# Video: Calculating the Electric Force on a Point Charge due to a Linear Charge Density

A thin rod has a linear charge density of 0.0022 c/m. What is the magnitude of the electric force on the point charge 𝑞 = 9.0 𝜇C, knowing that 𝑎 = 10 cm and 𝐿 = 40 cm?

09:37

### 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 𝑞 is equal to 9.0 microcoulombs, knowing that 𝑎 is equal to 10 centimetres and 𝐿 is equal to 40 centimetres?

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

We’ve been asked to find the magnitude or size of the electric force on the point charge 𝑞 is equal to 9.0 microcoulombs. In other words, what is the force exerted by this rod on this point charge? So let’s call this force 𝐹. And we don’t yet know what this is, so we’ll put a question mark. So how we’re gonna go about finding the value of 𝐹? Well first thing first, let’s 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 we’ve 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 we’ll do this in the limit that the width of one of these sections, which we’ll call d𝑥, because the rod is pointing in the 𝑥-direction. We’ll do this in the limit that d𝑥 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. We’ve been told in the question that the rod is thin. Therefore, it basically doesn’t have a width. But we’ve drawn it like this just to make things a little bit clearer.

And so if we make d𝑥 very very small, then we’ve got another point particle. And essentially, we’re 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 Coulomb’s law. Coulomb’s law tells us that the force between two charged point particles is equal to the charge of the first particle which we’ll call 𝑄 one multiplied by the charge of the second particle 𝑄 two divided by four 𝜋𝜀 nought, which are all just constants, and 𝑟 squared, where 𝑟 is the distance between the two point particles.

Now, we can use this equation in our scenario because we’re considering this point particle and any segment of the rod, which we’ve consider to also be a point particle. So let’s say that we’re considering this particular segment here. Let’s also say that it has an 𝑥-coordinate which we’ll call 𝑥, super creative but that’s what we’ll go with. In other words then, the segment that we’re considering is 𝑥 away from the origin or 𝑥 away from the left-hand end of the rod.

Now, we also know from the question that the length of the rod itself is 𝐿. And the distance between the right-hand end of the rod and the point particle is 𝑎. Therefore, the distance between the section of the rod that we’re considering and the point particle is equal to this distance, that’s 𝐿 plus 𝑎, minus this distance, which is 𝑥. In other words then, the 𝑟 in our equation that we’re considering, that’s the distance between our first point particle and our second point particle, is given by 𝐿 plus 𝑎 minus 𝑥. So at this point, we’ve worked out the value of 𝑟 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 𝑞. But what is the charge of the segment of the rod? Well, let’s say that the segment of the rod has a charge d uppercase 𝑄. And we can take this charge d𝑄 and link it to our charge for unit length 𝜆. Because we assume that the charge is evenly distributed throughout the rod and we can do this because we’ve been given a charge per unit length, we can then say that the charge d𝑄 divided by the length of the segment of the rod d𝑥 is equal to 𝜆 because, once again remember, 𝜆 is the charge per unit length. And the segment of the rod that we’re considering has a charge d𝑄 and a length d𝑥.

So d𝑄 divided by d𝑥 is equal to 𝜆. And, therefore, we can rearrange by multiplying both sides by d𝑥 to give us d𝑄 is equal to 𝜆d𝑥. 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 let’s write that down over here and then start plugging stuff into our force equation.

Now at this point, because we’re working with very small values such as d𝑥 and d𝑄, we can say that the force between the point particle and the segment that we’re considering or, in other words, the force exerted by the segment on the particle or by the particle on the segment is given by d𝐹 rather than 𝐹 for force because, remember, we’re considering the limit in which d𝑥 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 d𝐹 now is equal to the charge of the point particle, which we know to be 𝑞, multiplied by the charge of the segment that we’re considering, which we know to be d𝑄. And we’ll substitute the value of the d capital 𝑄 in in a second divided by four 𝜋𝜀 nought multiplied by the distance between the two, which happens to be 𝐿 plus 𝑎 minus 𝑥. 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 it’s 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 we’ve 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 𝜆 d𝑥 for d capital 𝑄. And so we’re left with the force expression d𝐹 is equal to 𝑞 multiplied by 𝜆 d𝑥 divided by four 𝜋𝜀 nought multiplied by 𝐿 plus 𝑎 minus 𝑥 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 d𝑥 becomes very very small. Because of this, well what we’ll 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 we’ll 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, we’re summing up the contributions from all of the segments in the rod. And the rod begins at 𝑥 is equal to zero and ends at 𝑥 is equal to 𝐿. So our limits are going to be 𝑥 is equal to zero and 𝐿. And the left-hand side simply integrates to 𝐹. That’s 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, let’s bring all of the constants from inside the integral to outside the integral. And so we’ve got the constants 𝑞, 𝜆, four, 𝜋, and 𝜀 nought which we’ve now brought outside the integral. And so we multiply all of these constants by the integral from 𝑥 is equal to zero to 𝐿 of one divided by 𝐿 plus 𝑎 minus 𝑥 whole squared with respect to 𝑥. Now, the integral evaluates to one divided by 𝐿 plus 𝑎 minus 𝑥. And of course, the limits are 𝑥 is equal to zero to 𝑥 is equal to 𝐿. To see this more clearly, attempt this integral with the substitution 𝑢 is equal to 𝐿 plus 𝑎 minus 𝑥. Give it a try.

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

This leaves us with 𝐿 plus 𝑎 in the numerator of the first fraction and 𝑎 in the numerator of the second. And happily, we’ve got 𝑎 multiplied by 𝐿 plus 𝑎 in the denominator of both fractions. This means we’ve got a common denominator. And so we can add the numerators of the two fractions. Therefore, what we get left with is 𝐿 plus 𝑎 minus 𝑎 all divided by 𝑎 multiplied by 𝐿 plus 𝑎. In other words, we’re left with 𝐿 divided by 𝑎 multiplied by 𝐿 plus 𝑎. So now, this is our expression for the force exerted by the rod on the point particle with charge lowercase 𝑞. And at this point, we know all of the quantities. We know 𝑞, 𝜆, 𝐿, and 𝑎. And we can look up the value of 𝜀 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, it’s 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 “micro” means 10 to the power of negative six. So we’re 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 𝑎 and 𝐿, 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 𝐹 as being 1424 newtons. However, it’s 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, we’ve reached the end of the question.