- #1
tsuwal
- 105
- 0
Homework Statement
This problem may be dull, I know, but maybe there is a hidden math trick that i don't know of. This picture sums up the problem.
So, you should prove by simplifing the integral that [itex]F^e[/itex], the eletric force applied between two spheres, onde with a charge [itex]q_1[/itex] and the other with the charge [itex]q_2[/itex] (distributed evenly in volume, with a charge density [itex]\rho[/itex]) doesn't depend on [itex]R_2[/itex], it only depends on [itex]q_1, q_2[/itex] and [itex]d[/itex]
Homework Equations
[itex]F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV [/itex]
The Attempt at a Solution
[itex]F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV =\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{\delta^{2}}drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d-rcos(\phi )^2+(rsin(\phi )^2))} drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d^2+r^2-2drcos(\phi ))} drd\phi[/itex]
How do you simplify this integral or at least show that the expression doesn't depend on [itex]R_2[/itex]? I tried to derivate with respect to [itex]R_2[/itex] but it didn't helped...