Video: Using the Neutral Point of an Electric Field to Deduce Missing Data about the Charge Configuration

Two fixed point charges each have magnitudes of 4.0 Γ— 10⁻⁢ C. The charges are located on the π‘₯-axis at the positions π‘₯ = 3.0 m and π‘₯ = βˆ’3.0 m. A charge 𝑄 is placed at the origin and the resultant electric field of the three charges is zero at a point on the 𝑦-axis where 𝑦 = 3.0 m. Find the magnitude of 𝑄.

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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 𝑄 is placed at the origin and the resultant electric field of the three charges is zero at a point on the 𝑦-axis where 𝑦 equals 3.0 meters. Find the magnitude of 𝑄.

As we get started, let’s 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 𝑦-directions in units of metres. We’re 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, there’s a charge at the origin and it’s labelled capital 𝑄. And it’s the magnitude of that charge we want to solve for.

The clue that helps us to solve for the magnitude of 𝑄 is that the electric field on the 𝑦-axis at a location 3.0 metres of that axis we’re told is equal to zero. And that’s the combined electric field from these three point charges. To begin solving for the magnitude of 𝑄, let’s start by recording the magnitude of the two fixed point charges. We’ll call that 𝑄 sub 𝑓. And then having recorded that, let’s clear some space on screen for our calculations.

One aspect of this problem scenario that may stand out is that we’re only told the magnitude of the two fixed point charges that are on the π‘₯-axis. We don’t know their sign. And actually, this may have an impact on the electric field at the location of interest, 3.0 metres up the 𝑦-axis. To see why, let’s 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, we’ll change the direction of the electric field that they create at our point of interest.

To see how this works, let’s create a small table on the left side of our screen. In this table, we’ll have a charge. We’ll call it 𝑄 sub 𝑓 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€” let’s call it 𝑄 sub 𝑓 two. If we were to write down all the possible sign combinations of these two charges, there will be four of them.

First, it’s possible that both these charges are positive. Then, it’s possible that they’re both negative. Then, it’s possible that 𝑄 sub 𝑓 one is positive and 𝑄 sub 𝑓 two is negative. And then, the last possibility is the opposite of that. And at this point, let’s extend our table and we’ll add a third column. This third column will be the electric field created at our point of interest 3.0 metres up the 𝑦-axis from just these two charges, 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 two. In other words, for now, we’re neglecting the presence of the charge 𝑄 at the origin.

Let’s now walk through this table row by row. And as we do, we’ll fill in the third column. That is, we’ll draw out the direction of the electric field that would form from 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 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 𝑄 sub 𝑓 one would create an electric field at our point of interest, which would go in this direction and 𝑄 sub 𝑓 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 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 were positive, our electric field would point upward.

Continuing on, let’s 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 𝑄 sub 𝑓 one will create a field at our point of interest in this direction and 𝑄 sub 𝑓 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 𝑄 sub 𝑓 one, the charge on the left, is positive, while 𝑄 sub 𝑓 two is negative. In this instance, the electric field created by 𝑄 sub 𝑓 one would point in this direction and that created by 𝑄 sub 𝑓 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. 𝑄 sub 𝑓 one has a minus sign and 𝑄 sub 𝑓 two has a plus sign. Now, the electric field at our point of interest from 𝑄 sub 𝑓 one would point in this direction and that from 𝑄 sub 𝑓 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 𝑄 at the origin, it couldn’t possibly make the electric field be zero at the point of interest. That’s because the charge 𝑄 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 𝑄 at that point is only vertical, that means it couldn’t 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 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 two cannot have opposite signs to them. Even though we’re 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 𝑄, the sum of those three fields at our point of interest is zero.

All that being said, let’s 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 𝑄, when we solve for the electric field a distance π‘Ÿ away from the charge, that’s equal to Coulomb’s constant 𝐾 times 𝑄 all divided by π‘Ÿ that distance squared.

In our scenario, we know that the electric field is zero at 3.0 metres up the 𝑦-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, let’s talk about the signs: plus between the first two terms and minus between the last two. We add the electric fields created by 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 two because they combine in the same direction.

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

Now, if we look at the line of action between 𝑄 sub 𝑓 one and our point of interest and 𝑄 sub 𝑓 two and that same point, we see that they make a 45-degree angle with that point. In other words, if we’re to draw in an angle and call it πœƒ between the line of action and the horizontal π‘₯-axis, πœƒ would be 45 degrees. This means we can rewrite our overall equation as zero is equal to the electric field from 𝑄 sub 𝑓 one times the sin of 45 degrees plus the electric field from 𝑄 sub 𝑓 two times that same trig angle minus 𝐸 sub 𝑄, 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, let’s start filling in for our electric field in terms of the equation for that field we’ve 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, we’ve called π‘Ÿ sub 𝑄 𝑓 one, π‘Ÿ sub 𝑄 𝑓 two, and π‘Ÿ sub 𝑄, respectively. Our goal is to work with this expression to solve for capital 𝑄, the magnitude of that charge is the answer to our question.

To start in that direction, we can notice that Coulomb’s constant 𝐾 appears in each term on the right side of our expression. This means we can divide both sides of our equation by 𝐾 and cancel it out. Next, let’s notice something about 𝑄 sub 𝑓 one and 𝑄 sub 𝑓 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‘Ÿ sub 𝑄 𝑓 one is equal to π‘Ÿ sub 𝑄 𝑓 two. And not only that, but the charges themselves 𝑄 sub 𝑓 one 𝑄 sub 𝑓 two are also equal and we know their magnitude.

The symbolic name we’ve given to these fixed charges is 𝑄 sub 𝑓. And let’s call the distance that these charges exist from our point of interest π‘Ÿ sub 𝑓. 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 𝑄 sub 𝑓 over π‘Ÿ sub 𝑓 squared times the square root of two minus 𝑄 over π‘Ÿ sub 𝑄 squared.

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

Looking again at our sketch, we can label the dashed lines π‘Ÿ sub 𝑓 since it’s that distance that represents the distance between our fixed charges and the place where the electric field is zero. Note that π‘Ÿ sub 𝑓 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‘Ÿ sub 𝑓 are 3.0 metres long and 3.0 metres long. So by the Pythagorean theorem, π‘Ÿ sub 𝑓 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‘Ÿ sub 𝑓 is equal to the square root of 18 metres. So that takes care of this terminal equation.

Now, what about π‘Ÿ sub 𝑄? Remember that π‘Ÿ sub 𝑄 is the distance from our point charge 𝑄 which is at the origin to the place where the electric field is zero. We’re told that this is 3.0 meters up the 𝑦-axis. Therefore, π‘Ÿ sub 𝑄 is 3.0 meters. And we’re now ready to plug in for 𝑄 sub 𝑓, π‘Ÿ sub 𝑓, and π‘Ÿ sub 𝑄. We see that 𝑄 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 𝑄 sub 𝑓. This is equal to 2.8 times 10 to the negative six coulombs. That’s 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.

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