Two point charges $Q_1$ and $Q_2$, are a distance a away from each other. Where is electric field zero? [closed]

Question

Problem: Two different point charges $Q_1$ and $Q_2$ are a distance $a$ away from each other. The point charges are both positive. Where is the electric field zero?

So far, I've found the total electric field for a point on the x-axis (assuming $Q_1$ is in the origin, and $Q_2$ is at (a,0)), by using Coulombs law and superposition. This gives me

$E_{tot} = E_1 + E_2 = k\left(\frac{Q_1}{x^2}-\frac{Q_2}{(x-a)^2}\right)\hat{x}$

Then setting $E_{tot} = 0$ and trying to solve for $x$. Now this is where I'm stuck, and even trying to use tools to solve for $x$ I don't arrive at anything similar to the answer which is

On the line between the charges, $\frac{a}{1+\sqrt{\frac{Q_2}{Q_1}}}$ from the charge $Q_1$

Am I using the correct method and just need to keep trying to solve for $x$, or am I missing something?

Answer 1

I will give you some hint in case you are interested in cracking the problem yourself (the way to learn) but leave the detailed answer below in case you really want it.

Note that you should also proof that the only point in space that has vanishing field lies in the line defined by the two charges. You can do that easily by writing down the field in the $y$ component and trying to see what happens with $x$ if $y\neq 0$.

You wrote the equation in the $x$ component correctly, now you should equal that expression you got to 0, since you want the two contributions to cancel each other. From then on is just some algebra.

One extra question for some fun: If you put some charge in the point where there is no field, will it remain there?

Now the problem is that: $$\vec E = \vec E^{(1)} + \vec E^{(2)} = \frac{Q_1}{|\vec r - \vec r_1|^3}\cdot (\vec r - \vec r_1) + \frac{Q_2}{|\vec r - \vec r_2|^3}\cdot (\vec r - \vec r_2)= \vec 0$$ This we can split into components $x$ & $y$. Component $y$.- $$ \frac{Q_1}{(x^2+y^2)^{3/2}}y = -\frac{Q_2}{((x-a)^2+y^2)^{3/2}}y$$ This can be fulfilled easily if $y = 0$ or $y \neq 0$ and $$ \frac{Q_1}{(x^2+y^2)^{3/2}} = -\frac{Q_2}{((x-a)^2+y^2)^{3/2}} $$ Component $x$.- $$ \frac{Q_1}{(x^2+y^2)^{3/2}}x= -\frac{Q_2}{((x-a)^2+y^2)^{3/2}}(x-a)$$ Using the result obtained for the component $y$ if $y \neq 0$ one gets: $$x = x-a$$ which is impossible. Then we have proven $y = 0$. Hence: $$ \frac{Q_1}{(x)^{2}}= sign(a-x)\frac{Q_2}{(x-a)^{2}} $$ So, if charges are of equal signs, $|x|<a$. If they are of opposite signs, $|x|>a$ (otherwise the $sign$ function gives a negative contribution and the two sides have opposite sign, which has no solution). From here one gets easily to: $$x = \frac{a}{1+\sqrt{\frac{Q_2}{Q_1}}}$$ if charges are of equal signs, or $$x = \frac{a}{1-\sqrt{\left|\frac{Q_2}{Q_1}\right|}}$$ which is a complete answer (the one that was given to you is incomplete, since it is only valid for charges of equal signs). Note that in the trivial case where charges are equal the field vanishes just at the middle. The answer to the extra question is "depends". Mathematically, it will, if you place it exactly there. However, if the charges are both negative, the field around the point tend to grow pointing away from the point. If the charges are positive, if you get a bit away from it the forces drive you back to it, so it will stay. In the $y$ direction similar arguments apply. Try to think yourself the case of oppositte signs.