Electric Field Zero: Charge Distribution on Y-Axis

[MarkFL]

Here is this week's POTW:

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A charge $-q_1$ is located at the origin of a $y$-axis and a charge $-q_2$ is located at $y=d$. At what point along this axis is the electric field zero?

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[MarkFL]

Congratulations to the following members for their correct submission:

• kaliprasad

My solution is as follows (kaliprasad used the same method):

Spoiler

Coulomb's law tells us that the magnitude of an electric field $E$ resulting from a point charge $q$ at a distance $r$ from the source is given by:

\(\displaystyle E=k_e\frac{|q|}{r^2}\) where $k_e$ is Coulomb's constant.

We require the magnitude of the vector sum of the two fields along the $y$-axis to be zero at $d$. For like charges, this will only occur in between the charges ($0<y<d$), where the directions of the two fields point in opposite directions. And so, we require:

\(\displaystyle k_e\frac{q_1}{y^2}=k_e\frac{q_2}{(d-y)^2}\)

This results in the following quadratic in $y$:

\(\displaystyle \left(q_2-q_1\right)y^2+2dq_1y-d^2q_1=0\)

When $q_1=q_2$ we get the linear equation:

\(\displaystyle 2dq_1y-d^2q_1=0\implies y=\frac{d}{2}\)

Otherwise, using the quadratic formula and discarding the negative root, we obtain:

\(\displaystyle y=\frac{-2dq_1+\sqrt{4d^2q_1^2+4\left(q_2-q_1\right)d^2q_1}}{2\left(q_2-q_1\right)}=\frac{-dq_1+d\sqrt{q_1q_2}}{q_2-q_1}=\frac{d\sqrt{q_1}\left(\sqrt{q_2}-\sqrt{q_1}\right)}{\left(\sqrt{q_2}+\sqrt{q_1}\right)\left(\sqrt{q_2}-\sqrt{q_1}\right)}=\frac{d}{1+\sqrt{\dfrac{q_2}{q_1}}}\)