A charge 17.7 × 10⁻⁴ C is distributed uniformly over a large sheet of area 200 m². The electric field intensity at a distance 20 cm from it in air will be [ε₀ = 8.85 × 10⁻¹² C²/N m²] (a) 5 × 10⁵ N/C (b) 6 × 10⁵ N/C (c) 7 × 10⁵ N/C (d) 8 × 10⁵ N/C

Answer: (a). σ = q/A = 17.7 × 10⁻⁴200 = 8.85 × 10⁻⁶ C/m² Large sheet: E = (σ)/(2ε₀) = 8.85 × 10⁻⁶2 × 8.85 × 10⁻¹², independent of distance. E = 5 × 10⁵ N/C So the answer is (a).

A charge 17.7 × 10⁻⁴ C is distributed uniformly over a large sheet of… | MHT CET 2023 (14 May, Shift 2)

Q59MHT CET 2023 · 14 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

A charge $17.7 \times 10^{-4}\text{ C}$ is distributed uniformly over a large sheet of area $200\text{ m}^2$ . The electric field intensity at a distance $20\text{ cm}$ from it in air will be $\left[\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2\right]$

Show answer and solution

Answer: (a)

$\sigma = \dfrac{q}{A} = \dfrac{17.7 \times 10^{-4}}{200} = 8.85 \times 10^{-6}\ \text{C/m}^2$
Large sheet: $E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$ , independent of distance.
$E = 5 \times 10^{5}$ N/C
So the answer is (a).

Report a mistake

Electric Flux & Gauss's Law question from MHT CET 2023 (14 May, Shift 2)

More questions on Electric Flux & Gauss's Law