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# A solid ball of radius R has a charge density ρ given by ρ=ρ0(1−r/R) for 0≤r≤R. The electric field outside the ball is:4rho0 R33 ∈0 r23 ρ0 R34 ∈0 r2ρ0 R312 ∈0 r2ρ0 R3∈0 r2

A
ρ0 R312 0 r2
B
ρ0 R30 r2
C
4rho0 R33 0 r2
D
3 ρ0 R34 0 r2
Solution
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#### ρ=ρ0(1−rR)dq=ρdvqin=∫dq=ρdvρ0(1−rR)4πr2dr∵dv=4πr2dr=4πρ0∫R0(1−rR)r2dr=4πρo∫Ror2dr−r2Rdr=4πρo[[r33]Ro−[r44R]Ro]=4πρo[R33−R44R]=4πρo[R33−R34]=4πρo[R312]q=πρoR33E.4πr2=(πρoR33∈o)⇒E=ρoR312∈or2

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