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Charge Q is distributed uniformly over a non conducting sphere of radius R. Find the electric p-0tential at distance r from the centre of the sphere r (r lt R). The electric field at a distance r from the centre of the sphere is given as (1)/(4pi epsilon_(0)). (Q)/(R^(3))hatr . Also find the potential at the centre of the sphere . |
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Answer» Solution :The ELECTRIC potential on the surface of such a sphere is `V(R) = (1)/(4pi epsilon_(0)).(Q)/(R)` We can use the equation `V(r) : V(R)= : vint_(R)^(r)vecE.dvecr` `:. V(r): (R) =: int_(R)^(r)(1)/(4piepsilon_(0))(Q)/(R^(3))RDR (hatr.hatr)` `( because dvecr = dr vecr)` `=: (Q)/(4pi epsilon_(0)R^(3))int_(R)^(r)rdr` `=: (Q)/(4piepsilon_(0)R^(2))[(r^(2))/(2)]_(R)^(r)` `=: (Q)/(4pi epsilon_(0)R^(3))[(r^(2))/(2) -(R^(2))/(2)]` `:. V(r)=V(R)+(Q)/(4piepsilon_(0)R^(3))[(R^(3))/(2)-(r^(2))/(2)]` `:. V(r)=(1)/(4piepsilon_(0))(Q)/(R)+(Q)/(8piepsilon_(0)R^(3))(R^(2)-r^(2))` `:. V(r) = (1)/(4piepsilon_(0))(Q)/(2R) (2+(1)/(R^(2))(R^(2)-r^(2)))` `=(1)/(4piepsilon_(0))(Q)/(2R) (2+(R^(2))/(R^(2))-(r^(2))/(R^(2)))` `:. V(r) = (1)/(4piepsilon_(0))(Q)/(2R) (3-(r^(2))/(R^(2)))=r lt R` `:. V(r)=(kQ)/(2R)[3-(r^(2))/(R^(2))]` At the centre of sphere r =0 `:. V (0) =(1)/(4pi epsilon_(0)).(3Q)/(2R) = (3kQ)/(2R)` |
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