Answer
See the detailed answer below.
Work Step by Step
We know that the electric field for a positive charge is given by
$$\vec E=\dfrac{kq}{r^2} \;\hat r$$
where $\hat r$ is the unit vector in the direction of the electric field.
So to find the electric field at the three given points, we need to sketch them as seen below.
Hence,
$$\vec E_A=
\dfrac{kq}{r^2}\;\cos\theta_A \;\hat i+\dfrac{kq}{r^2}\;\sin\theta_A\;\hat j$$
$$\vec E_A=
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{(4\times 10^{-2})^2}\cos0^\circ\;\hat i+\\
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{(4\times 10^{-2})^2}\;\sin0^\circ\;\hat j$$
$$\vec E_A=(\color{red}{\bf -6.74\times 10^4}\;{\rm N/c})\hat i$$
By the same approach,
$$\vec E_B=
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{(4\times 10^{-2})^2}\cos180^\circ\;\hat i+\\
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{(6\times 10^{-2})^2}\;\sin180^\circ\;\hat j$$
$$\vec E_B=(\color{red}{\bf 3.0\times 10^4}\;{\rm N/c})\hat i$$
And,
$$\vec E_C=
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{r_C^2}\cos\theta\;\hat i+\\
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{r_C^2}\;\sin\theta\;\hat j$$
where $\theta_C=180^\circ-\theta$, so $\sin(180^\circ-\theta)= \sin\theta=\dfrac{y}{r}$, and hence, $\cos(180^\circ-\theta)=-\cos\theta=-\dfrac{x}{r}$.
Recall that $r=\sqrt{x^2+y^2}$
$$\vec E_C=
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{r_C^2}\dfrac{-x}{r_C}\;\hat i+\\
\dfrac{(8.99\times 10^9)(-12\times 10^{-9})}{r_C^2}\dfrac{y}{r_C}\;\hat j$$
$$\vec E_C=
\dfrac{-x(8.99\times 10^9)(-12\times 10^{-9})}{r_C^3} \;\hat i+
\dfrac{y(8.99\times 10^9)(-12\times 10^{-9})}{r_C^3} \;\hat j$$
Hence,
$$\vec E_C=
\dfrac{-x(8.99\times 10^9)(-12\times 10^{-9})}{(x^2+y^2)^\frac{3}{2}} \;\hat i+
\dfrac{y(8.99\times 10^9)(-12\times 10^{-9})}{(x^2+y^2)^\frac{3}{2}} \;\hat j$$
Plug the known;
$$\vec E_C=
\dfrac{-(0.01)(8.99\times 10^9)(-12\times 10^{-9})}{(0.01^2+0.05^2)^\frac{3}{2}} \;\hat i+
\dfrac{(0.05)(8.99\times 10^9)(-12\times 10^{-9})}{(0.01^2+0.05^2)^\frac{3}{2}} \;\hat j$$
$$\vec E_C=(\color{red}{\bf 8.14\times 10^3}\;{\rm N/c})\hat i+(\color{red}{\bf -4.07\times 10^4}\;{\rm N/c})\hat j$$
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