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  • - [Instructor] So it turns out solving electric field

  • problems gets significantly harder

  • when there's multiple charges.

  • I mean, theoretically it shouldn't,

  • but people have a lot more problems

  • when there's multiple charges involved.

  • So say the question is this;

  • let's say we wanted to know what's the magnitude

  • and direction of the net electric field,

  • i.e. the total electric field,

  • created halfway between these two charges down here.

  • So you've got a positive eight nanocoulomb charge

  • and a negative eight nanocoulomb charge,

  • and they're separated by six meters

  • from the center to center distance.

  • But what we want to know is what's the total electric field

  • that they both create right there?

  • So each charge is going to create an electric field

  • at this point, and if you add up like vectors,

  • those electric fields, what total electric field

  • would you get?

  • Now at first you might think, well you should

  • just get zero, right?

  • It's very tempting to say that the electric field

  • is just gonna be zero there

  • because you've got a positive eight nanocoulomb charge

  • and a negative eight nanocoulomb charge

  • and those should just cancel, right?

  • But you have to be really careful,

  • turns out that's not true here,

  • this is not gonna be true.

  • And to see why, first you should just draw

  • what is the direction of each field at that point?

  • So this positive eight nanocoulomb charge is gonna

  • create a field at this point that goes radially away

  • from the positive charge, and so it's gonna go to the right.

  • And I'm not even looking, so when I'm trying to find

  • the electric field from this positive charge over here,

  • I'm not even paying attention to this negative charge,

  • I pretend like this negative charge

  • doesn't even exist.

  • Then I just ask what field would this

  • positive charge create?

  • It's still gonna create that field

  • whether this negative charge is over here or not.

  • And now I can do the same thing, I can ask

  • what field would this negative charge create?

  • And I'm gonna pretend like this positive charge

  • isn't even here.

  • So negative charges create a field that go radially in.

  • So over here radially in would point to the right.

  • So these don't cancel.

  • The negative charge created a field radially in,

  • that was to the right, the positive charge created a field

  • radially out of the positive charge,

  • and that was to the right.

  • So not only are these not gonna cancel,

  • these are gonna add up to twice the fields

  • cuz you're gonna add up these vectors,

  • you just add them up if they're in the same direction,

  • and you'll get two times the contribution

  • from one of them.

  • So it's not always the case, in other words

  • it's not always the case that a negative charge

  • and a positive charge have to cancel

  • their electric fields.

  • Those electric fields might point the same direction,

  • so you gotta be careful.

  • So how do we find this net electric field then,

  • what do we do?

  • Well we're gonna say that, all right, this electric field,

  • the first thing I can say is this net electric field

  • is just gonna point in the x direction.

  • So this is just really in the x direction,

  • all I really care about is the electric field

  • in this horizontal direction, and it's gonna be equal

  • to the sum of the electric fields

  • each charge creates there.

  • So we'll do the blue charge first, that's gonna be k

  • times the blue charge divided by r squared.

  • Then we'll do the yellow charge, it's gonna be

  • plus k, the charge of that yellow charge,

  • divided by r squared.

  • So we'll plug in some values here, this k is always

  • nine times 10 to the ninth,

  • and the q of this blue charge was positive eight

  • nanocoulombs, nano is 10 to the negative ninth,

  • I like using nano because then that negative nine

  • cancels with that positive nine.

  • And what distance do I put in here?

  • A lot of people wanna put in six,

  • but that's not what I want.

  • Think about it, I want the net electric field

  • halfway between the two charges,

  • so the r that I care about in this electric field formula

  • is the distance from the charge to the point

  • where I want to determine the electric field,

  • and in that case this is three meters.

  • So for this case, from the charge to the point

  • I'm concerned about finding the field

  • is three meters, not six meters.

  • If we were finding the force these charges exert

  • on each other, then I'd have to use six meters,

  • but that's not what I'm finding, I'm finding the field

  • each charge creates at this halfway point.

  • So I'm gonna plug in three meters down here,

  • and I can't forget to square it.

  • And now I have to be careful, just cuz my charge is positive

  • doesn't necessarily mean that the contribution

  • to the electric field is positive.

  • You have to check, you can't rely on the sign

  • of this charge to tell you whether the contribution's

  • positive or negative.

  • I've gotta look at what direction it points,

  • the direction this positive charge creates a field

  • is to the right.

  • Since that's typically the direction we call positive,

  • then I'm okay with calling this entire term here positive.

  • Then we're gonna have another term.

  • I'm gonna leave off the plus or minus cuz, I mean,

  • it might be plus, it might be minus,

  • we'll leave that off for a second,

  • we'll have to decide when we know what direction it goes.

  • So we do nine times 10 to the ninth,

  • and then the charge is negative eight nanocoulombs,

  • but I am not gonna plug in the negative sign.

  • Oops, and I left off coulomb on the other one here, sorry.

  • And then again, the distance I want is from the charge

  • to the point where we want to find the field,

  • and that again is three meters, and we can't forget

  • to square it.

  • So should this contribution be positive or negative?

  • I can't rely on the negative sign to tell me that,

  • I've gotta look at what direction it goes.

  • Since it goes to the right, that's the positive direction,

  • so this is gonna be plus, these add up,

  • these both go the same direction, the positive direction,

  • so the total net electric field is just gonna be

  • both of these added up.

  • So if I do this, if I square this three I'm getting nine,

  • and nine divided by nine is just one,

  • so I get eight newtons per coulomb,

  • and then this term is really the same thing,

  • nine is divided by nine so that goes away,

  • 10 to the ninth cancels with 10 to the negative ninth

  • and all I'm left with is this eight,

  • so it'd be plus eight newtons per coulomb.

  • So each charge is contributing eight newtons per coulomb

  • of electric field at this point

  • which means that the total net electric field

  • would just be 16 newtons per coulomb at that point.

  • That is the net electric field, that's the magnitude

  • of the net electric field at that point between them.

  • And which way does it go, what's the direction?

  • It goes to the right cuz both of these vectors

  • pointed to the right so the total is gonna be

  • twice as big as one of them and also to the right.

  • Now if you have a case like this and both terms,

  • you know both terms are gonna be equal,

  • you can just write one of them down and multiply by two,

  • you don't have to just add them both up,

  • but I wanted to show you this way so you could see

  • how everything works out.

  • And in the end we get 16 newtons per coulomb

  • for the total field which points to the right.

  • Now what if we changed this, what if we made this

  • instead of a negative eight nanocoulomb charge

  • we made this a positive eight nanocoulomb charge?

  • Well it would no longer create an electric field

  • that points to the right.

  • Positive charges create fields that point radially

  • away from them, so it would create its electric field

  • to the left, which means down here when we find

  • its contribution to the electric field