1
$\begingroup$

Problem 1.2 of the E&M text of Purcell & Morin reads:

"Two positive ions and one negative ion are fixed at the vertices of an equilateral triangle. Where can a fourth ion be placed, along the symmetry axis of the setup, so that the force on it will be zero? Is there more than one such place? You will need to solve something numerically."

The solution then says: "Let the sides of the triangle be 2 units long. Consider a point $P$ that lies a distance $y$ (so $y$ is defined to be a positive number) beyond the side containing the two positive ions. $P$ is a distance $y + \sqrt{3}$ from the negative ion and $\sqrt{1+y^2}$ from each of the positive ions. If the electric field equals zero at $P$, then the upward field due to the negative ion must cancel the downward field due to the two positive ions. This gives (ignoring the factor of ${e}/{4 \pi \epsilon_0}$)

$$\frac{1}{(y+\sqrt{3})^2} = 2 \cdot \frac{1}{1+y^2} \left( \frac{y}{\sqrt{1+y^2}} \right)$$

where the $y/ \sqrt{1 + y^2}$ factor in the first equality arises from taking the vertical component of the titled field lines due to the positive ions." This equation has to be solved numerically and there is only one solution, which turns out to be $y\approx 0.1463$.

To find the other point (beyond the negative ion) where the force vanishes, all we have to do is to exchange the $\sqrt{3}$ by $-\sqrt{3}$ in the above equation and it is claimed that the solution now is $y_1\approx 6.2045$ but it seems to me there are two other solutions to this equation, namely $y_2 = \sqrt{3}$ and $y_3 \approx 0.2486$.

I can accept that there are multiple solutions to the equation but how can you figure out that $y_1$ is the one that makes sense from the point of view of physics?

$\endgroup$
4
  • 1
    $\begingroup$ You are wrong for $y_2 = \sqrt{3}$. It's not even a solution of the equation with $\boldsymbol{-}\sqrt{3}$. Note that you have a zero denominator in the lhs of the equation with $\boldsymbol{-}\sqrt{3}$. This solution would coincide with the position of negative ion (impossible)...... $\endgroup$
    – Frobenius
    Aug 16, 2020 at 17:29
  • 1
    $\begingroup$ ....On the other hand you are right for the solution $y_3 \approx 0.2486$ ( $y_3 \approx 0.24858583$). This is a solution of the equation with $\boldsymbol{-}\sqrt{3}$. But you forget the assumption that $y$ refers to a point above the negative ion, so between the solutions of the algebraic equation acceptable are those for which $y>\sqrt{3}$. $\endgroup$
    – Frobenius
    Aug 16, 2020 at 17:30
  • 1
    $\begingroup$ omg I totally forgot about this condition… thank you for clearing things up for me! $\endgroup$
    – freddy90
    Aug 16, 2020 at 17:34
  • $\begingroup$ The solutions to the equation either with $\boldsymbol{+}\sqrt{3}$ or with $\boldsymbol{-}\sqrt{3}$ satisfy the condition the magnitude of the force from the negative ion to be equal to the magnitude of the force from the two positive ions, but not necessarily the condition these two forces to be opposite. So, for $y_3 \approx 0.2486$ these magnitudes are equal but the two forces point to the same direction. $\endgroup$
    – Frobenius
    Aug 16, 2020 at 17:49

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.