Question about resolving electrostatic force in 2 directions This may be a very stupid question but in the following scenario where particle A (fixed at (0, 0)) exerts an attractive force on particle B (free to move at (4, 3)), can I resolve the force in two directions x, y and then calculate the resultant force? 
If I resolve it into x, y directions, then the F_y is going to be bigger because of the shorter distance, and F_x is going to be smaller because of the larger distance. Then the vector sum of these two forces won't add up to the direction towards A:

Why is that? Is it not correct to resolve the forces in two directions and apply Coulomb's law independently and vector sum them?
 A: You must first calculate the Coulomb force based on the separation of the charges, then you can resolve it into $F_x, F_y$ components. If you don't use the total distance in the denominator of Coulomb's Law, you don't get the correct force magnitude.
You can do this using vector algebra or trigonometry. Using vector algebra we have $$\vec{F}_{BA}=\frac{kq_Aq_B}{r_{BA}^2}\hat{r}_{BA}=F_{BA}\hat{r}_{BA}$$
Now $r_{BA}^2=(x_B-x_A)^2+(y_B-y_A)^2$ and $\hat{r}_{BA}$=$\dfrac{(x_B-x_A)\hat{i}+(y_B-y_A)\hat{j}}{r_{BA}}.$
Combining these you can separate the force vector into the components. You probably know how do use trigonometry to find vector components. These two methods must give agreeing values.
See a longer version of this answer by me here.
A: No, it is not correct. Why is that?
Let us use the Coulumb's law with your "wrong" procedure. (Note that we are only interested in magnitudes here.)
$$F_x=k\frac{q_1q_2}{x^2}$$
$$F_y=k\frac{q_1q_2}{y^2}$$
where $k=\frac{1}{4\pi \epsilon_0}$, and $q_1$, $q_2$ are the (magnitudes of) charges. $x$ and $y$ are the x and y components of distance $r$, respectively.
What would be the magnitude of such force?
$$F=\sqrt{F_x^2+F_y^2}=kq_1q_2\sqrt{\frac{1}{x^2}+\frac{1}{y^2}}=kq_1q_2\frac{\sqrt{x^2+y^2}}{xy}=k\frac{q_1q_2r}{xy}$$
But the actual magnitude of force is just
$$F=k\frac{q_1q_2}{r^2}\neq k\frac{q_1q_2r}{xy}$$
Hence, there is a discrepancy in magnitudes $\implies$ your procedure is not suitable.

But even the direction is not correct:
The intuitive direction of the correct force is just $\arctan \frac{x}{y}$.
But using your two components of force, we get
$$\arctan\frac{F_x}{F_y}=\arctan\frac{y^2}{x^2}\neq \arctan \frac{x}{y}$$

As a sidenote, you can consider most extreme case: two charges could be on the x axis. In that case, the y component would not be defined.
