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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.

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  • $\begingroup$ I could swear that someone has asked this exact question before, but I can't find it now. It might have been deleted. $\endgroup$ Mar 9 at 21:06
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    $\begingroup$ 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? $\endgroup$ Mar 9 at 21:12

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

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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.

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