Force between a charged point and a charged sphere

Question

From coloumbs law we have that the charge between two points of charge is $F=\frac{kqQ}{r^2}$. What would be the charge on a point if one rather had a sphere with uniformly distributed charge inside of radius R exerting the force?

The distance between the charge and the center of the sphere is r.

The charge of the point is q

The charge distributed with the volume of the sphere is Q

k is given as $8.99*10^9 $

I am pretty sure this can be solved using a integral, but am wondering if my soloution is correct. I am thinking that one can use integrals to calculate the force from every point in the sphere, $Q_p$, summed up. Where $Q_p$ is $\frac{Q}{4/3*\pi*R^3}$. And where the force thus can be calculated using:

$$F=\int_{{r-R}}^{{r+R}}\int_{0}^{\sqrt{R^2-(x-r)^2}}2\int_{0}^{\pi}\frac{kqQ_{p}x}{(x^2+H^2)^{\frac{3}{2}}}d\theta dHdx$$

To set it more in context: I am trying to find the electrostatic force between a alpha-particle and a gold nuclei. Where I include the radius of the gold-nuclei.

Answer 1

$$\bf{F}=e\bf{E}$$ Using Gauss's law outside the sphere $$\oint_{\Sigma}\bf{E}\cdot\bf{dS}=E_{r}4\pi{r^{2}}=\frac{Q}{\epsilon}$$ Hence $$F=\frac{eQ}{4\pi\epsilon{r^{2}}}$$ So the result is still inverse square. Using Gauss's law inside the sphere $$\oint_{\Sigma}\bf{E}\cdot\bf{dS}=E_{r}4\pi{r^{2}}=\frac{Qr^{3}}{\epsilon{R}^{3}}$$ Hence $$F=\frac{eQr}{4\pi\epsilon{R}^{3}}$$ So the result is linear.