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Write the expression for electric field intensity at any point outside and inside due to a charged spherical shell. |
Answer» SOLUTION :Electric field due to a uniformly charged thin spherical shell : (i)When point P lies outside the spherical shell Suppose that we have to calculate electric field at the point P at a distance `r(rgtR)`from its centre. Draw the Gaussian surface through point P so as to ENCLOSE the charged spherical shell . The Gaussiansurface is spherical shell of radius r and centre O. Let `VECE`be the electric field at point P, then the electric flux through area element is `vecds`given by , `dphi=vecE*vecds` Since `vecds` is also along normal to the surface, `dphi=Eds` `THEREFORE` Total electric flux through the Gaussian surface isgiven by , `phi=oint_(s)Eds=Eoint_(s)ds` Now , `oint_(s)ds=4pir^(2)` `thereforephi=Exx4pir^(2)` ... (i) Since the charge enclosed by the Gaussian surface is q , according to Gauss theorem, `phi=(q)/(epsilon_(0))`...(ii) From equations (i) and (ii), we obtain `Exx4pir^(2)=(q)/(epsilon_(0))` `E=(1)/(4piepsilon_(0)).(q)/(r^(2))("for "rgtR)` (ii)When point P lies inside the spherical shell : In such a case , the Gaussian surface encloses no charge . According to Gauss law, `Exx4pir^(2)=0` i.e., E=0 `("for"r lt R)` Graph SHOWING the variation of electric field as a function of r : 
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