Unphysical Division By Zero for Zero Field Location

1. Sep 7, 2011

DocZaius

Unphysical Division By Zero for Trivial Case of Zero Field Location

1. The problem statement, all variables and given/known data

Q1 and Q2 are two positive charges a distance s apart. Find the distance x from Q1 where the field is zero.

2. Relevant equations

$E=\frac{kq}{r^2}$

3. The attempt at a solution

Let's put Q1 at the origin and Q2 to the right of it on the x axis.

$E_{1}$ is the contribution of $Q_{1}$ to the electric field
$E_{2}$ is the contribution of $Q_{2}$ to the electric field

$E_{1}=\frac{kQ_{1}}{x^{2}}$,$E_{2}=-\frac{kQ_{2}}{(s-x)^{2}}$

($E_{2}$ will have a negative sign in front of it since a positive charge's field will be negative on the x axis.)

Let us find at what x will $E_{total}=E_{1}+E_{2}=0$

$E_{1}+E_{2}=0$

$\frac{kQ_{1}}{x^{2}}=\frac{kQ_{2}}{(s-x)^{2}}$

$\frac{Q_{1}}{x^{2}}=\frac{Q_{2}}{x^{2}-2sx+s^{2}}$

$Q_{1}(x^{2}-2sx+s^{2})=Q_{2}x^{2}$

$(Q_{1}-Q_{2})x^{2}+(-2Q_{1}s)x+(Q_{1}s^{2})=0$

$x=\frac{2Q_{1}s\pm\sqrt{4Q_{1}^{2}s^{2}-4(Q_{1}-Q_{2})(Q_{1}s^{2})}}{2(Q_{1}-Q_{2})}$

This is the part I don't understand. According to this, if the charges are exactly equal, you divide by zero. However, if the charges are equal, the answer is obviously $x=\frac{s}{2}$.

Think of the most trivial case where Q1=Q2=s=1. x should obviously be 1/2. But with my answer, I must divide by zero. This is not an issue of a singularity where I would be trying to find a field on the location of the actual point-charge.

Can someone explain why equal charges lead to a division by zero?

Also please note that the form of the equation before the quadratic formula will yield the correct answer. However, I don't understand why solving for x leads to the division by zero problem.

Thank you.

Last edited: Sep 7, 2011
2. Sep 7, 2011

Staff: Mentor

When Q1 = Q2 the equation is no longer quadratic in x, and the Q's disappear entirely by mutual cancellation.
$$\frac{Q_1}{x^2} = \frac{Q_2}{(s - x)^2}$$
turns into
$$\frac{Q_1}{x^2} = \frac{Q_1}{(s - x)^2}$$
$$\frac{1}{x^2} = \frac{1}{(s - x)^2}$$
$$x^2 = s^2 - 2 s x + x^2$$
$$s^2 - 2 s x = 0$$
$$s - 2 x = 0$$
$$x = s/2$$

3. Sep 7, 2011

Pi-Bond

Also you need not necessarily resort to the Quadratic formula; your equation can also be written as

$\large \frac{s-x}{x} = \sqrt{\frac{Q_2}{Q_1}}$

Giving

$x=\large \frac{s}{\sqrt{\frac{Q_2}{Q_1}} + 1}$

4. Sep 7, 2011

DocZaius

Thus it seems like there is not just a general equation for what x is. One needs both the quadratic form, and the special case for equal charges. My next question is: What is it about this problem (other than the obvious way in which the algebra works out) that makes it so that we need a special case formula in addition to the general formula?

For example, there is a reason why finding out the field at the location of a point-charge leads to a division by zero. The math's red flag is associated with the physics' red flag in regards to that problem (physics does not have a qualitative way of explaining the field strength at a point charge, and the math backs that up). I don't see the analogous physics problem of the mid-point zero field for equal charges. The math's red flag here does not have an associated physics red flag. What is it about this problem in qualitative physics terms (again, in addition to the obvious way the algebra works out) that would account for this division by zero error?

Thanks.

Edit: Pi-Bond - thanks. Interesting. I wonder why the quadratic formula approach, which would seem to be valid, leads to a division by zero problem. Shouldn't all roads lead to the same answer as long as the methods are correct and valid?

Last edited: Sep 7, 2011
5. Sep 7, 2011

Delphi51

Fascinating! The quadratic formula is invalid when a=0. I'm tempted to do the quadratic formula derivation (using the completing the square method from grade 11) to see just where that happens.

6. Sep 7, 2011

SammyS

Staff Emeritus
Simplify that expression for the solution to x, resulting from using the quadratic formula.

$\displaystyle x=\frac{2Q_{1}s\pm\sqrt{4Q_{1}^{2}s^{2}-4(Q_{1}-Q_{2})(Q_{1}s^{2})}}{2(Q_{1}-Q_{2})}$
$\displaystyle =\frac{2Q_{1}s\pm2s\sqrt{Q_{1}^{2}-(Q_{1}-Q_{2})(Q_{1})}}{2(Q_{1}-Q_{2})}$

$\displaystyle =\frac{Q_{1}\pm\sqrt{Q_{1}^{2}-Q_{1}^2+Q_{2}Q_{1}}}{Q_{1}-Q_{2}}s$

$\displaystyle =\frac{Q_{1}\pm\sqrt{Q_{2}Q_{1}}}{Q_{1}-Q_{2}}s$​

Notice that for Q1 ≠ Q2, there are two solutions.

The one that makes for Q1 & Q2 having the same sign is the solution with the minus sign.

Therefore, we have: $\displaystyle x=\frac{Q_{1}-\sqrt{Q_{2}Q_{1}}}{Q_{1}-Q_{2}}s\ .$

Notice that for Q1 = Q2, this solution is of the form 0/0, which is indeterminate, rather than undefined.

Look at limQ_1 → Q_2 x .

7. Sep 8, 2011

Hurkyl

Staff Emeritus
Because the quadratic formula makes an assumption -- that the equation is quadratic -- and your problem does not satisfy the assumption. Your problem only satisfies that assumption when restricted to the case of unequal charges.

There are two other interesting points:
• The limit as the difference of charges goes to zero does converge to $x=1/2$
• In the projective real numbers, the corresponding equation does have a second solution when the charges are equal: $x = \infty$

Actually, your quadratic equation is already homogeneous if you treat s as another variable meant to be paired with x -- and the solutions are

$$(x : s) = (Q_1 \pm \sqrt{Q_1 Q_2} : Q_1 - Q_2)$$

The notation $(a:b)$ is commonly used for projective numbers. If $b\neq 0$, the notation corresponds to the ordinary number $a/b$. The remaining case is $(a:0) = \infty$. (a and b aren't allowed to both be zero)

(technicality -- the above solution assumes that at least one of $Q_1$ and $Q_2$ are nonzero)

Last edited: Sep 8, 2011