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Point charges π one equals 33.2 microcoulombs and π two equals 56.0 microcoulombs are placed 0.85 meters apart along a line πΏ.
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How far from π one along πΏ does the net electric field due to the charges equal zero?
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What is the magnitude of the electric field due to the charges at the midpoint of πΏ?
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Letβs start out by drawing out the line πΏ and the two point charges π one and π two on it.
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So, here is our line πΏ.
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And it extends, as far as we know, infinitely far in each direction.
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Along this line, there is charge π one and then a distance of 0.85 meters down the line, the charge π two.
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The first of our two questions says, how far from π one, the first charge, along πΏ does the net electric field due to the charges equal zero?
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At first, we may not be sure whether this means a distance to the right of π one or to the left.
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But, here, the charge values π one and π two given to us in the problem statement help us out.
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Notice that both of these charges are positive.
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Or at least we assume theyβre positive because weβre not told theyβre negative.
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Given a positive charge, represented here by a plus sign, the electric field created by this charge points radially outward from it.
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This means that both π two and π one create electric field lines that point away from themselves since theyβre both positive.
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Getting back to our question of whether the electric field is zero to the left or to the right of π one.
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The fact that both π two and π one created field lines that point away from themselves means that, to the left of π one, π oneβs electric field lines will point in this direction.
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And π twoβs electric field lines will point in the same direction.
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Since theyβre in the same direction, they canβt cancel one another out.
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But, on the other hand, on the right side of π one, π oneβs electric field lines will point this way.
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And π twoβs electric field lines will point opposite that.
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This tells us that the place where the electric field is zero must be in between π one and π two.
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That is, itβs to the right of π one.
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Knowing that, letβs recall the mathematical equation for the electric field created by a point charge.
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The electric field created by a point charge π, a distance π away from itself, is equal to π times Coulombβs constant π divided by π squared.
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In this first question, we know that the net electric field, that is, the combined field due to both π one and π two, must equal zero.
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So hereβs what we can write.
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We can say that πΈ sub π one, the electric field created by π one, minus πΈ sub π two, the field created by the second charge, is equal to zero.
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And if we call the distances for charge one and charge two π one and π two, respectively.
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Then, using our equation for electric field, we can write that ππ one over π one squared minus ππ two over π two squared is equal to zero.
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Looking at this equation, notice that a factor of π appears in both terms on the left side.
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This means that if we divide both sides of the equation by π, that factor cancels out.
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Next, letβs add this term, π two over π sub two squared, to both sides.
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When we do, we get this interesting equation.
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π one over π one squared is equal to π two over π two squared.
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Thatβs what the net electric field being zero means.
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Considering again this first question, we want to answer how far from π one along our line πΏ does this field zero out.
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In other words, at some location in between these two charges, letβs just pick that spot for illustration, the electric field is zero.
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And we want to find the distance between that point and π one.
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Notice though that, in our equation, that distance is represented by π one.
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And, similarly, the distance from the point where the electric field is zero over to π two is equal to π two.
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Seeing this on our diagram, another equation may stand out to us.
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And that is that if we add π one and π two together thatβs equal to the total distance separating π one and π two: 0.85 meters.
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So itβs π one we want to solve for.
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Thatβs the answer to this first question.
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And to do that, using this equation, weβll begin by solving not for π one but for π two.
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If we cross-multiply so that π two is isolated on one side of this equation, we find that π sub two squared is equal to π sub one squared times π two over π one.
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Or, taking the square root of both sides, π two equals π one times the square root of π two over π one.
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What weβre going to do next is take this expression for π two and substitute it in for π two in our equation which says that π one plus π two is equal to 0.85 meters.
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When we do that, notice that we now have an equation entirely in terms of the charges, which we know, and π one, which is what we want to solve for.
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To rearrange this equation and solve for π one, letβs start by factoring it out of both terms on the left side of the equation.
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And then, finally, weβll divide both sides of the equation by the expression in parentheses, one plus the square root of π two over π one.
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Now to solve for π one, itβs just a matter of plugging in for π two and π one.
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And recall that weβre told π two and π one in our problem statement.
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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.
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And, finally, when we go to calculate π one, we find that, to two significant figures, itβs 0.37 meters.
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Thatβs the distance from charge π one along line πΏ at which the electric field is zero.
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Thanks to the influence of the electric fields from these two charges.
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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 π one and π two.
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We know this field magnitude wonβt be zero.
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And to solve for its numerical value, letβs refer again to our equation for the electric field created by a point charge.
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This field at the midpoint, which we can call πΈ sub π, will be equal to the electric field caused by charge π two minus the field caused by charge π one.
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The reason for this minus sign is simply that we want our overall electric field at the midpoint to be positive.
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And we know that π two is greater than π one and therefore will create a stronger field at that midpoint.
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So then, working off of our electric field relationship, the electric field at the midpoint between our two charges is equal to Coulombβs constant times π two divided by π sub π squared, where π sub π is half the distance between π one and π two, minus π times π one over π sub π squared.
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Looking at the right side of this equation, notice that we have factors of π over π sub π squared in both terms and can therefore factor them out.
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This is what our expression simplifies to.
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And, as a next step, letβs start finding out the numerical values of these various terms on the right-hand side.
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First, starting with Coulombβs constant π, this constant is approximately 8.99 times 10 to the ninth newton-meters squared per coulomb squared.
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Then, π sub π, halfway the distance between π one and π two, must be that total distance, 0.85 meters, divided by two.
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And, finally, π two and π one, the charge values, are given to us.
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Letβs write an expression for πΈ sub π then with all these values plugged in.
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When we do substitute in these numbers, notice what happens to the units in the expression.
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One factor of the unit of charge, coulombs, cancels out.
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And both factors of distance, in meters, cancel out.
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When we enter this expression on our calculator, we find a result of 1.1 times 10 to the sixth newtons per coulomb.
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Thatβs the electric field magnitude midway between the two charges π one and π two.