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Really simple problem here but I am having trouble understanding why I should get one particular result. So there are 2 charges separated by a distance $l$ which are respectively defined as $q_1 = 8q$ and $q_2 = -2q$. The problem is in 1D, and the positive charge is on the left of the negative charge. The problem statement asks to calculate the points where the electric field and potential are 0. Now, for the first question my math gives:

a) $E_1 (x) + E_2(x) = 0 \implies \frac{8q}{(l+x)^2} = \frac{2q}{x^2} \implies \sqrt{\frac{8q}{2q}} = \frac{(l + x)}{x} \implies \pm2x = x + l \implies x_1 = l \: \vee \: x_2 = -l $

Solution $x_2$ makes no physical sense, so it is discarded. This reasoning turns out to give the correct result, so I assume it is correct. However, notice that I could only take the square root because the charges had opposite signs. My first question is, if I were to set up this math for the exact same problem, except the charges were equal, how could I solve it? The square root wouldn't be an option in that case.

On to the second question:

b) $ V_1(x) + V_2(x) = 0 \implies \frac{q_1}{q_2} = \frac{(l + x_1)}{x_1} \implies 4x_1 = l + x_1 \implies x_1 = \frac{l}{3} $

This answer turns out to be right too, but incomplete: in fact, my textbook suggests I should also apply this reasoning:

$ \frac{8q}{4 \pi \epsilon_0 x_2} - \frac{2q}{4 \pi \epsilon_0 (l-x_2)} = 0 \implies x_2 = \frac{4}{5} l $

Which basically means changing frame of reference and searching for zero potential points in between the two charges. I really don't understand how is that logic right. Since that point is in between two opposite sign charges, as soon as I put a test charge there, that charge will move towards the negative sign charge and away from the positive sign charge, which means that there is electric potential energy there, therefore electric potential. So I don't see how potential in that point could ever be 0. What am I missing?

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    $\begingroup$ From a) is it seems the origin of coordinates taken is at charge 2 (charge 1 is then at $x=-l$ coordinate and distance from charge 1 is $\vert x-(-l) \vert = \vert x+l \vert$. Check your math, the solution to a) doesn't appear to be right. If the first charge is the origin of the coordinates, second charge is at $x=l$, distance from charge 2 is then $\vert x-l \vert$ (and don't omit absolute values!). It doesnt matter how you choose the origin of coordinates, it should yield the same physical position, but you need to be clear about where the origin is. $\endgroup$ Oct 1 at 16:49
  • $\begingroup$ The origin of coordinates in point a) is taken at charge 2, in fact the distance at the denominator does show $(l + x)^2$, in which I have omitted the absolute value because it's a square. It seems to me our calculations yield the same result $\endgroup$
    – peppece
    Oct 1 at 17:29
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    $\begingroup$ In a) $-2x= l+x$ yields $x=-l/3$ (but this doesn't change the correct solution $x=l$). In fact you should clearly distinguish the different domains $-L<x<0$ and $x>0$ on which you search for a solution since the equations to solve are not necessarily the same on these 2 domains . My comment on the absolute values was more for the question related to potentials. Also the second equation with potentials seems to be with origin taken at the charge 1 which may add some confusion. $\endgroup$ Oct 1 at 21:08
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Potentials are always defined up to a constant, so solving for positions with 0 potential doesn't bear much physical sense. The physical meaning of the potential is its local variation (gradient) giving the electrical field.

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Okay, to solve your first problem: It is important to remember that E (electric field strength) is a vector. In your example, you used the correct formula:

$E_1 (x) + E_2(x) = 0$

When you then say:

$\frac{8q}{(l+x)^2} = \frac{2q}{x^2} $

You are, in effect, acknowledging that the field caused by the 2q charge is ‘negative’ at the point we are talking about (in accordance with the sign convention, which I’m assuming is: right = positive).

That is not always the case. If we look at the space between the two charges, the fields are in the same direction i.e. both positive.

In light of that, when you have two like charges, the point at which the electric field will be 0, is in between the charges, as it is in this region where the fields are in opposite directions. If you take a positive charge as an example (outwardly radiating field), the two positive charges will be outwardly radiating their field in eachother’s directions (when you are looking at the line joining them). In that case, the one field will be given a negative sign (to show its direction) and so your working will still be valid.

For your second question, your textbook is correct. Despite the fact that your logic is correct, the point of 0 potential doesn’t mean that a positive test charge won’t move when placed at that point (that point would be the one you calculated in the previous question). The reason the test charge can move at 0 potential, is because it can move to a lower (-ve) potential.

I hope that helps :)

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  • $\begingroup$ I did assume field is positive to the right; the reason I acknowledged the field by the 2q charge to be negative is that at point a) I have taken x = 0 at the negative charge, so with x > 0 the fields have opposite directions (both charges are to the left of this position, so the negative one "pulls" and the positive one "pushes" the test particle). $\endgroup$
    – peppece
    Oct 3 at 10:12

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