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The potential inside a charged spherical ball is given by ϕ = ar² + b where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:

1. −6aϵ0
2. −24πaϵ0
3. 6aϵ0
4. 24πaϵ0

Correct answer: −6aϵ0

Solution

phi = a r^2 + b -> E = -dphi/dr = -2 a r. Using Gauss in spherical form, rho = e0*(1/r^2) d/dr(r^2 E) = e0*(1/r^2)(-6 a r^2) = -6 a e0. The charge density is -6 a e0.

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