The electric field intensity on the surface of a charged solid sphere of radius ' r ' and volume charge density ' ρ ' is given by ( ε₀ = permittivity of free space) (a) zero (b) (ρ r)/(3ε₀) (c) (1)/(4πε₀)(σ)/(r) (d) (5ρ r)/(6ε₀)

Answer: (b). Gauss's law: E · 4π r² = (ρ · 4/3π r³)/(ε₀) E = (ρ r)/(3ε₀) So the answer is (b).

The electric field intensity on the surface of a charged solid sphere… | MHT CET 2021 (24 Sep, Shift 1)

Q82MHT CET 2021 · 24 SEP · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field intensity on the surface of a charged solid sphere of radius ' $r$ ' and volume charge density ' $\rho$ ' is given by ( $\epsilon_0 =$ permittivity of free space)

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Answer: (b)

Gauss's law: $E \cdot 4\pi r^2 = \dfrac{\rho \cdot \frac{4}{3}\pi r^3}{\epsilon_0}$
$E = \dfrac{\rho r}{3\epsilon_0}$
So the answer is (b).

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