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Unphysical Division By Zero for Trivial Case of Zero Field Location
Q1 and Q2 are two positive charges a distance s apart. Find the distance x from Q1 where the field is zero.
[itex]E=\frac{kq}{r^2}[/itex]
Let's put Q1 at the origin and Q2 to the right of it on the x axis.
[itex]E_{1}[/itex] is the contribution of [itex]Q_{1}[/itex] to the electric field
[itex]E_{2}[/itex] is the contribution of [itex]Q_{2}[/itex] to the electric field
[itex]E_{1}=\frac{kQ_{1}}{x^{2}}[/itex],[itex]E_{2}=-\frac{kQ_{2}}{(s-x)^{2}}[/itex]
([itex]E_{2}[/itex] 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 [itex]E_{total}=E_{1}+E_{2}=0[/itex]
[itex]E_{1}+E_{2}=0[/itex]
[itex]\frac{kQ_{1}}{x^{2}}=\frac{kQ_{2}}{(s-x)^{2}}[/itex]
[itex]\frac{Q_{1}}{x^{2}}=\frac{Q_{2}}{x^{2}-2sx+s^{2}}[/itex]
[itex]Q_{1}(x^{2}-2sx+s^{2})=Q_{2}x^{2}[/itex]
[itex](Q_{1}-Q_{2})x^{2}+(-2Q_{1}s)x+(Q_{1}s^{2})=0[/itex]
Now using the quadratic formula:
[itex]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})}[/itex]
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 [itex]x=\frac{s}{2}[/itex].
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.
Homework Statement
Q1 and Q2 are two positive charges a distance s apart. Find the distance x from Q1 where the field is zero.
Homework Equations
[itex]E=\frac{kq}{r^2}[/itex]
The Attempt at a Solution
Let's put Q1 at the origin and Q2 to the right of it on the x axis.
[itex]E_{1}[/itex] is the contribution of [itex]Q_{1}[/itex] to the electric field
[itex]E_{2}[/itex] is the contribution of [itex]Q_{2}[/itex] to the electric field
[itex]E_{1}=\frac{kQ_{1}}{x^{2}}[/itex],[itex]E_{2}=-\frac{kQ_{2}}{(s-x)^{2}}[/itex]
([itex]E_{2}[/itex] 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 [itex]E_{total}=E_{1}+E_{2}=0[/itex]
[itex]E_{1}+E_{2}=0[/itex]
[itex]\frac{kQ_{1}}{x^{2}}=\frac{kQ_{2}}{(s-x)^{2}}[/itex]
[itex]\frac{Q_{1}}{x^{2}}=\frac{Q_{2}}{x^{2}-2sx+s^{2}}[/itex]
[itex]Q_{1}(x^{2}-2sx+s^{2})=Q_{2}x^{2}[/itex]
[itex](Q_{1}-Q_{2})x^{2}+(-2Q_{1}s)x+(Q_{1}s^{2})=0[/itex]
Now using the quadratic formula:
[itex]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})}[/itex]
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 [itex]x=\frac{s}{2}[/itex].
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.
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