Saved Bookmarks
| 1. |
Obtain the expression for electric field due to an uniformly charge spherical shell. |
|
Answer» Solution :Electric Field DUE to a uniform charged spherical shell : Consider a UNIFORMLY charged spherical shell. Radius - R Total charge - Q (a) At a point outside the shell (`r gt R`) : P is a point ourside the shell at a distance r from the centre. The charge is uniformly distributed on the surface of the sphere. If `Q gt 0`, field point radially outward. If `Q lt 0,` field point readially inward. Applying Gauss LAW `ointvecE.vec(dA) = Q/epsi_(0)"...(1)"` `vecE` and `vec(dA)` are in the same direction. Hence E`oint "dA"=Q/epsi_(0)` But `oint`dA = total area of Gaussian SUFACE `= 4 pi r^2` Substituting in (1) `E.4 pir^2 =Q/epsi_0 (or)E=1/(4 piepsi_0)Q/r^2` In vector from `vecE= 1/(4 pi epsi_0) Q/r^2 hatr` (b) At a point on the suface of the spherical shell ( r = R). Electric field at point on the spherical shell, is r = R `vec E = Q /(4 pi epsi_(0) R^2) hatr` (c) At a point inside the shell ( ` r gt R`): Consider a point P inside the shell at a distance r from the center. `ointvecE.vec(dA) = Q/epsi_(0)"...(1)"` `E. 4 pi r^2 = Q/epsi_0` SINCE Gaussian surface encloses no charge, so Q = 0 `:. "E =0"` |
|