Video: Calculating The Electric Field Due to Two Charges

Point charges 𝑄₁ = 33.2 πœ‡C and 𝑄₂ = 56.0 πœ‡C are placed 0.85 m apart along a line 𝐿. How far from 𝑄₁ along 𝐿 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 𝐿?

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Video Transcript

Point charges 𝑄 one equals 33.2 microcoulombs and 𝑄 two equals 56.0 microcoulombs are placed 0.85 meters apart along a line 𝐿. How far from 𝑄 one along 𝐿 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 𝐿?

Let’s start out by drawing out the line 𝐿 and the two point charges 𝑄 one and 𝑄 two on it. So, here is our line 𝐿. And it extends, as far as we know, infinitely far in each direction. Along this line, there is charge 𝑄 one and then a distance of 0.85 meters down the line, the charge 𝑄 two. The first of our two questions says, how far from 𝑄 one, the first charge, along 𝐿 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 𝑄 one or to the left. But, here, the charge values 𝑄 one and 𝑄 two given to us in the problem statement help us out. Notice that both of these charges are positive. Or at least we assume they’re positive because we’re not told they’re 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 𝑄 two and 𝑄 one create electric field lines that point away from themselves since they’re both positive.

Getting back to our question of whether the electric field is zero to the left or to the right of 𝑄 one. The fact that both 𝑄 two and 𝑄 one created field lines that point away from themselves means that, to the left of 𝑄 one, 𝑄 one’s electric field lines will point in this direction. And 𝑄 two’s electric field lines will point in the same direction. Since they’re in the same direction, they can’t cancel one another out. But, on the other hand, on the right side of 𝑄 one, 𝑄 one’s electric field lines will point this way. And 𝑄 two’s electric field lines will point opposite that.

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

So here’s what we can write. We can say that 𝐸 sub 𝑄 one, the electric field created by 𝑄 one, minus 𝐸 sub 𝑄 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‘Ÿ one and π‘Ÿ two, respectively. Then, using our equation for electric field, we can write that π‘˜π‘„ one over π‘Ÿ one squared minus π‘˜π‘„ two over π‘Ÿ 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, let’s add this term, 𝑄 two over π‘Ÿ sub two squared, to both sides. When we do, we get this interesting equation. 𝑄 one over π‘Ÿ one squared is equal to 𝑄 two over π‘Ÿ two squared. That’s what the net electric field being zero means. Considering again this first question, we want to answer how far from 𝑄 one along our line 𝐿 does this field zero out. In other words, at some location in between these two charges, let’s just pick that spot for illustration, the electric field is zero. And we want to find the distance between that point and 𝑄 one. Notice though that, in our equation, that distance is represented by π‘Ÿ one. And, similarly, the distance from the point where the electric field is zero over to 𝑄 two is equal to π‘Ÿ two.

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

What we’re going to do next is take this expression for π‘Ÿ two and substitute it in for π‘Ÿ two in our equation which says that π‘Ÿ one plus π‘Ÿ 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‘Ÿ one, which is what we want to solve for. To rearrange this equation and solve for π‘Ÿ one, let’s start by factoring it out of both terms on the left side of the equation. And then, finally, we’ll divide both sides of the equation by the expression in parentheses, one plus the square root of 𝑄 two over 𝑄 one.

Now to solve for π‘Ÿ one, it’s just a matter of plugging in for 𝑄 two and 𝑄 one. And recall that we’re told 𝑄 two and 𝑄 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‘Ÿ one, we find that, to two significant figures, it’s 0.37 meters. That’s the distance from charge 𝑄 one along line 𝐿 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 𝑄 one and 𝑄 two. We know this field magnitude won’t be zero. And to solve for its numerical value, let’s refer again to our equation for the electric field created by a point charge. This field at the midpoint, which we can call 𝐸 sub π‘š, will be equal to the electric field caused by charge 𝑄 two minus the field caused by charge 𝑄 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 𝑄 two is greater than 𝑄 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 Coulomb’s constant times 𝑄 two divided by π‘Ÿ sub π‘š squared, where π‘Ÿ sub π‘š is half the distance between 𝑄 one and 𝑄 two, minus π‘˜ times 𝑄 one over π‘Ÿ sub π‘š squared.

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

And, finally, 𝑄 two and 𝑄 one, the charge values, are given to us. Let’s write an expression for 𝐸 sub π‘š 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. That’s the electric field magnitude midway between the two charges 𝑄 one and 𝑄 two.

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