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So this is a problem I already know how to solve, but I feel like my method is really inefficient and I'm wondering if there's a more intuitive, less tedious way to arrive at this conclusion.

If we have two charges separated by the distance $a$. One charge has charge $+4Q$, the other has charge $-Q$. The right charge $-Q$ is sitting on the origin.

                      (y)
                       ^
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----(4Q)<_____a_____>(-Q)-------------------> (x)
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The problem is to find where the electric field is zero.

My solution:

I know that the electric field from a point charge is $kq/r^2$, so I tried setting them equal to zero:

$$\frac{k4q}{(a+x)^2} - \frac{kq}{x^2} = 0$$

Solving this for $x$ with Solve[4/(a+x)^2 - 1/x^2 == 0, x], I get the solutions

$$x = -\frac{a}{3}$$ $$x = a$$ $$(a\neq0)$$

We can disregard the negative solution since the equation we solved only gives valid solutions from the right of the x axis.

But this means that I need to solve two more equations to be sure that I didn't miss any solutions.

For the space between the two charges (clearly no solution, as they have different charges): $$\frac{k4q}{(a+x)^2} + \frac{kq}{x^2} = 0$$

For the space on the left of the two charges: $$-\frac{k4q}{(a+x)^2} + \frac{kq}{x^2} = 0.$$

The final answer is that the electric field is zero at $$x=a,$$ but it took three equations to get there. Is there a simpler way to do this that I'm missing? Thank you!

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  • $\begingroup$ Why are you neglecting $x=-a/3$ for the first solution? And why are you changing signs of $q$? Shouldn't you change the sign of $x$? $\endgroup$ – Kyle Kanos Sep 25 '13 at 14:36
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Your equation $\frac{k4q}{(a+x)^2} - \frac{kq}{x^2} = 0$ is equivalent to : $\frac{(a+x)^2}{x^2} = 4$, so $(1 + \frac{a}{x})^2 = 4$, so $1 + \frac{a}{x} = \pm2$, so $\frac{a}{x} = -3$ or $\frac{a}{x} = 1$, so $x = -\frac{a}{3}$ or $x=a$

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