so for a positive charge it's going to move from a high potential or high v

to a low v it will accelerate in that direction due to electric field

but a negative charge will accelerate from a low v to a high potential

due to electric field now in this problem the electric field is accelerating

the positive charge from point a to point b and we know that an electric field will

accelerate a positive charge from high potential to low potential so if the 50 volts is considered the

high potential the low potential must be 30. if it was 70 it would be going from low

to high and that high to low so conceptually that's how you can determine that this is indeed the

correct answer and of course if you use the fact that the voltage change is negative 20 that

helps to see that it's going from high to low as well but in the event that you didn't take

the negative sign into or into an account just understanding this concept will

help you to see that the positive charge is going to move towards a lower voltage so conceptually you would know that you

need to subtract 20 and not add 20 to get the answer if we were dealing with a negative

charge we would add 20. the negative charge will go from a low potential to a high potential but

the positive charge will go from a high potential to a low potential when accelerated by an electric field

number three a positive charge is released from rest at point a

as shown in the figure below as it accelerates toward point b is the charge electric potential energy

increasing or decreasing what would you say well we know that there's going to be an

electric field that will emanate from the positively charged plate and it's directed towards the negatively

charged plate now we also know that the charge

is going to fill a force that will accelerate it in the same direction

as the electric field so it's going to accelerate towards point b and then towards point t

now how can we describe what's happening with regard to the charges potential energy

well since there's a force that's accelerating

the charge towards point b there must be an acceleration based on newton's second

law then that force is equal to ma so if there's a force there's an acceleration

now what's happening to the speed of the charged particle if it's accelerated

accelerating towards the right its velocity is increasing and therefore the magnitude of the speed is going to

increase as well so v is increasing for this particular particle

if the speed is increasing what's going to happen to the kinetic energy which basically answers part b

as the speed increases the kinetic energy of this positive charge is going to go

up which means that the change in kinetic energy is positive

now if the kinetic energy increases what's going to happen to the potential energy

as the kinetic energy goes up the potential energy goes down we don't have any non-conservative

forces acting on a positive charge the only force acting on it is the electric force

due to electric field which is a conservative force like gravity so therefore the mechanical energy of the

positive charge will remain constant the mechanical energy

is the sum of the kinetic energy and the potential energy

so if the mechanical energy is constant due to the lack of any non-conservative forces acted on the charge

as ke goes up pe must go down

so since the potential energy decreases the change in the potential energy will be negative

now what about part c is the work done on this particular charge is it positive

or negative work is equal to the change in kinetic energy

so if the change in kinetic energy is positive the work done on this positive charge is

also going to be positive now let's get rid of this

there's another way in which we can analyze the work done if it's positive or negative

keep in mind work is force times displacement if the force vector

and the displacement vector are in the same direction the work done is going to be positive

if the force vector and the displacement vector are in opposite directions the work done is negative

now if the force vector is perpendicular to the displacement vector so if there are right angles

then the work done is zero so let's use this to confirm our answer in this example

the charge fills the force that's accelerating it towards the right so that's the force vector it's

heading due east now what about the displacement vector where is the charge moving is it moving

to the left or to the right well we know it's going to move towards a negatively charged plate in the

direction of the electric field so the displacement vector for this charge is towards the right

because the force and the displacement vector are in the same direction

the work done on that charge is positive its kinetic energy will increase now for part d

we're going to repeat parts a through c so the new situation is that the positive charge

is going to start at point b so it's it's at point b and at this

instant it's moving to the left

so going back to part a actually let's do part b first is the kinetic energy increasing or decreasing

so let's think about what's happening here the charge is moving

towards point a however the electric field is still towards the right so therefore it's

still going to fill an electric force towards the right so what does it mean when f and v are

opposite in direction let's talk about that when the force vector and the velocity

vector are in the same direction this means that the object is

accelerated it's speeding up the force vector and the acceleration

vector are in the same direction so think of f and a as being similar now if the force vector

and the velocity vector are opposite to each other the object is slowing down

it's decelerating towards the left it's accelerating towards the right if you look at just

the positive sign but the signs of acceleration and velocity are different therefore it's going to be

slowing down so make sure you understand that if the force and velocity vectors are in the same direction

the object is going to be speeding up if they're in opposite directions the object is slowing down

and then if the force and velocity vectors are perpendicular to each other

the object is neither speeding up or slowing down but it's going to turn and it's going to do so at constant

speed which means that we're going to have circular motion now in this example the force

and the velocity vectors are opposite to each other which means it's going to be slowing down

the speed is going to be decreasing if it's slowing down that means that the kinetic energy is decreasing

therefore the change in kinetic energy is negative now if the kinetic energy is decreased and

that means that the potential energy the electric potential energy is increasing so the change

in electric potential energy is going to be positive now what about the work done

since the change in kinetic energy is negative the work done

on this particle or on this positive charge is going to be negative as well

because this force it's not accelerating the charge it's slowing it down it's decreasing its

kinetic energy therefore because the electric force is decreasing the kinetic energy of the charge we can

say that it's doing negative work on it it's slowing it down not speeding it up now let's look at force and displacement

for the positive charge the force vector is due

east but because the velocity vector is due west that means that it's moving to the left so it's going to have a

displacement in the negative x direction at least for a time until it changes direction

but while it's moving to the left the displacement vector is directed to the left

whenever the force and displacement vectors are opposite to each other we can see that the work done is going to

be negative therefore the electric field is not speeding up the particle it's slowing it

down it's doing negative work on it number four how much work is required to move a

negative 50 micro coulomb charge from an electric potential of negative 50 volts

to 250 volts let's call this point a point b

so this time the electric field vector is going to be pointing towards the left away from the positive charge

towards the negatively charged plate now the positively charged plate is going to have a higher potential it's

going to be at an electric potential of 250. the negatively charged plate it makes

sense that it's going to be at a lower potential or an electric potential of negative 50.

the voltage between these two is the difference of those two numbers

but before we calculate voltage let's write the equation for work work is going to be negative q

times the voltage now the charge the negative charge is going to fill a

force that will accelerate it in the direction opposite to the electric field it's going to accelerate towards the

positively charged plate so the negative charge is going from a to b

so b is the final position a is the initial position so if it's going from a to b

the change in voltage or the sign of delta v will be positive or negative

the negative charge is going from a low potential to a high potential so because the potential is increasing

the voltage should have a positive sign so it's being accelerated across a

positive voltage delta v is going to be vb minus va

so the final potential that it experiences 250 the initial potential is negative 50.

so 250 minus negative 50 that's 250 plus 50 says 300. so we have a negatively charged particle

with a charge of negative 50 micro coulombs and it's accelerating through a

potential difference of positive 300 volts

so remember an electric field will accelerate a negatively charged particle

from a low potential to a high potential

however an electric fill will accelerate a positive charge from a high potential to a low potential so if we had a

positive charge it would go towards the low potential plate or negative 50.

but since we're dealing with a negative charge it's going to go from low v to high v

now these two negative signs will cancel so to get our to get our answer it's going to be 50

times 10 to the negative 6 times 300 and so the work is going to be .015 joules

now the answer is positive the question is does the sign of the work make sense

the work done on this negative charge should it be positive or negative since we got a positive answer from the

formula chances are this is correct but let's analyze it in what direction is the electron moving

from left to right or right to left we know that it's going from a to b from a low potential to a high potential

so therefore the displacement vector is towards b because it's moving in that direction

now because the force and the displacement vector are parallel to each other in the same direction

the work done by the electric field on the negative charge is positive

these two vectors are parallel to each other in the same direction now let's analyze it from the other

angle we know that this charge is accelerating towards the right therefore it's

speeding up if the speed is increasing the kinetic energy is increasing which

means the change in kinetic energy is positive therefore

the work done on that charge is positive as well we can say that the electric field

does positive work on the negative charge because it's increasing its kinetic energy it's making it move

faster so now let's move on to the next part

part b if the charge accelerates from rest what is the final speed if it has a mass

of 0.01 kilograms i mean 0.01 grams now we know that work is equal to the change in kinetic energy

and the change in kinetic energy is the final kinetic energy minus

the initial kinetic energy kinetic energy depends on the speed there is no kinetic ener there is no

kinetic energy if a particle or an object is not moving so notice that it accelerates from rest

therefore the initial speed is zero which means that the initial kinetic energy

is zero so if this is zero we could say that the work

is equal to the final kinetic energy so let's go ahead and use that formula so for this particular example because

it accelerates from rest work is equal to the final kinetic energy

and the final kinetic energy is one-half mv squared so the work is .015 joules

and then the mass is 0.01 grams but we need to convert that to kilograms to convert grams to kilograms

is we can do in mind there's a thousand grams per one kilogram

so he needs to divide .01 divided by a thousand is one times ten to the negative five

kilograms which is point zero zero zero zero one

so now let's solve for v if we take point zero one five divided by point five

and then take that result divided by one times ten to negative five

you'll get that three thousand is equal to v squared now let's take the square root of both

sides the square root of 3 000 is 54.77 so this is going to be the final speed

of the negative charge as it accelerates from an electric potential of negative 50 to

250. you