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Question

Calculate electric field intensity due to uniformly charged non-conducting sphere:(a) outside the sphere(b) at the surface of sphere(c)inside the sphere(d) at the centre of the spherePlot at graph between electric Held intensity and distance.

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Solution

Electric field intensity due to a uniformly charged non-conducting spherewhen charge is given to non-conducting sphere , it uniformly spreads through out its volume . if q is the charge given and R is the radius of the sphere, then the volume charge density.p=q43πR3or p=3q4πR3............(1)(a) Outside the sphere : in this case taking O as center and r as radius a spherical Gaussian surface is drawn . the point P will be situated at this surface.the direction of →E will be outwards directed due to symmetry . the charge enclosed by the Gaussian surface.Σq=qTherefore from Gauss's lawϕE=qϵ0..........(2)From the defination of the electric flux,ϕE=∮S→E→ds=∮SEdscos00=∮SEdsor ϕE=E∮Sds ( as E will be same at each points of Gaussian surface ) or ϕE=E4πr2...........(3) Now equating equations (2) and (3) we get E×4πr2=qϵ0or E=14πϵ0qr2..........(4)From equation (1) p=3q4πR2⇒q=43πR3p∴ From equation (4) E=14πϵ0 frac4πR3p3r2or E=pR33ϵ0r2or E=p3ϵ0(R3r2)...........(5) in vector form→E=P3ϵ0(R3r2)^r→E=p3ϵ0(R3r2)^r...........(6) where ^r = unit vector in direction OP(b) When point is situated on the surface of the sphere ; in this case we can consider the surface of the surface as Gaussian surface . therefore the whole charge of the sphere will again be charge enclosed.thus by putting r=R equation we get E=P3ϵ0(R3R2)or E=pR3ϵ0..........(7)In vector from →E=pR3ϵ0^r...........(8) (c) When point P is inside the sphere (r<R)Here also we draw a spherical Gaussian surface taking O as center and r as radius . point P will be situated at this surface. now the charge enclosed by this Gaussian surface.Σq=p43πr3Substituting the value of p form equation (1) σq=3q4πR3×4πr33or Σq=qr3R3Therefore from Gauss's law,ϕE=Σqϵ0=qr3ϵ0R3or ϕE=qr3ϵ0R3..........(9) From the defination of the electric flux,ϕE=∮S→E→ds=∮SEdscos00(for spherical surface ) =E∮Sdsor ϕE=E×πr2........(10) Now equating equation (9) and (10) we haveE×4πr2=qr3ϵ0R3∴E=14πϵ0qr3r2R3or E=14πϵ0qrR3.........(11) Again from the equation (1) q=43πR3p∴ From equation (11) E=14πϵ0rR3×43πR3por E=p3ϵ0r .............(12) In vector form ,→E=p3ϵ0r^r...........(13) (d) At the centre of the sphere : for central point r=0∴ From equation (12) E=0.........(14) The sketch showing the variation of E with r is given figure

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