What is the electric field vector at the midpoint 𝑀 given that 𝑞 equals 6.0 microcoulombs and 𝑎 equals 20.0 centimeters as shown?
Looking at this figure, we see the midpoint 𝑀, midpoint of the line on which it lies. At either end of this line are the charges plus 𝑞. And these charges are separated by a distance 𝑎 from a third charge of two 𝑞. We want to solve for the electric field vector. That is, the electric field magnitude as well as direction at point 𝑀 due to the influence of these three charges. All three charges are positive. And we can recall that, for a positive charge, the electric field that it creates is radially outward from that charge. This tells us something about the electric field directions from each of the three charges at the midpoint 𝑀.
To see what these directions are on our figure, let’s create a color code for these three charges. We’ll say that our top charge of 𝑞 is color-coded green. Our bottom charge of 𝑞 is color-coded gold. And that the charge of two 𝑞 is color-coded magenta. Knowing that each of these charges will create a radially outward-pointing electric field, we can say that the field from our top charge 𝑞 at point 𝑀 will point in this direction.
Then, if we consider the field created by our bottom charge 𝑞, that will actually point in exactly the opposite direction. More than that, because these two charges have the same magnitude and are separated from the midpoint 𝑀 by the same exact distance, that means that the magnitude of these electric fields at point 𝑀 are the same. In other words, they perfectly cancel one another out. As far as the electric field at that specific location goes, it’s as though these two charges don’t even exist. Which means that it’s only the electric field of our charge two 𝑞 which is experienced at the midpoint 𝑀. That field points roughly in this direction.
We now know generally what the electric field looks like at the midpoint 𝑀. But we want to solve for it numerically, its magnitude as well as its direction. To solve for these values, let’s begin by recalling the equation for the electric field created by a point charge. Given a point charge 𝑞, if we multiply that charge by Coulomb’s constant 𝑘 and then divide it by the distance from the charge to the point where we’re calculating the field squared, then that’s the magnitude of the electric field created by 𝑞 at distance 𝑟 away. All this means that to calculate the electric field magnitude at the midpoint 𝑀, we’ll need to know the distance between our two 𝑞 charge and 𝑀. We can call that distance 𝑑. And we’ll start to use some of the geometry of this triangle on our figure to solve for it.
If we knew the distance from 𝑀 to one of the two charges 𝑞, then we could use the fact that this line segment is at right angles to the dashed line representing our distance 𝑑. This information along with the fact that we know the length of 𝑎, it’s 20.0 centimeters, will let us solve for 𝑑. But what is that distance from 𝑀 to one of the charges 𝑞? To solve for it, let’s start by considering a larger distance. We’ll solve for the whole length of the line that 𝑀 is midpoint of.
We can see that this line forms the hypotenuse of a right triangle where the lengths of the other two legs of the triangle are 𝑎. Since this is a hypotenuse, let’s call this distance ℎ. And we could use the Pythagorean theorem to solve for it. We can write that the hypotenuse ℎ squared is equal to the sum of each of the side lengths squared, 𝑎 squared plus 𝑎 squared in this case. A squared plus 𝑎 squared is two 𝑎 squared. And if we then take the square root of both sides of our equation, we see that ℎ is equal to the square root of two times 𝑎. So then, we’ve solved for ℎ.
But remember, it’s not this specific value we want to know, but rather half of this distance. That distance, after all, forms the leg of our other right triangle that we want to use to solve for 𝑑. Since the distance from the midpoint 𝑀 to one of the two charges 𝑞 is equal to ℎ over two, we’ll divide both sides of this equation to solve for that length. ℎ over two, then, is the square root of two over two times 𝑎. And notice that this distance we’ve solved for is equal to the length of the leg of another right triangle. In this right triangle, though, the hypotenuse is 𝑎. And the two other sides are 𝑑 and, as we’ve solved, the square root of two over two times 𝑎. This means, if we apply the Pythagorean theorem to this smaller right triangle, we would write it as the quantity square root of two over two times 𝑎 squared plus 𝑑 squared, where 𝑑 is the side length we’re interested in, is equal to the hypotenuse 𝑎 squared.
To start solving for 𝑑, let’s square this term, the square root of two over two times 𝑎. That simplifies to 𝑎 squared divided by two. And if we then subtract this term, 𝑎 squared divided by two, from both sides of the equation, we find that 𝑑 squared is equal simply to 𝑎 squared over two. Or that 𝑑 is equal to 𝑎 over the square root of two. Which, by the way, is equivalent to the square root of two divided by two times 𝑎.
Now that we’ve solved for the distance 𝑑, we have a value to plug in for 𝑟 in our equation for electric field. Here’s how we can write that equation as it applies in our case. First, we’ll write that the magnitude of the electric field vector at point 𝑀 is equal to Coulomb’s constant times the charge two times 𝑞, all divided by the distance 𝑑 squared, the square root of two over two times 𝑎 squared. Now, as we saw before, if we square out the square root of two over two times 𝑎, we find the result of 𝑎 squared over two. Or, simplifying this overall expression, four times 𝑘 times 𝑞 over 𝑎 squared. This of course is an expression for the magnitude of the electric field. But, remember, we want to know both its magnitude as well as its direction. And, what is its direction?
Well, if we look back at our sketch, we can see the direction of the arrow we’ve drawn in is at 45 degrees to the horizontal. In other words, if we were to draw the angle in between 𝑑 and our horizontal line, then that angle would be 45 degrees. In order to help clarify this a bit, let’s lay down an 𝑥𝑦-coordinate plane, where the origin is at our charge two 𝑞. Drawing the axes this way, our charge two 𝑞 is at the origin of this coordinate system. And we see that motion to the right is motion in the positive 𝑥-direction. While motion vertically upward is in the positive 𝑦.
Looking at the electric field vector formed at point 𝑀, we see that it has both an 𝑥- and a 𝑦-component to it. In fact, this vector is evenly split between these two directions, because it’s at a 45-degree angle from zero. This means if we were to draw out the 𝑥-component of our electric field vector, it will look like this. And the 𝑦-component would look like this. And again, as we noted, this angle between that vector and the horizontal is 45 degrees.
Looking at this right triangle, we can see then that the vertical section of this displacement is given by the sin of 45 degrees. While the horizontal displacement is given by the cosine of that angle. All this means that if we go back over to our expression for electric field magnitude, we can change this into an expression for the electric field vector. And we can do it by including this directional information, cos of 45 degrees times the magnitude of our electric field vector in the 𝑖 hat, that is 𝑥-direction, and the sin of 45 degrees times the magnitude of our vector in the 𝑗 hat, or 𝑦-direction. With this direction information included, we’ve now dropped our absolute value bars. And we truly have an electric field vector expression.
The last step for us then is to plug in for the different values 𝑘, 𝑞, and 𝑎 and simplify as far as we can. Speaking of simplification, the cos of 45 degrees and the sin of 45 degrees are equal to exact values. They are both equal to the square root of two over two. And notice that in both these terms, we have a factor of one half, which can cancel with one of the twos in this four. So our expression reduces to two times 𝑘 times 𝑞 over 𝑎 squared multiplied by the quantity square root of two 𝑖 plus square root of two 𝑗.
Speaking of 𝑘 and 𝑞 and 𝑎, what are those values? Well, for 𝑞 and for 𝑎, we’re given the values in the problem statement. 𝑞 is 6.0 microcoulombs. And 𝑎, as we said, is 20.0 centimeters. And 𝑘, which is Coulomb’s constant, is a value we can look up. We can approximate this value as 8.99 times 10 to the ninth newton-meters squared per coulombs squared. Knowing all this, we’re now ready to plug in and solve for our electric field vector.
We plug in for the values of 𝑘 and 𝑞, which we write as 6.0 times 10 to the negative six coulombs, as well as 𝑎, a distance which we convert to units of meters. It’s 0.20 meters. With all these values in place, here’s how we can write out our calculated answer. We can say it’s 3.8𝑖 plus 3.8𝑗, all times 10 to the sixth newtons per coulomb. In other words, the electric field magnitude in the 𝑥-direction is the same as that in the 𝑦-direction, 3.8 times 10 to the sixth newtons per coulomb. This is the electric field vector at the midpoint 𝑀.