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Obtain the expression of electric field by thir spherical shell with uniform charge distribution at a point outside it. |
Answer» Solution :Let `sigma`be the UNIFORM surface charge density ofthin spherical shell of radius R.  Consider a point P OUTSIDE the shell with radius vector `vecr`. To calculate `vecE`at P, we take the Gaussian surface to be a SPHERE of radius t and with centre 0 passing through P. The ELECTRIC field at each point of the Gaussian surface has the same magnitude E and is ALONG the radius vector at each point. Thus, `vecE`and `DeltavecS`at every point are parallel and the flux through each element is `EDeltaS`. Summing over all `DeltaS`the flux through the Gaussian surface is `E xx 4pir^(2)`. The charge enclosed is `sigma xx 4piR^(2)` By Gauss.s law, `E xx 4pir^(2) = sigma/epsilon_(0).4piR^(2)` `therefore E = (sigmaR^(2))/(epsilon_(0)r^(2)) =q/(4piepsilon_(0)r^(2)) =(kq)/r^(2)` `therefore vecE = (sigmaR^(2))/(4pi epsilon_(0)r^(2))hatr = (kq)/r^(2).hatr` |
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