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Question

# A ball of radius R carries a positive charge whose volume density depends only on a separation r from the ball's centre as ρ=ρ0(1−r/R), where ρ0 is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find:(A) the magnitude of the electric field strength as a function of the distance r both inside and outside the ball;(b) the maximum intensity Emax and the corresponding distance rm.

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Solution

## (a) We assume the ball to be divided into infinite number of concentric thin shell of thickness dr. Let us assume such a shell at a radial distance r.Volume of this shell =pdV=4πr2ρdr=4πr2ρ0(1−rR)drNet charge enclosed in the sphere of radius rq=∫pdv=r∫04πρ0r2(1−rR)dr=4πρ0(r33−r44R)Electric field at radial distance rE=q4πε0r2=14πε0r2×4πρ0(r33−r44R)=ρ0r3ε0(1−3r4R)Now for those points outside the sphere, we have to take into account the total charge contained in the sphere of radius R.Total charge Q=R∫0ρ0(1−rR)4πr2dr=πρ0R33Electric field, using Gauss's laws, at a point which is at distance r(>R) from the centre;∫E.ds=Qε0⟹E=Q4πε0r2=ρ0R312ε0r2(b) Again let us consider the expression for electric field within the sphere of radius R:E=ρ0r3ε0(1−3r4R)For maximum electric field:dEdr=0⟹1−6r4R=0⟹r=2R3Putting, r=2R3, value of maximum electric field, Emax=ρ0R9∈0

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