- #1

matpo39

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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