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

Answer: (d). Gauss's law on the surface: E · 4π r² = (σ · 4/3π r³)/(ε₀) (here σ is the volume charge density). E = (σ r)/(3ε₀) So the answer is (d).

The electric field intensity on the surface of a solid charged sphere… | MHT CET 2025 (21 Apr, Shift 2)

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Electric Flux & Gauss's Law

The electric field intensity on the surface of a solid charged sphere of radius $\mathbf{r}$ and volume charge density $\sigma$ is ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (d)

Gauss's law on the surface: $E \cdot 4\pi r^2 = \dfrac{\sigma \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$ (here $\sigma$ is the volume charge density).
$E = \dfrac{\sigma r}{3\varepsilon_0}$
So the answer is (d).

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Electric Flux & Gauss's Law question from MHT CET 2025 (21 Apr, Shift 2)

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