Comprehensive Guide to Electric Fields: Concepts, Calculations, and Applications

in this video we're going to focus on electric fields let's begin our discussion

with the formula that will help us to calculate the electric field

the electric field is equal to the electric force that is acting on

a tiny test charge divided by

the magnitude of that test charge so it's f over q the force is measured in newtons

the charge q is measured in coulombs so electric field has the units newtons per coulomb

now the electric field is a vector much like force is a vector but the good thing about the electric

field is it tells you how the electric force is going to be

acting on a positive or negative charge

so let's talk about positive charges first what happens

if we put a positive test charge in an electric field and that test charge is going to fill a

force that will accelerate it in the same direction as the electric field now what about a negative test charge

what's going to happen if we put it in an electric field a negative test charge

will fill a force that will accelerate it in the opposite direction to the

electric field so make sure you understand that positive charges

they accelerate in the same direction as the electric field the force and electric field vectors will be in the

same direction but for negative charges

they will fill a force that will accelerate it in the opposite direction of the electric field

now as we said before an electric field can exert a force on

any type of charge a positive charge or a negative charge but it turns out that charges

can also create their own electric fields the electric field created by a positive

charge extends in all directions away from the positive charge

the electric field created by a negative charge extends in all directions

toward the negative charge so it's going inward towards the negative charge in all directions

now let's talk about the equation that will help us to calculate the electric field

created by a point charge so let's say we have okay let me draw a better circle

let's say we have this charge which we'll call big q and we want to calculate the electric

field at some point a

how can we do that and let's say a is some distance r well we know that

the electric field at point a will be going in this direction all you need to do is draw an arrow

from the positive charge to the point of interest and that will give you the direction

of the electric field now let's say if we place a tiny positive charge

at point a so let me just erase point a so this tiny positive charge will be considered our test charge the reason

why it has to be tiny because if it's large enough it will affect

q big q will be repelled by little q but if the magnitude of little q is very

very small it won't affect big q as much and thus it won't affect the electric

field that is created by big q as much so that this equation will work if we choose a very tiny positive test charge

now according to columbus law whenever you have two charges next to each other they will

exert a force in each other two like charges will fill a force that will repel them

and we can calculate that force using this equation it's equal to k times q1 which we'll call big q

times q2 which we'll call little q over r squared where r

is the distance between the two charges so what we're going to do now is we're

going to take this equation and substitute it for f in that equation

but first i'm going to rewrite this equation as f over 1 times 1 over q

that's the same as f divided by q so now let's replace f with what we see here

so this is going to be f i mean it's no longer f but it's going to be k big q times little q

over r squared so this whole thing is f and then we're going to multiply it by 1

over q and so we could cancel out little q the test charge

thus we get the equation that gives us the electric field or the magnitude of the electric field

for a point charge capital q so it's k q over r squared

so if we have our charge q and we wish to calculate the electric field

at point a we could use this formula k

is 9 times 10 to the 9 newtons

times square meters over square coulombs q is the magnitude of the charge in

columns now remember one micro coulomb is one times ten to the minus six

coulombs a nano clue is 10 to the negative nine clues and a

milli clue is 10 to the minus three columns and r is going to be the distance

in meters now let's say we have a positive charge here

and we have point a let's say point b

point c and point d determine the electric field

created by the charge q at points a b c and d

so at point a the electric field in order to draw it we're going to draw starting from the

positive charge towards point a and so the electric field is going north

for point b we need to draw towards point b so the electric field the direction

is west for point d it's going to be at an angle so it appears to be going in the

the northwest direction and for point c just draw towards point c

so it's going in uh oh wait i have to make a correction that is not northwest that is north east

and this is southeast now let's do something similar but with a negative charge

so for the sake of practice go ahead and determine the direction of the electric field

at the following points so point a point b

point c and point d so this time the electric field is going

to point toward the negative charge so we're going to draw it from the point to the negative charge

so this electric field vector is going west

and then here we're going to draw it from v towards the negative charge so this is going in the south direction

and then d to the negative charge here it's going east and then from c to the negative charge

this electric field vector is pointed in the northwest direction so that's how we can determine the

direction of any electric field vector that is created by some type of charge either a

positive or negative charge and we could determine the direction of that vector at any point

using examples such as these but now let's focus on some word problems

number one a force of 100 newtons is directed north on a negative 20 micro coulomb point

charge what is the magnitude and direction of the electric field at this point

so feel free to pause the video and work on this example problem so let's begin by drawing a picture

so let's draw our negative point charge now there is an electric force that is

acting on this charge and that electric force is directed north what is the direction of the electric

field now if you recall if we have an electric field pointing

east and if we were to place a positive test charge in that electric field

it will fill a force in the same direction as the electric field

but a negative charge will fill a force that will accelerate in the opposite direction

so if the electric field is due east

the electric force acting on the negative charge will be west

so if the force is north what is the direction of the electric

field actin on this charge q it has to be in the other direction

it has to be south so that is the direction of the electric field

it's direct itself now if we want an angle we can draw this this is 0 90

180 270. so the electric field is directed along

the negative y-axis so we could say that it's at an angle

of 270 degrees relative to the positive x-axis

and that is in a counterclockwise direction so those are the ways in which we can

describe the direction of the electric field we could say it's

due south or it's at an angle of 270 degrees so now that we have the direction

of the electric field let's calculate the magnitude of the electric field

so we could use this formula we can take the electric force and divide it by the magnitude of the charge

the force acting on it is 100 newtons the magnitude of the charge we don't

need to worry about the negative sign since we already know the direction the charge is 20 micro coulombs

and we can replace micro with 10 to the six i mean 10 to the minus six

now let's do some algebra let's see if we can get this answer without the use of a calculator

we could divide a hundred by twenty how many twenties would make up a hundred bucks that's gonna be five

twenty so a hundred divided by twenty is five now the ten to the minus six if we move

it to the top the negative exponent will become a positive exponent so the negative six

will change to positive six and thus the answer is going to be five times ten to the sixth

and the units in newtons per coulomb so that is the electric field that's acting on this negative charge

number two a positive charge of 50 micro coulombs is placed in an electric field of 50 000

newtons per column directed upward what mass should the charge have to remain

suspended in the air well let's talk about how we can create such an electric field

we can create this electric field using a battery and two

parallel metal plates so this is the electrical symbol of a battery

we're going to connect it to these two plates this is the negative side of the battery

and this is the positive side so this plate is going to acquire a positive charge

and this plate is going to acquire a negative charge so if we choose a high enough voltage

and if the distance between the two plates is just right we can adjust it such that we get an

electric field of 50 000 newtons per coulomb now we want the electric field to be

directed upward so we want the positively charged plate to be at the bottom the negatively

charged plate to be at the top remember the electric field always extends away from a positive charge and points

towards a negative charge so this is how we can create a uniform electric field using two parallel plate

i mean using two parallel metal plates now let's place our positive charge

in the middle between these two plates if this charge has the right mass it can actually remain suspended in the

air now let's talk about it so we have an electric field that is

directed upward and it's acting on its positive charge what will be the direction of the

electric force on its charge the electric force will be the same

direction as the electric field for a positive charge so the electric force will be directed

upward now gravity gravity likes to bring things down

so gravity is going to exert a weight force on this positive charge

bringing it down in the negative y direction in order for this charge to remain

suspended in the air the electric force that wants to accelerate the charge

towards the negatively charged pay that has to be exactly equal to the gravitational force or the weight force

that wants to bring the positive charge down towards the

towards this plate so if we can get these two forces to equal each other

then the positive charge will remain suspended in the air so let's write

the forces in the y direction the sum of the forces in the y direction is equal to the electric force this is

positive and the weight force is negative because it's going in a negative y direction

and we want the the sum of the forces in the y direction to be zero so that there's no net acceleration so it

remains suspended in the air moving this term to the other side we see that the electric force has to

equal the weight force now the electric field is equal to f over q

multiplying both sides by q we can see that f the electric force is

the electric field times q so let's replace the electric force with e times q

now the wave force is simply mg so now we have everything that we need in order to calculate the mass of this

charge let's divide both sides by g so the mass of the charge is going to be

equal to the electric field times the magnitude of the charge divided by the gravitational

acceleration the electric field is 50 000 newtons per coulomb

the magnitude of the charge we're dealing with uh a 50 microclimate charge so it's going to be 50

times 10 to the negative 6 coulombs and then we're going to divide that by the gravitational acceleration of 9.8

meters per second squared so it's 50 000 times 50 times 10 to the minus 6 divided by 9.8

and you should get .255 kilograms

so an electric field of 50 000 newtons per coulomb can suspend

a positive charge with a mass of 0.255 kilograms or 255 grams it can suspend it in the air

if the mass is greater than this number then the charge will fall down if it's too light if the mass is less

than this number then the charge will accelerate towards the negatively charged plate

so the mass has to be at the right number in order for it to remain suspended

number three an electron is released from rust in a uniform electric field and accelerates

to the east at a rate of four times ten to the sixth meters per second squared

what is the magnitude and direction of the electric field so let's draw a picture

so first let's draw our electron and then it is

accelerating towards the east now according to newton's second law the net force is equal to the mass times the

acceleration the net force is in the same direction as the acceleration

so the electric force exerted by this electron due to the electric field is going to be due east as well

now if the electric force is east what is the direction of the electric field

it's going to have to be west let me put it here when dealing with a negative charge the

direction of the force and electric field they will be opposite to each other

so now that we have the direction of the electric field let's focus on getting the magnitude of

the electric field so from the last problem we saw that the electric force

is equal to the electric field times the charge and

using newton's second law we can replace the force with the mass times the acceleration

so now we can calculate the electric field if we divide both sides by q so for this problem

the magnitude of the electric field is going to be the mass times the acceleration which is the force

divided by the charge so the mass is what is the mass of an electron if you

look it up the mass of an electron is 9.11 times 10

to the negative 31 kilograms the acceleration given to us in this problem

is 4 times 10 to the 6 meters per second squared

and then we're going to divide that by the magnitude of the charge so what is the charge of an electron

the charge of an electron is negative 1.602 times 10 to the negative 19 coulombs

so these are some numbers that you want to be familiar with so let's put that here

so let's go ahead and plug in these numbers by the way don't worry about the

negative sign for q it's not going to be relevant here so the magnitude of the electric field

is going to be 2.27 times 10 to the negative 5 newtons per coulomb

so this is the answer for this problem and the direction is

west so make sure you write down these numbers

the mass of an electron as we've considered is 9.11 times 10 to the negative 31 kilograms

the mass of a proton is 1.67 times 10

to the negative 27 kilograms the charge of an electron

it's going to be a negative 1.602 times 10 to the negative 19 coulombs

the charge of the proton it's going to have the same magnitude but the opposite charge is going to be

positive 1.602 times 10 to the negative 19 coulombs

so those are some numbers that you want to be familiar with when working on problems associated with

electric fields if you're dealing with protons and electrons number four

a 40 micro coulomb point charge is placed at the origin calculate the magnitude and direction of

the electric field created by the point charge at the following locations

so let's draw the point charge first now point p is five meters away

from the point charge along the x-axis so this is point p and this is five meters away

what is the electric field at point p well we know the direction the direction of the electric field

is going to be east if we draw it from the positive charge

towards point p now to calculate the magnitude of the electric field it's going to be k

q divided by r squared so k is 9 times 10 to 9 and then it's

newtons times square meters over square columns

and i'm running out of space so i'm not going to put the units here i'm just going to write 9 times 10 to the nine

q is 40 micro clones so 40 times 10 to the minus 6 coulombs

r is in meters r is the distance between a charge and the point of interest

so that's five meters so it's going to be nine times 10 to the 9

times 40 times 10 to the minus 6 over 5 squared and so the electric field is going to be

14 400 newtons per coulomb

so that's the magnitude of the electric field at point p and this is the direction

for those of you who want to understand how the units work here it is so k is in newtons

times square meters over square coulombs q is in coulombs and r is in meters so we have square meters

as you can see square meters cancel now coulomb squared we can write that as coulomb times coulomb

so we can cancel one of the coulomb units thus we're left with newtons per coulomb

which is what we have here so that's it for part a so it's 14 400 newtons per coulomb directed east

now let's move on to part b so let's calculate the electric field at point s

so q is at the origin point s is three meters

east of q and then four meters north from that point

so s is at this position here the electric field

can be drawn from q to s so the electric field is going in that direction it's going in the

the northeast direction let's calculate the magnitude of the electric field

so let's use this formula again it's going to be k q

over r squared so k doesn't change it's a constant it's 9 times 10 to 9.

q is still the same it's 40 times 10 to the minus

but r is different r is no longer the value that we have here

but in actuality it turns out r is the same r

is the distance between the charge and the point so we need to use the pythagorean theorem to calculate the

hypotenuse of that right triangle so this is a b and this is c so c squared is equal to a squared plus

b squared a is three b

is four three squared is nine four squared is sixteen

and then nine plus sixteen is twenty five taking the square root of both sides

we get 5 again so by coincidence the electric field is going to have the same magnitude

as it did in part a which was uh 14 400

newtons per coulomb what's going to be different though is the direction of the electric field

vector how can we determine the direction we know it's somewhat in the northeast

direction but not necessarily at a 45 degree angle what we need to do is calculate theta

so perhaps you remember from trigonometry sokotova

if we focus on the torah part that tells us tangent tangent of theta is equal to the

opposite side divided by the adjacent side so tangent of the angle theta is going

to be equal to four over three to calculate theta we need to take the arc tangent

or the inverse tangent of four over three go ahead and type in your calculator and

make sure it's in degree mode so arc tangent 4 3 is

53.1 degrees so that is the angle with respect to

the positive x-axis it's 53.1 so we could say that this is the

electric field vector at an angle of 53.1 degrees with respect to the x-axis or we could say

53.1 degrees north of east because here we're starting from east

and we're heading towards the north direction so it's 53.1 north of east

so that's it for part b so that's how we can calculate the magnitude of the electric field vector

and also its direction using arctangent number five an electron initially at rest

is placed in an electric field of two times ten to the four newtons per coulomb directed to the west

the distance between the plates is one centimeter what is the acceleration of the electron

due to the electric field so the electric field will emanate away from the positive charge and will point

towards the negative charge so as we can see the electric field is directed west

now what effect will it have on the electron a negatively charged particle will fill

a force that will accelerate it in the opposite direction to the electric field so the electron is going to shoot out of

this uh between these two parallel plates let's calculate the acceleration

we know the force acting on a charged particle is equal to the electric field times the charge of that particle

and since this is the only force acting on the electron in the x direction then that force

is going to be the electric force so we can replace the net force with m a based on newton's

second law so m a is equal to e times q and to solve for a we're going to divide

both sides by m so for part a the acceleration is going to be the force which is e

times q divided by the mass so we have an electric field of 2 times

10 to the 4 newtons per coulomb and the charge of an electron is 1.602

times 10 to the negative 19 coulombs we're not going to worry about the negative sign the mass of an electron is

9.11 times 10 to the negative 31 kilograms so let's go ahead and plug this in

so the acceleration is 3.517 times 10 to the 15

meters per second squared so that's going to be the acceleration when the electron

leaves the negatively charged plate so now let's move on to part b

what would be the speed of the electron after it leaves the hole so how can we get that

so now we need to go back to kinematics the electron is initially at rest so v initial is zero

we're looking for the final speed so we'll put a question mark and we know the distance between the

plates it's approximately it's one centimeter

so what kinematic formula has acceleration v initial v final and d it's going to be this one

v final squared is equal to v initial squared plus 2 a d

so to solve for the final speed we simply need to take the square root of both sides

so v final is equal to this the initial speed is zero this is going to be two

times the acceleration which is 3.517 times 10 to the 15.

and the distance between its plates is one centimeter so if we convert one centimeter into meters

we need to divide by 100 and it's going to be 0.01 meters and so we're going to get the square

root of 7.034 times ten to the thirteen and so the final speed

is about eight million and three hundred eighty six thousand eight ninety four point

five so we can round that to let's say 8.39

and this is times 10 to the sixth meters per second so that's how fast the electron is going

to be moving when it leaves the hole number six a 200 micro coulomb charge is placed at

the origin and a negative 300 micro coulomb charge is placed one meter to the right of it

what is the magnitude and direction of the electric field midway between the two charges and then for the second part

30 centimeters to the right of the negative charge so let's start with the first part of

the problem let's begin by drawing a picture so this is going to be the first charge

we'll call it q1 and the second charge q2 so these two charges are separated by a

distance of one meter and we want to determine

the magnitude and the direction of the electric field midway between the two charges

so that's going to be at this point how can we do that well we need to determine the direction

of each electric field at that point q1 is going to create an electric field

called e1 which will be directed east now q2 will create an electric field e2 which starts from the point but points

towards the negative charge and that's going to be directed east as well

so remember the electric field created by a positive charge extends away from the positive charge

but the electric field created by a negative charge points towards the negative charge

so at the center e1 and e2 they are in the same direction so the net electric field

is going to be e1 plus e2 along the x-axis or the horizontal axis both of these are positive because

they're going in the positive x direction e1 is k

times q1 over r1 squared e2 is k times q2 over r2 squared

now what's r1 and r2 so r1 is the distance between q1 and the point of interest

r2 is the distance between q2 and the point of interest

so r1 and r2 they're both half of point of one meter which means that r1 and r2 they're both equal to 0.5

so because r1 and r2 are the same we can simply call it r so let's replace r1 with r

and let's replace r2 with r as well so now we could simplify this equation

by factoring out the gcf the greatest common factor which is going to be k

over r squared and then we're left with q1 plus q2 so this is the formula that we could use

to calculate the net electric field for this particular part in the problem now let's go ahead and plug in the

numbers so it's going to be k which is 9 times 10 to the 9

over r squared r is 0.5 and then times q1 q1 is 200

times 10 to the negative six now for q2 we're going to use a positive value not

negative 300 times 10 to the negative six because we already know the direction of

e2 it's going to the right and it's going to have a positive value because it's heading

in the eastward direction so whenever you're calculating the magnitude for electric field or electric

force you don't need to include the negative charge

you could just find the direction based on where the arrow is going so let's replace q2 with 300

times 10 to the negative six coulombs so 200 plus 300 this becomes 500 so it's

9 times 10 to the 9 times 500 times 10 to negative 6

divided by 0.5 squared thus the net electric field is 18 million which is 1.8

times 10 to the 7 newtons per coulomb so that's the answer for part a

now let's move on to part b so let's redraw the picture for that so here we have our positive charge

and the negative charge so we got q1 q2 and they're separated by distance of

one meter but 30 centimetres to the right or 0.3 meters

we're going to have our new point of interest and let's call this

point b so we want to determine the net electric field at point b

so e1 the electric field created by q1 if we draw it from q1 to point b we could see that it's going east

now if we draw an electric field from point b to q two because it's a negative charge it needs

to go towards a negative charge it's going to the left now which of these two electric fields

is greater e1 or e2 what would you say notice that point b is closer to the negative charge

than it is to the positive charge so e2 is going to have a greater effect than uh i mean q2 is going to have a

greater effect on point b than q1 because it's closer so if you go back to the equation for

electric field there's two things that the electric field depends on

the magnitude of the charge and the inverse square of the distance but the distance is square

so the distance has a greater impact than the charge but also

q2 has a bigger charge in q1 so those are two factors that favor q2 over q1 point b is closer to q2

and q2 has a greater uh charge magnitude than q1 so therefore we can conclude that e2 is

going to be bigger than e1 now the net electric field is going to be positive e1 because it's

going along the positive x-axis and then plus negative e2 because that's moving towards the west or the

negative x-axis now if e1 is bigger than e2 the net electric field will be positive

if e2 is greater than e1 it's going to be negative and we've confirmed that e2 is going to be bigger so we should get a

negative value if we get a positive value for the net electric field we did something wrong

so let's go ahead and do the math so then that electric field is going to be k

q 1 over r 1 squared plus k

q q2 over r2 squared so this time r1 and r2 will be different

so we can't simplify this process by factoring so let's plug in the numbers this is k

q1 is 200 oh this should be a negative sign by the way

based on what we have here so q1 is 200 times 10 to negative 6. now r1

r1 is the distance between q1 and point b so r1 is going to be the sum of one and

point three thus r1 is 1.3 meters squared and then minus

now because we've considered the direction of e2 we've assigned a negative value

we don't need to plug in this negative value for q2 we've already taken that into an account

so it's to be minus k and then q2 we're going to use the positive value of q2

300 times 10 to the minus 6 and r2 that's the distance between q2 and point b

that's 0.3 meters so let's calculate e1 first let's focus on this fraction

nine times ten to the nine times two hundred times ten to the negative six divided by one point three squared

that's one point zero six five times ten to the seven

newton newtons per coulomb now focusing on e2

that's gonna be nine times ten to nine times three hundred times ten to negative six

divided by point three squared so this is 3

times 10 to the 7 newtons per coulomb so we can see that this number is bigger than that number

so the net electric field is going to be negative 1.065

minus 3. wait something is wrong let me double check my work

this should be times ten to the six not ten to the seven that's one million sixty five thousand

and eighty eight so that's one point zero six five times ten to six e2 is 30 million which is three times

ten to seven so now if we subtract those two numbers we get this answer

negative two point eight nine

times so it's 28 million nine hundred thirty five thousand

so it's negative two point eight nine times ten to the seven newtons per coulomb

so we can see why it's negative e2 is significantly larger than e1

and so that's it for this problem number seven the electric field at point x

two meters to the right of a certain positive charge is 100 newtons per coulomb

what will be the magnitude of the new electric field if the magnitude of the positive charge doubles in value

so let's draw a picture first so here is our positive charge and let's say this is

point x and the distance between these two is two meters

now at that point the electric field is a hundred newtons per coulomb when the magnitude of the charge is

we'll call it q but what happens if we double the magnitude of the charge

so the electric field is k q over r squared if you double q

the electric field is going to double a quick and simple way to get the answer is to

plug in 1 for everything that that doesn't change q doubles so we're going to plug in 2

r remains the same so the electric field is going to double it's going to go from 100 to 200.

now what about part b let's say the distance between the charge and point x doubles

so let's say point x is now over here and the magnitude is q what will be the new electric field

so this time q doesn't change k is the same so we're going to replace it with a 1. everything that doesn't

change replace it with 1. now the distance doubles two squared is four so the electric

field is going to be one-fourth of its original value one-fourth of a hundred or hundred

divided by four that's 25 so what you need to take from this is

that the electric field is weaker at a greater distance away from the charge

the closer you move towards the point charge the greater the electric field will be

so as the distance increases the electric field decreases but as the distance from the point

charge decreases the electric field increases there's two ways in which you can

increase the electric field you can increase the magnitude of the charge

which will cause the electric field to go up or

you could reduce the distance between the point of interest and the charge and that will also increase the electric

field now let's move on to part c the distance between the charge

and point x reduces by a factor of three so what's going to be the the magnitude of the new electric field in that case

so we're bringing it a lot closer to q so here's the new uh position of x

so here the distance doubled to 4 meters but now it's going to be reduced by factor three so it's two thirds

of a meter so let's use this formula again for part c k and q doesn't change

r is now one third of its original value one times one is one one squared is one three squared is nine

now we need to multiply the top and the bottom by nine one times nine is nine

one-ninth times nine the nines cancel we get one so the electric field is going to

increase by a factor of nine a hundred times nine is nine hundred so as you can see as we get closer

to the point charge the magnitude of the electric field greatly increases

now what happens if we triple the magnitude of the charge but at the same time reduce the distance

to one-fourth of its original value so i won't draw a new picture for this let's just get the answer

so k doesn't change q triples

and the distance is reduced to one fourth of its value so we have one times three which is

three one squared is one four squared is sixteen

so now i'm going to multiply the top and the bottom by sixteen so it's going to be three times sixteen

which is forty eight so the electric field will increase by a magnitude of forty eight

so the original electric field was a hundred if we multiply that by 48 the new electric field will be 4 800

newtons per coulomb so this right here is the answer to part d so the reason why it's so high is

because we've increased the charge which causes e to go up and at the same

time we reduce the distance which greatly increase the value of e

so that's it for this problem number eight two identical point charges with a

magnitude of a hundred microclimbs are separated by distance of one meter as shown below

part a at what point will the net electric field be equal to zero will it be to the left will it be

between q1 and q2 or to the right so let's identify three points of interest

the first point will be somewhere to the left which we'll call point a

the second point will be somewhere in the middle between q1 and q2 likely the midpoint

and then c will be to the right now to draw the electric field vector created by q1

we need to draw a line from q1 to point a this is going to be e1

and for e2 we're going to draw it from q2 to point a

so at point a both electric fields are moving to the left

so there's going to be a net electric field at point eight it's not going to be zero but here's a question for you

which electric field will be greater e1 or e2 now remember

these charges are identical so the magnitude of the charge is the same the only thing that's different is the

distance q1 is closer to point a than q2 so e1 is going to be bigger

than e2 so this is e1 e2 is going to be a smaller vector nevertheless the net electric field at

point a will be directed west now what about at point c to draw e1 we're going to draw

a line from q1 to point c and e2 will be from q2 to point c now which one is bigger

q2 is closer to point c than q1 so e2 is gonna be bigger at point two i mean e2 is going to be bigger at point

so for e1 we're going to draw a small vector and for e2 we're going to draw a bigger

vector nevertheless the net electric field at point c will be directed east

now what about at point b e1 is going to be directed away from q1 but towards point b

e2 will be directed away from q2 but towards point b and q1 and q2 have the same charge and

at the midpoint at point b they will be equally distant from point b so if the charges are the same and the

distances are the same the magnitude of e1 and e2 will be the same but they're opposite in direction which means that

e1 and e2 cancels so at point b the net electric field is zero

so let's assume that q1 is the origin it's at position zero point b will be at 0.5 meters

so at 0.5 meters relative to the first charge then that electric field

will be zero e1 and e2 will cancel completely so that's the answer for part a

at point b or at the midpoint between q1 and q2 the net electric field will be

zero now what about part b if the charge on q2

doubles to 200 micro coulombs where along the x-axis relative to the

first charge will the net electric field be equal to zero

and the distance between these two is still the same you think the net electric field will be

equal to zero between q1 and b or between b and q2 now q2 is bigger

than q1 so at point b where they're equidistant e1

is going to be a smaller vector than e2 e2 is going to be twice as large in order

to make these vectors equal we need to increase e1 and decrease e2 if we can't change q1 and q2 the only

thing we could change is location we need to move closer to q1 as we move closer to q1 e1 is going to get bigger

e2 will get smaller and at some point they're going to equal each other so we're going to place point p

between q1 and b somewhere between q1 and b the net electric field will be equal to

zero and we need to find that point so that point is going to be r1 r2

is the distance between q2 and point p now let's call r1x if r1 is equal to x what's r2

notice that the total distance is 1 so r2 is going to be 1 minus x

if you add x and 1 minus x you're gonna get one so at this point

all we need to do to get the answer for part b is

get the value of x because x represents the distance relative to the first charge where the

net electric field will be equal to zero so how can we calculate x the net electric field at point b

remember at point b we have e1 which is going towards the right and e2 i mean at point not b but point p

at point p we have e1 going to the right and e2 is going to be going to the left so because e1 is going to the right it's

positive e2 is going to the left it's negative now we want

point p is defined as the point where the net electric field is zero so e1 minus e2 will be zero if we add e2 to

both sides we'll get that e1 is equal to e2 they have to be the same in magnitude

but opposite direction e1 is k q1

over r1 squared e2 is k q2

over r2 squared now let's divide both sides by k so we can cancel

that term let's replace r1 with x so we have q1 over x squared and let's

replace r2 with one minus x don't forget to square it

now let's cross multiply so here we're going to have q 2 times x squared

and then this is going to be q 1 times 1 minus x squared now let's replace q2 with

its value and let's keep the unit micro coulombs so we have 200 micro coulombs times x

squared q1 is a hundred micro coulombs

and what we're going to do at this point is we're going to divide both sides by 100

micro coulombs so the unit micro coulombs will cancel on the left side

on the left we have two hundred over one hundred which is two so we get two x squared is equal to one

minus x squared now we don't need to foil 1 minus x squared

what we could do is take the square root of both sides so we're going to have the square root

of 2 and the square root of x squared is just x the square root of 1 minus x squared is

just 1 minus x now the square root of 2 is 1.414 so now we need to do some algebra

let's add 1x to both sides let me write that better so there's a coefficient of one here

as we add one x to both sides my handwriting is just not working today i don't know why one point four one x

plus one x is two point four one x and we can bring down the one on the right side

so we have two point four one x i mean two point four one four x is equal to one so to get x by itself we

need to divide both sides by two point four one four

so x is going to be one divided by two 2.414 and so we get 0.414

so at 0.414 meters to the right of q1 which is point p at that point

the net electric field will be equal to zero so that's how we can calculate the exact

location along the x axis where the net electric field will be

equal to zero if q1 and q2 have different magnitudes of charge you