0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ If the point charge resides inside the sphere, the force exerted will be zero. If charge is outside, use Gauss law, and if charge is on the sphere, use boundary condition and get the force. This is an Electrostatics question. You can put it up in Physics branch of stackexchange. $\endgroup$ – user3001408 Feb 20 '17 at 13:20
  • $\begingroup$ If the force is outside how would you use Gauss law and boundary conditions? $\endgroup$ – User123456789 Feb 20 '17 at 13:27
0
$\begingroup$

$$\bf{F}=e\bf{E}$$ 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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ there is always a discontinuity of field at the sphere because of presence of charge. $\endgroup$ – user3001408 Feb 20 '17 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.