Confusion about $x$ & $y$ components of Coulombic Force? I am new to this website and have a question about components of Coulomb's force. I know that we can find the $x$-component and $y$-component of a coulombic force by taking $|\mathbf{F}|\cos(\theta)$ and $|\mathbf{F}|\sin(\theta)$ respectively, where $|\mathbf{F}|$ is the magnitude of the total force.
My question is this: If the equation for finding the coulombic force magnitude is $$F=\frac{kq_1q_2}{r^2}$$ why is it that $x$-component of the Coulombic force cannot be calculated with just
$$F_x=\frac{kq_1q_2}{r_x^2}\ ?$$
and similarly, the $y$-component with $$F_y=\frac{kq_1q_2}{r_y^2}\ ?$$
I tried it myself for a made-up scenario where I have an alpha particle ($q_1=2$) $1$ meter away from a gold nucleus ($q_2=79$) with an angle of $\frac{\pi}{4}$... the $F_x$ I find using the above (wrong) method of $$F_x=\frac{kq_1q_2}{r_x^2}$$ comes out to $2.84084\cdot 10^{12}$, whereas if I do it correctly using $|\mathbf{F}|\cos(\frac{\pi}{4})$,  I get $F_x=1.004\cdot 10^{12}$.
The reason I am confused by this is because the fact that we cannot use the wrong method seems to suggest that the force felt in the $x$-direction is somehow influenced by the distance the particle is away from the nucleus in the $y$-direction, and that seems really odd to me. How can this be?
please be nice; I am new to physics and would really like to understand.
 A: An inverse-square force such as the Coulomb force can be expressed vectorially as
$$\mathbf{F}(x,y,z)=\frac{\gamma}{r^2}\hat{\mathbf{r}}$$where $\gamma$ is some constant, $r$ the radial distance from the origin, and $\hat{\mathbf{r}}$ the unit vector pointing from the origin to the point at which we're evaluating this force. Since $|\hat{\mathbf{r}}|=1$ by definition, we have
$$|\mathbf{F}| = \frac{\gamma}{r^2}$$ as you're aware. To decompose $\mathbf{F}$ in cartesian coordinates, we use
$$\hat{\mathbf{r}} = \frac{\mathbf{r}}{|\mathbf{r}|}$$ and so
$$\hat{\mathbf{r}} = \frac{1}{\sqrt{x^2+y^2+z^2}}\begin{bmatrix}x\\y\\z\end{bmatrix}$$
Using this the components of $\mathbf{F}$ come out to be
$$F_x=\frac{\gamma ~x}{(x^2+y^2+z^2)^{3/2}}$$
$$F_y=\frac{\gamma ~y}{(x^2+y^2+z^2)^{3/2}}$$
$$F_z=\frac{\gamma ~z}{(x^2+y^2+z^2)^{3/2}}$$
A: Coulomb force is a central potential; it depends on the distance of 2 charged particles. The direction of the force is the direction of the vector which connects the 2 charges.
You can analyze the Coulomb force in x and y coordinates, and by applying Newton's law you can find the acceleration of the particles in the x and y direction, from which, given initial conditions, you can find the equation of motion in the x and y direction.
