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Consider two point charges $+Q$ and $-Q$, which are fixed a distance $d$ apart. Can you find a location where a third positive charge $Q$ could be placed so that the net electric force on this third charge is zero?

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closed as off-topic by JMac, garyp, stafusa, Kyle Kanos, John Rennie Dec 13 '18 at 6:29

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    $\begingroup$ Can YOU find such a location? What analysis have you performed $\endgroup$ – Bob D Dec 12 '18 at 21:15
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No there is no such point present in or outside the plain but you can take a point at infinity where field is tending to 0

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So let me start out by saying that I am interpreting this question as saying a "fixed distance" means that +q and -q are fixed (anchored in place) and the third Q (positive Q) is free to move. And also to clarify it is asking where the net electric force acting on Q is = 0, not electric field.

For physics, drawing a picture/diagram is always a must. So I would start by drawing a few simple circles to represent the charges.

There is an attractive force trying to pull +q and -q together. If you put a +Q in between it will be repelled away by the +q and pulled into the -q. Aka there is a force on it, it's going to move. There is no way to put it in between these anchored charges and have the net force = 0 (it doesn't move) on your added Q.

Let's assume that +q and -q are in line with each other (say they are both centered on the x-axis). Let's say that +q is to the right of -q. You need to define your coordinate axis, so let's make +q = (x=0). So forces pointing to the right are positive and forces to the left are negative. And any "x" to the left of +q is negative.

Because +q and -q are fixed, the positive charge Q can be placed on the x-axis to the right of +q (so +q is between Q and -q).

+q will want to repel Q (pushing right) and -q will want to attract Q (pulling left). There is a distance, no matter the magnitude of Q, that it can be placed to the right of +q where the opposing forces on Q cancel each other out.

From there you would want to use Coulomb's equation: $$ F = \left(\frac{1}{4\pi *\epsilon_0}\right)\left(\frac{q_1*q_2}{(r)^2}\right)$$

We defined forces pointing right to be + and pointing left negative. We want the net force (sum of forces) to = 0:

$$ \left(\frac{1}{4\pi*\epsilon_0}\right)\left(\frac{q^+*Q}{(r_{(q^+,Q)})^2}\right) - \left(\frac{1}{4\pi*\epsilon_0}\right)\left(\frac{q^-*Q}{(r_{(q^-,Q)})^2}\right) = 0 $$

Canceling/dividing out like terms (as you can see the magnitude of Q does not matter) we get: $$ \left(\frac{q^+}{(r_{(q^+,Q)})^2}\right) - \left(\frac{q^-}{(r_{(q^-,Q)})^2}\right) =0$$

We know from +q to -q is a distance of d, so from +q to Q it will be (d + [the distance between +q and Q]):

$$ \left(\frac{q^+}{(r_{(q^+,Q)})^2}\right) - \left(\frac{q^-}{(d+r_{(q^+,Q)})^2}\right) =0$$

From there it's just algebra to solve for any variable you need, depending on what the question/teacher is asking for exactly.

Depending on which magnitude of charge is stronger will dictate how far Q is away from +q (how far it is being pushed away form +q). The charges don't have to be in line with each other, but remember these forces are vectors (they have direction), so when you start introducing angles, things get more complicated and you'll learn about that as you go.

Also note that in determining the fact that the forces are attractive and repulsive and pointing in opposite directions we have accounted for the negative charge. So if you're given some values for, like a 2nd part of the question, put the absolute value of that charge in for -q. In fact many teachers recommend using absolute values for charges and then determining if there is an attractive or repulsive charge in the end by thinking through it.

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