### Video Transcript

Two fixed point charges each have
magnitudes of 4.0 times 10 to the negative six coulombs. The charges are located on the
π₯-axis at the positions π₯ equals 3.0 meters and π₯ equals negative 3.0 meters. A charge π is placed at the origin
and the resultant electric field of the three charges is zero at a point on the
π¦-axis where π¦ equals 3.0 meters. Find the magnitude of π.

As we get started, letβs first
sketch in the location of these three charges. We have here a set of coordinate
axes showing displacement in the horizontal π₯- and vertical π¦-directions in units
of metres. Weβre told that one of the two
fixed point charges is at a location on the π₯ axis at 3.0 meters and the other is
on the same axis at negative 3.0 meters. Along with that, thereβs a charge
at the origin and itβs labelled capital π. And itβs the magnitude of that
charge we want to solve for.

The clue that helps us to solve for
the magnitude of π is that the electric field on the π¦-axis at a location 3.0
metres of that axis weβre told is equal to zero. And thatβs the combined electric
field from these three point charges. To begin solving for the magnitude
of π, letβs start by recording the magnitude of the two fixed point charges. Weβll call that π sub π. And then having recorded that,
letβs clear some space on screen for our calculations.

One aspect of this problem scenario
that may stand out is that weβre only told the magnitude of the two fixed point
charges that are on the π₯-axis. We donβt know their sign. And actually, this may have an
impact on the electric field at the location of interest, 3.0 metres up the
π¦-axis. To see why, letβs consider the
electric fields that are created by a positive charge represented by this plus sign
and a negative charge represented by this minus sign.

If we were to draw in the electric
field lines coming from this positive charge, ignoring the presence of the minus
charge, the net field would point really away from the charge and would point
outward. But on the other hand, if we would
have drawn the electric field lines created by this negative charge, ignoring the
presence of the positive charge, then this field would also point radially. But it would point in toward the
minus sign, instead of away from it with the plus. What this means is that depending
on whether our two fixed point charges are positive or negative, weβll change the
direction of the electric field that they create at our point of interest.

To see how this works, letβs create
a small table on the left side of our screen. In this table, weβll have a
charge. Weβll call it π sub π one. And that charge represents our left
fixed charge at a location on the π₯-axis of negative 3.0 metres. And the other of these two fixed
charges β the one at positive three metres β letβs call it π sub π two. If we were to write down all the
possible sign combinations of these two charges, there will be four of them.

First, itβs possible that both
these charges are positive. Then, itβs possible that theyβre
both negative. Then, itβs possible that π sub π
one is positive and π sub π two is negative. And then, the last possibility is
the opposite of that. And at this point, letβs extend our
table and weβll add a third column. This third column will be the
electric field created at our point of interest 3.0 metres up the π¦-axis from just
these two charges, π sub π one and π sub π two. In other words, for now, weβre
neglecting the presence of the charge π at the origin.

Letβs now walk through this table
row by row. And as we do, weβll fill in the
third column. That is, weβll draw out the
direction of the electric field that would form from π sub π one and π sub π two
having these signs to their charges. Beginning with the first row, we
imagine that both these charges have a positive sign to them. If we glance over at the way a
positive sign creates an electric field, we see that field points outward from the
charge.

This means that π sub π one would
create an electric field at our point of interest, which would go in this direction
and π sub π two would create a field at that same point which would roughly go in
this direction. If we add these two vectors
together, the resultant has an upward vertical component and no horizontal
component. Those components cancel out. So if both π sub π one and π sub
π were positive, our electric field would point upward.

Continuing on, letβs look at the
second row of our table, imagining that these two charges were both negative. We see that in this case both
charges would create an electric field that points in toward the charge. So π sub π one will create a
field at our point of interest in this direction and π sub π two will create a
field in this direction. If we add these two fields
together, we do get a resultant component in the vertical direction, this time
downward and once again no horizontal component. So we draw that in our table. If both charges are negative, then
the electric field created by the two has a net downward direction and no horizontal
component.

Now, we move on to imagine that π
sub π one, the charge on the left, is positive, while π sub π two is
negative. In this instance, the electric
field created by π sub π one would point in this direction and that created by π
sub π two would point in this direction. If we add these vectors together,
we see that now they only have a horizontal component and no vertical component. In this case, the electric field at
our point of interest would point left to right.

And then lastly, we reverse the
signs of each charge. π sub π one has a minus sign and
π sub π two has a plus sign. Now, the electric field at our
point of interest from π sub π one would point in this direction and that from π
sub π two would point about like this. Adding them together, we again get
a resultant that only has a horizontal component, no vertical. And in this case, that vector
points from right to left.

Now that our table is done, take a
look at the last two rows, in particular, the electric field direction resulting
from these charge combinations. If the electric field created by
the two fixed point charges only had a horizontal component, then that would mean
that regardless of the sign or value of our charge π at the origin, it couldnβt
possibly make the electric field be zero at the point of interest. Thatβs because the charge π β
thanks to its location relative to the point of interest β can only create a
vertically aligned electric field at the point of interest. But if the field created by π at
that point is only vertical, that means it couldnβt cancel out the horizontal
component created in these last two scenarios.

But since we know for a fact that
the electric field is indeed zero at that point, thanks to the influence of these
three point charges, then we know that π sub π one and π sub π two cannot have
opposite signs to them. Even though weβre only told the
magnitude of these two charges, we know that in the end they must either both have a
positive charge or both have a negative charge. And that is so when we add in the
electric field created by π, the sum of those three fields at our point of interest
is zero.

All that being said, letβs now
start to calculate the electric field created from each of these point charges. And to do it, we can recall the
mathematical equation for the electric field created by a point charge. Given a point charge π, when we
solve for the electric field a distance π away from the charge, thatβs equal to
Coulombβs constant πΎ times π all divided by π that distance squared.

In our scenario, we know that the
electric field is zero at 3.0 metres up the π¦-axis thanks to the influence of three
point charges. With these electric fields from our
three point charges written out, there are two things we want to explain about this
equation. First, letβs talk about the signs:
plus between the first two terms and minus between the last two. We add the electric fields created
by π sub π one and π sub π two because they combine in the same direction.

On the other hand, this field is
opposed by the field created by our point charge π at the origin. So thatβs one thing. And then, the second thing is
notice in our subscript that we have parentheses π£ in our first two terms for the
fields created by our fixed point charges. Thatβs a notational shorthand to
let us know that weβre only considering the vertical component of the fields created
by these charges. And why weβre doing that? Itβs because as we saw earlier, the
horizontal component must cancel out.

Now, if we look at the line of
action between π sub π one and our point of interest and π sub π two and that
same point, we see that they make a 45-degree angle with that point. In other words, if weβre to draw in
an angle and call it π between the line of action and the horizontal π₯-axis, π
would be 45 degrees. This means we can rewrite our
overall equation as zero is equal to the electric field from π sub π one times the
sin of 45 degrees plus the electric field from π sub π two times that same trig
angle minus πΈ sub π, the electric field created by our point charge at the
origin. And either recalling this value or
looking it up, we can remember that the sin of 45 degrees can be represented exactly
as the square root of two over two.

We now have an expression for the
net electric field at our point of interest which is entirely in the vertical
direction. As a next step, letβs start filling
in for our electric field in terms of the equation for that field weβve written
down. When we do that, note that for the
distances between our charges and the point of interest for the electric field is
zero, weβve called π sub π π one, π sub π π two, and π sub π,
respectively. Our goal is to work with this
expression to solve for capital π, the magnitude of that charge is the answer to
our question.

To start in that direction, we can
notice that Coulombβs constant πΎ appears in each term on the right side of our
expression. This means we can divide both sides
of our equation by πΎ and cancel it out. Next, letβs notice something about
π sub π one and π sub π two, specifically, their distances away from the point
of interest for the electric field is zero. If we look at our sketch, we can
see the those distances between those fixed charges and where the electric field is
zero are the same. That is, π sub π π one is equal
to π sub π π two. And not only that, but the charges
themselves π sub π one π sub π two are also equal and we know their
magnitude.

The symbolic name weβve given to
these fixed charges is π sub π. And letβs call the distance that
these charges exist from our point of interest π sub π. Using that convention, we can see
that we now have two identical terms in our equation and we can therefore add them
together. When we do, notice that the factor
of two that would result out from can be cancelled with the factor of one-half that
appears in each term. Our overall equation simplifies so
that it now reads zero is equal to π sub π over π sub π squared times the square
root of two minus π over π sub π squared.

Since we want to solve for π,
letβs add this term π over π sub π squared to both sides of equation which
effectively cancels that term from the right side. And then, weβll multiply both sides
of the equation by π sub π squared. We now have an expression for the
quantity we want to solve for, π. To figure it out, all we need to do
is plug in for π sub π, π sub π, and π sub π. We already know π sub π, 4.0
times 10 to the negative six coulombs.

Looking again at our sketch, we can
label the dashed lines π sub π since itβs that distance that represents the
distance between our fixed charges and the place where the electric field is
zero. Note that π sub π is the
hypotenuse of our right triangle. This means we can use the
Pythagorean theorem to solve for it. The two legs of this triangle other
than π sub π are 3.0 metres long and 3.0 metres long. So by the Pythagorean theorem, π
sub π squared is equal to 3.0 meters squared plus 3.0 meters squared. That gives us 18 meters squared on
the right side. And if we then take the square root
of both sides of this expression, we see that π sub π is equal to the square root
of 18 metres. So that takes care of this terminal
equation.

Now, what about π sub π? Remember that π sub π is the
distance from our point charge π which is at the origin to the place where the
electric field is zero. Weβre told that this is 3.0 meters
up the π¦-axis. Therefore, π sub π is 3.0
meters. And weβre now ready to plug in for
π sub π, π sub π, and π sub π. We see that π is equal to 4.0
times 10 to the negative six coulombs divided by the square root of 18 meters
quantity squared multiplied by the square root of two multiplied by 3.0 meters all
squared.

This overall expression reduces to
the square root of two divided by two multiplied by π sub π. This is equal to 2.8 times 10 to
the negative six coulombs. Thatβs the magnitude of the charge
we would need to place at the origin in order to make the electric field 3.0 metres
above that point equal to zero.