a

Uniform Electric Field (2 of 9) Motion of Charged Particles Perpendicular to the Field

okay in today's video we are going to be

going over the motion of charged

particles in electric fields and we're

gonna be talking about charged particles

that are moving across or perpendicular

to the electric field if you want to

know about charged particles that are

moving parallel or along the electric

field then link to those videos in the

upper right hand corner up here and you

can see that information also and this

video you're talking about the velocity

the acceleration and the displacement

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very much let's get started

okay we have electric fields here's our

electric field we have a positively

charged plate with negative charge plate

the electric field goes from the

positive to the negative and we're gonna

send in this positively charged particle

from left to right in this direction now

here it's outside the field it's gonna

be coming from the left and it's already

has some initial velocity in the X

direction something else gave it that

initial velocity in the X direction and

when it enters that field then it's

gonna fly off in that direction and make

that curve parabolic path look like that

okay it makes that curve parabolic path

when it does that it's moving in the X

direction and in the Y direction at the

same time the sum of the motion in the X

and the sum of the motion in the Y makes

it have that curved para Bloch path like

that okay now let's talk about first

before we talk about what's doing in the

X and the y let's talk about what's

happening what forces are acting on that

object as it moves through that electric

field okay so it's moving like this has

this curved path at some point it was

somewhere around here and anywhere it is

in that path there's only one force

acting on it and that is the electric

force and the electric force is down in

the negative direction that's the only

force it's just like in mechanic's and

gravitation when you kick something or

throw something and it's flying through

the air the only force that's acting on

is the force of gravity when it's up

there flying through the air it's only

acting I'm being acted upon by the

force of gravity a lot of people think

well it's moving from left to right it's

moving in this direction there must be

some force that's causing it to move

from left to right well there was some

force over here that got it going in the

in the first place but once it's going

as Newton's first law that objects in

motion stay in motion

obviously at rest stay at rest unless

acted upon by an unbalanced force

there's no forces so therefore we know

that the initial and the final velocity

in the X direction are going to be the

same the only force is the negative is

the electric force in the negative y

direction okay now let's talk about

what's doing the X Direction first we

kind of went over that there's no forces

acting on it in the X direction there

before we can say the forces are

balanced and therefore it's moving at a

constant velocity and most importantly

you need to remember that if it's moving

at a constant velocity and it has

balanced forces then there's no

acceleration in the X direction there's

no acceleration and as I said that means

that the final velocity in the X

Direction is going to be equal to the

initial velocity in the X Direction the

velocity in the X Direction doesn't

change okay you need to remember that

that's important and then because

there's no excell acceleration we can

use this nice simple equation that says

that velocity is equal to the distance

divided by the time okay so if we wanted

to find the time it takes to go through

if we know the distance across and we

know the initial velocity then we can

find the time that's gonna be important

because the time in the X and the time

in the wire are gonna be the same which

I'll talk about in just a moment okay so

that's the X direction in the Y

direction there is a force it's the

electric force that's acting the

negative Y direction that means the

forces are unbalanced and that means it

is accelerating not experiencing

constant velocity in the Y direction

it's accelerating and actually the

initial velocity as it enters in the Y

direction is zero and then as it starts

to move down as increasing it's

accelerating in the Y direction okay and

therefore in the except for if we wanted

to calculate the acceleration we can use

this equation to calculate the

acceleration that's one of the two

common equations you would use to

calculate the acceleration okay we'll

talk about the other equation when we do

example in just a moment okay so and

then it's important to remember as I

mentioned that the time to go in the Y

is the same as the time to go in the X

and we could often calculate the X

Camino the initial velocity we know the

distance across the electric field we

can calculate the time using this

equation the time in the X and then in

order to get the distance or something

else that it moves in the Y we can say

or we know that the time in the Y is

equal to the time in the X the time to

go through the distance that goes in the

x direction is equal to the time it

takes to go its distance in the Y

direction and we have X plus y equals

that position right there okay so okay

so this is the example problem we're

gonna go through for this video and it

says here with a velocity of 9.0 times

10 to the 6 meters per second an

electron enters a homogeneous electric

field between the parallel plates of a

capacitor and it's gonna be traveling

perpendicular to the electric field so

these orange lines are the plates of the

capacitor that is an electric field

which goes from the positive to the

negative and that electron has to be

coming in here with some initial

velocity this is our electron and it's

going to be traveling with some initial

velocity and it's going to be traveling

perpendicular to the electric field and

it's gonna take this parabolic curve

path like that approximately it says the

plates of the capacitor are square and

have a length of 65 millimeters and they

have a distance between them of 50

millimeters and the potential difference

between the plates is 150 volts and we

want to be able to answer I think I have

five questions here we want to know what

is the acceleration of the electron in

the Y direction we want to know how much

times it takes the electron to travel

through the capacitor we wonder how far

from the x axis the electron will be

when it leaves the place the capacitor

what will the velocity be the electron

in the Y direction when it leaves the

capacitor it leaves the electric field

and we also want to determine the angle

I think we're gonna do the magnitude of

the velocity of the electron that it

makes with the x-axis when it leaves

that electric field so we want to answer

those five questions I think we had

here's the information here's our

diagram that we can refer to and the

first thing is we want to know what is

the acceleration of the electron in the

Y direction now there's kind of two

equations that we can use this one of

the

this is the other one you can see in

this problem okay

we're given that U and D so the distance

between the plates and the potential

difference so we're going to use this

equation we can easily calculate e the

electric field strength because we know

the potential difference is equal to e

times D we could solve that for e which

would give us the potential difference

times the distance and we could

substitute that in here but we have

everything we need to use this equation

so we're something gonna write that

equation down we're gonna plug the

values in now I didn't think I'd give it

to you but you know it's an electron and

the charge on an electron is one point

six times ten to the minus nineteen

coulombs the potential difference is 150

the mass of an electron is nine point

one times ten to minus 31 kilograms and

that's the distance between the plates

it says here 50 millimeters please

remember that is 50 times 10 to minus 3

it must be in meters okay you can't just

put 50 in there all right if we do that

then we get an acceleration of about

five point three times ten to the 14

meters per second squared

acceleration okay next one is how much

time now we have to remember this is

important when the electron flies in

from the left-hand side here it's

already moving with some initial

velocity when it enters the electric

field okay it's gonna be moving

perpendicular to the electric field and

we want to know how much time it's gonna

take to move through the capacitor now

in the x-direction there's no forces

acting on that electron there's only a

force the electric force acting on that

electron in the Y direction in the X

direction there is going to be no forces

there's gonna be a constant velocity and

no acceleration so we can simply use

this equation because we know the

distance is sixty five millimeters and

we want to figure out the time because

we're given the initial velocity that

means the time is equal to going to be

equal to the distance divided by the

velocity the distance is sixty five

times ten to the minus three meters the

initial velocity in the X Direction is

9.0 times 10 to the 6 meters per second

it's kind of a constant velocity in the

X direction so we can use this equation

when we have no

acceleration and we get that the time

that it would take to travel through

that capacitor is 7.2 times 10 to the

minus 9 seconds and I just want to point

out because we're going to use this in

the next slides that this is the time we

calculated to go in the x-direction

but we know when the electron flies

through that capacitor through that

electric field it's moving in the X and

the y direction at the same time so

therefore we know that the time it takes

to go in the X Direction is going to be

equal to the time it takes to go in the

Y direction so if we need the time it

takes to go in the Y direction we can

simply use what we calculated for the

time and X which is this time right here

okay don't forget that that is an

important point so we want to know how

far from the x axis the electron is when

it leaves the plates of the capacitor

now for some people this little computer

that says how far from the x axis now

what we're really finding is how far did

it move along the y axis the distance

away from the x axis is in the Y

direction so we want to actually find

the distance in the Y direction people

get a looking through sleeping okay how

far is there like along the x axis or

something like that but what we're

really finding is the distance in the Y

which is equal to the initial velocity

in the Y because this is y times the

time it takes to go in the Y which we

calculated in the previous slide we know

the X now we know the Y time one half ay

T squared and this has to be the

acceleration also in the Y direction

we're finding the value for the Y so

we've got to use all the values for the

Y now we know when the electron comes in

it's moving in this direction only

therefore the initial velocity in the Y

Direction important in the Y Direction

is zero so our equation simplifies to

Delta y equals one-half a T squared now

we want to find the distance so we can

simply plug those values in it's

one-half the acceleration which we

calculated in the Y direction I think we

did that on the first slide the time

this is what we calculated in the

previous slide remember we calculated

the time in the X but we know the time

in the X is equal to the time in the Y

so it's the same time and therefore you

got to square that and we get the time

soomi we get the distance that it moves

away from the x-axis or the distance

that it moves along the y-axis there's

one point three seven times ten to the

minus two meters all right just like

that all right

any other questions I think we can go on

okay Dee what's the velocity electron in

the Y direction when it leaves let's

just point out once again that I've lost

in the X is constant so we know the

velocity in the X plane at least what we

want to know the velocity and the Y so

you're gonna have an equation for that

and that's that the final velocity in

the Y squared is equal to the initial

velocity in the X entombing also in the

y squared plus 2a Delta Y all right all

of this is for the Y we once again know

that in the Y direction it didn't have

an initial velocity the initial velocity

was zero so equation simplifies to this

the final velocity squared in the Y

Direction is equal to 2ei Delta Y this

is the acceleration in the Y direction

we're going to take the square root of

both sides and we get the final velocity

in the Y is equal to that which we just

plug the values in this is true because

there's a two here this is the

acceleration in the Y direction this is

the distance it's going to go away from

the x axis which is the distance in the

Y direction and we can take the square

root of all that and we get that the

final velocity in the Y Direction is

equal to three point eight times ten to

the six and meters per second remember

this is the initial and the final in the

X because there's no acceleration this

is the final velocity in the Y direction

the initial velocity in the Y Direction

was zero okay let's see I think we have

this one last problem which is two parts

because we're gonna find the magnitude

and the direction of the velocity and I

think we're gonna find the magnitude

first so this says determine the angle

of the velocity of the electron that the

book determine the angle that the

velocity electron makes with the x-axis

when it leaves the electric field okay

here's the electron here's the electron

we know that these are our final

velocities okay remember this is not

changed from the X this has we just

calculated so we're gonna draw a vector

that represents each of those that's the

final velocity in the X

that's the final velocity in the Y now I

made them somewhat proportional because

this is approximately half of this a

little less than half but I tried to

make it about half so you can see these

are the velocities and that represents

some motion that's going to be in this

direction because we can take that

vector for the Y and move it and you can

see we have a right triangle now that

green line the hypotenuse of that right

triangle it's gonna be the sum of these

two vectors and that's our final

resultant velocity we can use the

Pythagorean theorem V squared is equal

to the F final x squared final y squared

we can take the square root of both

sides and we get that the final velocity

the resultant velocity is the square

root sum of the square roots of the

other two final velocities and that

means that the final velocity the

magnitude of that vector is 9.8 times 10

to the 6 meters per second okay and that

should make sense because this is nine

this is three point eight it's a little

longer so the Velata and that one has to

be the longest side of that triangle

okay now the last thing is we're gonna

find the angle find the angle we need

our trig functions we know now all three

sides but we're kind of given or we had

previously calculated these two sides

too oftentimes we just use the tangent

remember the tangent is the opposite

over the adjacent sohcahtoa call

opposite over adjacent and this says the

arc tangent inverse tangent of the angle

this is the angle we're trying to find

here this is supposed to be a lambda

it's supposed to be a lambda for the

angle plug the values in the inverse

tangent of the angle is equal to the

ratio of those two sides the opposite

over the adjacent and then you get zero

point four four two and then if you take

the arc or the inverse tangent of that

you get that that angle is approximately

23 degrees okay so there you go we did

ABCDE five different things I think

that's pretty complete thank you very

much for watching I hope you found the

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