Question

Two particles, each of positive charge $q$, are fixed in place on a $y$ axis, one at $y=d$ and the other at $y=-d$.
Graph $E$ versus $\alpha$ for the range $0<\alpha <4$.

Answer 1

Let $q_1$ denote the charge at $y=d$ and $q_2$ denote the charge at $y=-d$.

The individual magnitudes $|{\overrightarrow{E}}_1|$ and $|{\overrightarrow{E}}_2|$ are figured, where the absolute values signs for

$q$ are unnecessary since these charges are both positive.

The distance from $q_1$ to a point on the $x$ axis is the same as the distance from $q_2$ to a point on the $x$ axis:

$r=\sqrt{x^2+d^2}$.

By symmetry, the $y$ component of the net field along the $x$ axis is zero.

The $x$ component of the net field, evaluated at points on the positive $x$ axis, is
$E_x=2\left(\frac{1}{4\pi {\epsilon }_0}\right)\left(\frac{q}{x^2+d^2}\right)\left(\frac{x}{\sqrt{x^2+d^2}}\right)$
where the last factor is $\cos \theta =x/r$ with $\theta$ being the angle for each individual field as measured from the $x$ axis.
The graph of $E=E_x$ versus $\alpha$ is shown below.

For the purposes of graphing, we set $d=1m$ and $q=5.56\times {10}^{-11}C$.