A conducting sphere of radius 0.1 m has uniform charge density 1.8 /m² on its surface. The electric field in free space at radial distance 0.2 m from a point on the surface is ( ε₀ = permittivity of free space) (a) 6 × 10⁻⁶ε₀ Vm⁻¹ (b) 6 × 10⁻⁸ε₀ Vm⁻¹ (c) 2 × 10⁻⁷ε₀ Vm⁻¹ (d) 1 × 10⁻⁷ε₀ Vm⁻¹

Answer: (c). Outside a charged sphere: E = (σ 4π R²)/(4πε₀ r²) = (σ R²)/(ε₀ r²), with r from the centre. r = 0.1 + 0.2 = 0.3 m. E = 1.8 × 10⁻⁶ × 0.01ε₀ × 0.09 = 2 × 10⁻⁷ε₀ V m ⁻¹ So the answer is (c).

A conducting sphere of radius 0.1 m has uniform charge density 1.8… | MHT CET 2023 (14 May, Shift 1)

Q70MHT CET 2023 · 14 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A conducting sphere of radius $0.1\text{ m}$ has uniform charge density $1.8\ \mu\text{C/m}^{2}$ on its surface. The electric field in free space at radial distance $0.2\text{ m}$ from a point on the surface is ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (c)

Outside a charged sphere: $E = \frac{\sigma 4\pi R^2}{4\pi\varepsilon_0 r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}$ , with $r$ from the centre.
$r = 0.1 + 0.2 = 0.3$ m.
$E = \frac{1.8 \times 10^{-6} \times 0.01}{\varepsilon_0 \times 0.09} = \frac{2 \times 10^{-7}}{\varepsilon_0}$ V m $^{-1}$
So the answer is (c).

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Electric Flux & Gauss's Law question from MHT CET 2023 (14 May, Shift 1)

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