Main content

### Course: Class 12 Physics (India) > Unit 1

Lesson 5: Electric field due to point charges- Magnitude of electric field created by a charge
- Electric field due to a point charge
- Electric field (vector) due to a point charge
- Net electric field from multiple charges in 1D
- Superposition principle (basic)
- Find where a third charge would be in equilibrium
- Electric field due to two charges on line joining them
- Net electric field from multiple charges in 2D
- Superposition principle (intermediate)

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Net electric field from multiple charges in 1D

In this video David solves an example problem to find the net electric field created by multiple charges at a point in between them. Created by David SantoPietro.

## Want to join the conversation?

- does it mean that the magnitude of net electric field created halfway between two charges of same magnitude and charge be always zero?irrespective of distance between them?Then how do they repel?(19 votes)
- Think of it this way. Two people are pushing each other with the same force. If you go and stand in the middle of them you wont have to apply any sort of force to stand. But the guys on the side, who are actually doing all the pushing will have stop themselves from moving backward with their legs. That's why the charges actually repel if no external force is holding the two charges together.(6 votes)

- Isn't electric field a vector quantity? Then how is it possible to add them up using algebraic sum? shouldn't vector sum be used?(4 votes)
- You are right to assume that it is a vector quantity. The reason we are adding them or subtracting them algebraically is that we have already figured out the direction of the vectors before we write them down. Thus we can decide if the vectors have a positive sign (pointing to the right or up if we have decided that these are the positive directions for example) or a negative sign (pointing to the left or down).(17 votes)

- Why is the negative charge pointing to the right? Shouldn't it be pointing left, since negative and positive charges attract?(2 votes)
- You are right that they attract but when talking about Electric Fields, think about them as roads. When dealing with a negative Q, electric fields always point
*toward*the charge and positive Q field always point*away*from the charge.(7 votes)

- What does one mean by the point that the two same charges are placed in same line and another charge is brought exactly between the charges such that the system is in an equilibrium?(2 votes)
- My understanding of the question would be this...

equilibrium = these three charges have no acceleration. They dont experience a net force. The forces within the system is balanced.

More detail...

Take two positive charges. A distance r between them. They will feel a mutual force of repulsion.

Now place a negative charge half way between them.

Now the negative charge feels two (equal) forces; one from each of the two positive forces.

Each of the positive forces continue to feel the positive (Repulsive) force from the other positive charge. PLUS the attractive force from the negative charge.

[ Hint: Think about what happens to the forces when r is very large and also when very small]

So; i think the question is asking this... what must the distance r be (the distance between the positive charges) so that all these forces cancel out? ie the system is in equilibrium

Alternatively, the question may be asking you to find r/2. (the distance from centre to outer charges) but the maths is about the same..

Hope that makes sense

and do let us know when you get your answer

[For a mathematical / visual solution You might try drawing graphs of forces/ distance and total (resultant forces) for each particle]

:-)(6 votes)

- What would happen if both charges were negative? They would both point inward, so do they cancel?(2 votes)
- The same thing would happen as in the case for positively charged particles.

But this time each negative charge's electric field will point inwardly towards itself.Since both charges are put at opposite ends,then each charge's electric field will be opposite with respect to the other,thus counteracting each other's effect on the particle placed halfway between them.(3 votes)

- Is there any other unit or name for newtons per coulombs?(1 vote)
- N/C is the same as volt/meter. Both measure electric field strength(3 votes)

- is the direction of the electric field created by a negative charge radially outward in the vicinity of another negative charge?(1 vote)
- Electric field direction is always determined by the direction of the force that would be felt by a small positive charge placed in the field at that spot.(3 votes)

- At6:47, how when the direction is right it is +ve , when left it is -ve ?(1 vote)
- Great question,

The + sign indicates the field ( at some point ) is pointing to the right.

The - sign indicates the field ( at some point ) is pointing to the left.

This comes in incredibly handy when using mathematics to work out electric field problems. Instead of writing 'the electrical field at this point is pointing to the right'

or 'this electrical field is pointing to the left at this particular point' we can express both statements with one of two symbols.(3 votes)

- Two equal positive charges are held in place at a fixed distance. If you put a third positive charge midway between these two charges, its electrical potential energy of the system (relative to infinity) is zero because the electrical forces on the third charge due to the two fixed charges just balance each other.IS THIS TRUE OR FALSE(2 votes)
- It is only true when when the charges of the first 2 charges are equal and opposite [i.e. one being negative and other positive, but with the same magnitude value {eg. +2 and -2 or +8 and -8}].. like in the above given case.

But, if you are to take charges that are not equal and opposite, it is false(1 vote)

- why here you write 10 to power -9 with both 8(1 vote)
- He wrote -9 because the value of nanocoulomb is 10^-9. If it is 8C he would not have written -9(3 votes)

## Video transcript

- [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 we'd have to include it
as a negative contribution cuz it's pointing in
the negative direction. Even though it's a positive
charge, the contribution it gives to the total
electric field is negative cuz it points in the negative direction. And that would give me zero,
so if I had this a positive this whole thing would add
up to zero cuz I'd have eight and then minus eight and I'd get zero newtons per coulomb, so the electric field would completely cancel
right in the middle. So what I'm saying is you
have to be very careful with your negative
signs, don't just assume these contributions are
always gonna add up. You can find each one always
plugging in the charges as positive even if they're negative and then decide should I add or subtract these contributions based on whether they go to the right or to the left. If they point to the right
you'd choose a positive in front of this term since it points in the positive x direction. And if they point to the
left you're gonna choose a negative in front of this term because it would point in
the negative x direction. So recapping, to find
the total electric field from multiple charges,
draw the electric field each charge creates at the
point where you want to determine the total electric field, use this formula to get the
magnitude of the contribution from each charge, then decide
whether those contributions should be positive or
negative based not on the sign of the charge but the
direction the field is pointing from that charge, add up
the two contributions, and that'll give you
the total electric field at that point.