The electrostatic potential inside a charged spherical ball is given by = ar^2 + b where r is the distance from the centre; a, b are constants. Then t - Physics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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The electrostatic potential inside a charged spherical ball is given by ϕ = a r^{2} + b where r is the distance from the centre; a, b are constants. Then the charge density inside ball is:

Solution & Step-by-step Explanation

Electric field E = - \frac{d ϕ}{d r} = - 2 a r .Using Gauss's Law in differential form (\nabla \cdot E = \frac{ρ}{ϵ _{0}}):In spherical coordinates for radial field: \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} E) = \frac{ρ}{ϵ _{0}} $ ρ = ϵ_{0} \frac{1}{r ^{2}} \frac{d}{d r} [r^{2} (- 2 a r)] = ϵ_{0} \frac{1}{r ^{2}} \frac{d}{d r} (- 2 a r^{3}) $ ρ = ϵ_{0} \frac{1}{r ^{2}} (- 6 a r^{2}) = - 6 a ϵ_{0}

Try it yourself before checking the explanation above.

The electrostatic potential inside a charged spherical ball is given by ϕ = a r^{2} + b where r is the distance from the centre; a, b are constants. Then the charge density inside ball is:

- 6 a ϵ_{0} r

- 24 πa ϵ_{0} r

- 6 a ϵ_{0}

- 24 πa ϵ_{0}

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