- #1
matpo39
- 43
- 0
ok i was flying through the homework no prob. then i hit this problem and i got an answer but i don't think its right.
A uniformly charged sphere of radius R and volume charge density [tex] \rho_0[/tex] is adjacent to a uniformly charged infinite plane of surface charge density [tex] \sigma_0[/tex]. the charge densities are related by
[tex] \sigma_0=\frac{\rho_0R}{2}[/tex]
the center of the sphere is a distance d from the plane. Find two points, one inside the sphere and one outside the sphere where the electric field is oriented away from the plane at a 45 degree angle with respect to the z axis.[note these points are not on the axis] (in the figure the infinite plane lies in the xy plane )
well i started this off by finding the electric field inside the sphere
[tex]\vec{E_s}=\frac{\rho_0R}{3\epsilon_0}[/tex]
i then found the charge of the infinite plane via the pill box gaussian surface and came up with
[tex]\vec{E_p}=\frac{\sigma_0}{2\epsioln_0}=\frac{\rho_0R}{4\epsilon_0}[/tex]
[tex]\vec{E_s}+\vec{E_p}=\frac{7\rho_0R}{12\epsilon_0}[/tex]
and breaking to components i got
[tex]\frac{7\rho_0R}{12\epsilon_0}(cos45+sin45)[/tex]
and by a similar approch i got
[tex][\frac{\rho_0R}{\epsilon_0}(\frac{R^2}{3r^2}+\frac{1}{4})](cos45+sin45)[/tex]
like i said i don't think this is right, so if some one could help me out a bit that would be great
thanks
A uniformly charged sphere of radius R and volume charge density [tex] \rho_0[/tex] is adjacent to a uniformly charged infinite plane of surface charge density [tex] \sigma_0[/tex]. the charge densities are related by
[tex] \sigma_0=\frac{\rho_0R}{2}[/tex]
the center of the sphere is a distance d from the plane. Find two points, one inside the sphere and one outside the sphere where the electric field is oriented away from the plane at a 45 degree angle with respect to the z axis.[note these points are not on the axis] (in the figure the infinite plane lies in the xy plane )
well i started this off by finding the electric field inside the sphere
[tex]\vec{E_s}=\frac{\rho_0R}{3\epsilon_0}[/tex]
i then found the charge of the infinite plane via the pill box gaussian surface and came up with
[tex]\vec{E_p}=\frac{\sigma_0}{2\epsioln_0}=\frac{\rho_0R}{4\epsilon_0}[/tex]
[tex]\vec{E_s}+\vec{E_p}=\frac{7\rho_0R}{12\epsilon_0}[/tex]
and breaking to components i got
[tex]\frac{7\rho_0R}{12\epsilon_0}(cos45+sin45)[/tex]
and by a similar approch i got
[tex][\frac{\rho_0R}{\epsilon_0}(\frac{R^2}{3r^2}+\frac{1}{4})](cos45+sin45)[/tex]
like i said i don't think this is right, so if some one could help me out a bit that would be great
thanks