### Video Transcript

Point charges π one equals 33.2
microcoulombs and π two equals 56.0 microcoulombs are placed 0.85 meters apart
along a line πΏ. How far from π one along πΏ does
the net electric field due to the charges equal zero? What is the magnitude of the
electric field due to the charges at the midpoint of πΏ?

Letβs start out by drawing out the
line πΏ and the two point charges π one and π two on it. So, here is our line πΏ. And it extends, as far as we know,
infinitely far in each direction. Along this line, there is charge π
one and then a distance of 0.85 meters down the line, the charge π two. The first of our two questions
says, how far from π one, the first charge, along πΏ does the net electric field
due to the charges equal zero?

At first, we may not be sure
whether this means a distance to the right of π one or to the left. But, here, the charge values π one
and π two given to us in the problem statement help us out. Notice that both of these charges
are positive. Or at least we assume theyβre
positive because weβre not told theyβre negative. Given a positive charge,
represented here by a plus sign, the electric field created by this charge points
radially outward from it. This means that both π two and π
one create electric field lines that point away from themselves since theyβre both
positive.

Getting back to our question of
whether the electric field is zero to the left or to the right of π one. The fact that both π two and π
one created field lines that point away from themselves means that, to the left of
π one, π oneβs electric field lines will point in this direction. And π twoβs electric field lines
will point in the same direction. Since theyβre in the same
direction, they canβt cancel one another out. But, on the other hand, on the
right side of π one, π oneβs electric field lines will point this way. And π twoβs electric field lines
will point opposite that.

This tells us that the place where
the electric field is zero must be in between π one and π two. That is, itβs to the right of π
one. Knowing that, letβs recall the
mathematical equation for the electric field created by a point charge. The electric field created by a
point charge π, a distance π away from itself, is equal to π times Coulombβs
constant π divided by π squared. In this first question, we know
that the net electric field, that is, the combined field due to both π one and π
two, must equal zero.

So hereβs what we can write. We can say that πΈ sub π one, the
electric field created by π one, minus πΈ sub π two, the field created by the
second charge, is equal to zero. And if we call the distances for
charge one and charge two π one and π two, respectively. Then, using our equation for
electric field, we can write that ππ one over π one squared minus ππ two over
π two squared is equal to zero. Looking at this equation, notice
that a factor of π appears in both terms on the left side. This means that if we divide both
sides of the equation by π, that factor cancels out.

Next, letβs add this term, π two
over π sub two squared, to both sides. When we do, we get this interesting
equation. π one over π one squared is equal
to π two over π two squared. Thatβs what the net electric field
being zero means. Considering again this first
question, we want to answer how far from π one along our line πΏ does this field
zero out. In other words, at some location in
between these two charges, letβs just pick that spot for illustration, the electric
field is zero. And we want to find the distance
between that point and π one. Notice though that, in our
equation, that distance is represented by π one. And, similarly, the distance from
the point where the electric field is zero over to π two is equal to π two.

Seeing this on our diagram, another
equation may stand out to us. And that is that if we add π one
and π two together thatβs equal to the total distance separating π one and π two:
0.85 meters. So itβs π one we want to solve
for. Thatβs the answer to this first
question. And to do that, using this
equation, weβll begin by solving not for π one but for π two. If we cross-multiply so that π two
is isolated on one side of this equation, we find that π sub two squared is equal
to π sub one squared times π two over π one. Or, taking the square root of both
sides, π two equals π one times the square root of π two over π one.

What weβre going to do next is take
this expression for π two and substitute it in for π two in our equation which
says that π one plus π two is equal to 0.85 meters. When we do that, notice that we now
have an equation entirely in terms of the charges, which we know, and π one, which
is what we want to solve for. To rearrange this equation and
solve for π one, letβs start by factoring it out of both terms on the left side of
the equation. And then, finally, weβll divide
both sides of the equation by the expression in parentheses, one plus the square
root of π two over π one.

Now to solve for π one, itβs just
a matter of plugging in for π two and π one. And recall that weβre told π two
and π one in our problem statement. If we let the exponents and units
cancel out, as they do since they appear both in numerator and denominator, this
fraction becomes 56.0 divided by 33.2. And, finally, when we go to
calculate π one, we find that, to two significant figures, itβs 0.37 meters. Thatβs the distance from charge π
one along line πΏ at which the electric field is zero. Thanks to the influence of the
electric fields from these two charges.

In the next part of our question,
we want to solve for the magnitude of the electric field due to the charges not at
this location but at the midpoint between π one and π two. We know this field magnitude wonβt
be zero. And to solve for its numerical
value, letβs refer again to our equation for the electric field created by a point
charge. This field at the midpoint, which
we can call πΈ sub π, will be equal to the electric field caused by charge π two
minus the field caused by charge π one.

The reason for this minus sign is
simply that we want our overall electric field at the midpoint to be positive. And we know that π two is greater
than π one and therefore will create a stronger field at that midpoint. So then, working off of our
electric field relationship, the electric field at the midpoint between our two
charges is equal to Coulombβs constant times π two divided by π sub π squared,
where π sub π is half the distance between π one and π two, minus π times π
one over π sub π squared.

Looking at the right side of this
equation, notice that we have factors of π over π sub π squared in both terms and
can therefore factor them out. This is what our expression
simplifies to. And, as a next step, letβs start
finding out the numerical values of these various terms on the right-hand side. First, starting with Coulombβs
constant π, this constant is approximately 8.99 times 10 to the ninth newton-meters
squared per coulomb squared. Then, π sub π, halfway the
distance between π one and π two, must be that total distance, 0.85 meters,
divided by two.

And, finally, π two and π one,
the charge values, are given to us. Letβs write an expression for πΈ
sub π then with all these values plugged in. When we do substitute in these
numbers, notice what happens to the units in the expression. One factor of the unit of charge,
coulombs, cancels out. And both factors of distance, in
meters, cancel out. When we enter this expression on
our calculator, we find a result of 1.1 times 10 to the sixth newtons per
coulomb. Thatβs the electric field magnitude
midway between the two charges π one and π two.