### Video Transcript

A thin rod has a linear charge
density of 0.0022 coulombs per metre. What is the magnitude of the
electric force on the point charge ๐ is equal to 9.0 microcoulombs, knowing that ๐
is equal to 10 centimetres and ๐ฟ is equal to 40 centimetres?

Okay, so in this question, weโve
got thin rod here. And weโve been told that that rod
has a linear charge density which weโll call ๐ of 0.0022 coulombs per metre. As well as this, we know that weโve
got a point charge which weโve drawn very large here so we can see it clearly. But we know itโs a positive
charge. And it has a magnitude ๐ of 9.0
microcoulombs. Finally, weโve been given
information about the length of the rod ๐ฟ and the distance between the right-hand
end of the rod and the charge particle which is ๐. So weโve been told that ๐ is equal
to 10 centimetres and ๐ฟ is equal to 40 centimetres.

Weโve been asked to find the
magnitude or size of the electric force on the point charge ๐ is equal to 9.0
microcoulombs. In other words, what is the force
exerted by this rod on this point charge? So letโs call this force ๐น. And we donโt yet know what this is,
so weโll put a question mark. So how weโre gonna go about finding
the value of ๐น? Well first thing first, letโs zoom
in to our setup a little bit. So here is a bit more of a close-up
of our rod and our point particle. Now at this point, what we can do
is to split up our thin rod into very small sections.

Once weโve done this, what we can
do is to work out the force exerted by a random section of the rod on this point
particle. And then generalize so we can work
out the force exerted by any other section on the point particle and then add up the
effect of each one of these sections. And weโll do this in the limit that
the width of one of these sections, which weโll call d๐ฅ, because the rod is
pointing in the ๐ฅ-direction. Weโll do this in the limit that d๐ฅ
is very very small. And the reason we do that is
because then we can treat this section and every other section as a point particle
itself because, remember, this width of the rod is negligible. Weโve been told in the question
that the rod is thin. Therefore, it basically doesnโt
have a width. But weโve drawn it like this just
to make things a little bit clearer.

And so if we make d๐ฅ very very
small, then weโve got another point particle. And essentially, weโre making a
charge rod out of lots of different point particles and seeing the effect of all of
this point particles lined up together on this point particle. Now, this is very useful because we
can recall Coulombโs law. Coulombโs law tells us that the
force between two charged point particles is equal to the charge of the first
particle which weโll call ๐ one multiplied by the charge of the second particle ๐
two divided by four ๐๐ nought, which are all just constants, and ๐ squared, where
๐ is the distance between the two point particles.

Now, we can use this equation in
our scenario because weโre considering this point particle and any segment of the
rod, which weโve consider to also be a point particle. So letโs say that weโre considering
this particular segment here. Letโs also say that it has an
๐ฅ-coordinate which weโll call ๐ฅ, super creative but thatโs what weโll go with. In other words then, the segment
that weโre considering is ๐ฅ away from the origin or ๐ฅ away from the left-hand end
of the rod.

Now, we also know from the question
that the length of the rod itself is ๐ฟ. And the distance between the
right-hand end of the rod and the point particle is ๐. Therefore, the distance between the
section of the rod that weโre considering and the point particle is equal to this
distance, thatโs ๐ฟ plus ๐, minus this distance, which is ๐ฅ. In other words then, the ๐ in our
equation that weโre considering, thatโs the distance between our first point
particle and our second point particle, is given by ๐ฟ plus ๐ minus ๐ฅ. So at this point, weโve worked out
the value of ๐ in our force equation and so we can plug in the values of the
charges of the point particle and the segment of the rod.

Now, we know that, in this case,
the charge of the point particle is lowercase ๐. But what is the charge of the
segment of the rod? Well, letโs say that the segment of
the rod has a charge d uppercase ๐. And we can take this charge d๐ and
link it to our charge for unit length ๐. Because we assume that the charge
is evenly distributed throughout the rod and we can do this because weโve been given
a charge per unit length, we can then say that the charge d๐ divided by the length
of the segment of the rod d๐ฅ is equal to ๐ because, once again remember, ๐ is the
charge per unit length. And the segment of the rod that
weโre considering has a charge d๐ and a length d๐ฅ.

So d๐ divided by d๐ฅ is equal to
๐. And, therefore, we can rearrange by
multiplying both sides by d๐ฅ to give us d๐ is equal to ๐d๐ฅ. In other words then, the charge on
the segment of the rod is equal to the charge per unit length multiplied by the
length of the segment of the rod. So letโs write that down over here
and then start plugging stuff into our force equation.

Now at this point, because weโre
working with very small values such as d๐ฅ and d๐, we can say that the force
between the point particle and the segment that weโre considering or, in other
words, the force exerted by the segment on the particle or by the particle on the
segment is given by d๐น rather than ๐น for force because, remember, weโre
considering the limit in which d๐ฅ is very very small. And so the force between a
particular segment of the rod and the point particle is also going to be very small,
infinitesimal even, as Snagglepuss would say.

Anyway, so d๐น now is equal to the
charge of the point particle, which we know to be ๐, multiplied by the charge of
the segment that weโre considering, which we know to be d๐. And weโll substitute the value of
the d capital ๐ in in a second divided by four ๐๐ nought multiplied by the
distance between the two, which happens to be ๐ฟ plus ๐ minus ๐ฅ. And we square this distance. So this is our expression for the
force exerted between this segment of the rod and this point particle. And itโs important to remember that
this is the electric force exerted between the two objects but is also, therefore,
the force exerted by one object on the other or by the other on the first one.

And since weโve been asked to find
the force exerted on the point particle, this is still the same force. Now at this point, we can
substitute in ๐ d๐ฅ for d capital ๐. And so weโre left with the force
expression d๐น is equal to ๐ multiplied by ๐ d๐ฅ divided by four ๐๐ nought
multiplied by ๐ฟ plus ๐ minus ๐ฅ whole squared. Now at this point, we can sum up
all the contributions from all of the segments of the rod in the limit that d๐ฅ
becomes very very small. Because of this, well what weโll be
doing is finding the force exerted on this point particle by this segment which we
were considering plus this segment plus this segment plus this segment plus this
segment plus all of the segments in the rod.

And what this means in simpler
words is that weโll be integrating both sides of our equation. But then what are the limits of the
integral on the right hand side of our equation now? Well, weโre summing up the
contributions from all of the segments in the rod. And the rod begins at ๐ฅ is equal
to zero and ends at ๐ฅ is equal to ๐ฟ. So our limits are going to be ๐ฅ is
equal to zero and ๐ฟ. And the left-hand side simply
integrates to ๐น. Thatโs the force exerted by the rod
on the point particle. This means that all we need to do
is to evaluate the integral on the right-hand side of the equation.

But before we do that, letโs bring
all of the constants from inside the integral to outside the integral. And so weโve got the constants ๐,
๐, four, ๐, and ๐ nought which weโve now brought outside the integral. And so we multiply all of these
constants by the integral from ๐ฅ is equal to zero to ๐ฟ of one divided by ๐ฟ plus
๐ minus ๐ฅ whole squared with respect to ๐ฅ. Now, the integral evaluates to one
divided by ๐ฟ plus ๐ minus ๐ฅ. And of course, the limits are ๐ฅ is
equal to zero to ๐ฅ is equal to ๐ฟ. To see this more clearly, attempt
this integral with the substitution ๐ข is equal to ๐ฟ plus ๐ minus ๐ฅ. Give it a try.

Anyway, so this is what the
integral evaluates to. And this is what we get when we
substitute ๐ฅ is equal to ๐ฟ and ๐ฅ is equal to zero. But then the denominator here is ๐ฟ
plus ๐ minus ๐ฟ which just becomes ๐. And here, the denominator is ๐ฟ
plus ๐ minus zero which just becomes ๐ฟ plus ๐. Now at this point, we can just
simplify the expression in the parentheses a little bit more by multiplying the
first fraction by ๐ฟ over ๐ on the numerator and the denominator and by multiplying
the second fraction by ๐ on the numerator and the denominator.

This leaves us with ๐ฟ plus ๐ in
the numerator of the first fraction and ๐ in the numerator of the second. And happily, weโve got ๐
multiplied by ๐ฟ plus ๐ in the denominator of both fractions. This means weโve got a common
denominator. And so we can add the numerators of
the two fractions. Therefore, what we get left with is
๐ฟ plus ๐ minus ๐ all divided by ๐ multiplied by ๐ฟ plus ๐. In other words, weโre left with ๐ฟ
divided by ๐ multiplied by ๐ฟ plus ๐. So now, this is our expression for
the force exerted by the rod on the point particle with charge lowercase ๐. And at this point, we know all of
the quantities. We know ๐, ๐, ๐ฟ, and ๐. And we can look up the value of ๐
nought which happens to be 8.85 times 10 to the power of negative 12 per metre cubed
per kilogram second to the power of four amp squared.

But before we sub in all these
values, itโs important to remember to convert them all to standard units. Now, the SI unit for charge is
coulombs. So we need to convert 9.0
microcoulombs to coulombs. Now, we can recall that the prefix
โmicroโ means 10 to the power of negative six. So weโre left with 9.0 times 10 to
the power of negative six coulombs. And of course, a centimetre is a
hundredth of a metre. So for the values of ๐ and ๐ฟ, we
get 0.1 metres and 0.4 metres, respectively. Now, we can plug in all of our
values to the expression on the bottom of the screen, which looks a little bit like
this. And if we wanted to, we could go
through all of the units to make sure they were cancelled out to leave us with the
unit of force.

But in the end, we get a value for
๐น as being 1424 newtons. However, itโs important to recall
that the information given to us in the question gave us quantities to two
significant figures as the minimum number of significant figures. Therefore, we need to give our
answer to two significant figures as well. And so giving our answer to two
significant figures and converting it to standard form for fun, we can say that the
value of the force exerted by this rod on the point particle is 1.4 times 10 to the
power of three newtons. That is a massive force by the
way. But at this point, weโve reached
the end of the question.