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State Gauss 'law in electrostatics. Consider an overall neutral sphere of radius R. This sphere has a point charge +Q at its centre and this positive charge is surrounded by a uniform density rho of negative charges up to a radius R. Use Gauss, law to obtain expression for the electric field , of this sphere , at a point distant r, from its centre ,wherer lt R,r gt R Show that these two expressions give identical results, for the electric field, at r= R. |
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Answer» Solution :Consider a sphere of radius R and having point CHARGE +Q at its CENTRE point. Negative CHARGES density of sphere be `RHO. ` then total negative charge on sphere ` =(4)/(3)pi R^(3) rho =-Q` ` rArr "" rho =-(3Q)/( 4 pi R^(3)) ` For a point `P_1 ` situated at a distance r (where r `lt R ) `frm its centre, considering a sphere of radius.r. as the Gaussian SURFACE , we have ` rArr "" phi _in =int oversetto (E) . oversetto (ds) = E . 4 pi r^(2) (1)/(in_0)("charged enclosed")` `"" (1)/( in_0)[Q +rho .(4)/(3)pi r^(3) ]` `rArr ""E. 4 pi r^(2)=(1)/( in_0)[Q -(3Q)/( 4 pi R^(3)).(4)/(3)pi r^(3) ]=(1)/(in_0)[Q -(Qr^(3))/( R^(3)) ]=(Q)/( in_0)[1-(r^(3))/( R^(3)) ]` `rArr ""E= ( Q)/( 4 pi in _0)[(1)/( r^(2)) -(r)/( R^(3)) ]` For r= R, we have ` E_("surface")=0.` (b) For a point `P_2` situated at a distance r (where `r gt R) ` we again consider a spere of radius r as the Gaussian surface. Now total charges enclosed in the surface. `=Q +rho .(4)/(3)pi R^(3)=Q- ( 3Q)/( 4 pi R^(3) ) .(4)/(3) pi R^(3) =Q -Q = 0 ` Thisshown that `phi_in =E. 4pi r^(2)=(1)/(in_0) ` (charge enclosed)`=(1)/(in_0)(0)=0 ` `rArr "" E= 0 ` and field at surface of sphere `E_("surface") =0` . 
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