# 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.

• I could swear that someone has asked this exact question before, but I can't find it now. It might have been deleted. Mar 9 at 21:06
• If you did it the way you propose, then the force between two charges sitting on the $x$-axis would have an infinite $y$-component, since they would have $r_y = 0$. Does that seem plausible? Mar 9 at 21:12

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 $$F_x=\frac{\gamma ~x}{(x^2+y^2+z^2)^{3/2}}$$ $$F_y=\frac{\gamma ~y}{(x^2+y^2+z^2)^{3/2}}$$ $$F_z=\frac{\gamma ~z}{(x^2+y^2+z^2)^{3/2}}$$