In the figure, a very large plane sheet of positive charge is shown. P1 and P2 are two points at distance l and 2l from the charge distribution. If σ is the surface charge density, then the magnitude of electric fields E1 and E2 at P1 and P2 respectively are: (a) E₁ = σ/ε₀, E₂ = σ/2ε₀ (b) E₁ = 2σ/ε₀, E₂ = σ/ε₀ (c) E₁ = E₂ = σ/2ε₀ (d) E₁ = E₂ = σ/ε₀

Answer: (c). Infinite sheet: E = (σ)/(2ε₀), independent of distance. So E₁ = E₂ = (σ)/(2ε₀). So the answer is (c).

In the figure, a very large plane sheet of positive charge is shown… | JEE Main 2022 (25 Jun, Shift 1)

Q73JEE Main 2022 · 25 JUN · SHIFT 1Easy

Electric Flux & Gauss's Law

In the figure, a very large plane sheet of positive charge is shown. P1 and P2 are two points at distance l and 2l from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields E1 and E2 at P1 and P2 respectively are:

Show answer and solution

Answer: (c)

Infinite sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$ , independent of distance.
So $E_1 = E_2 = \dfrac{\sigma}{2\varepsilon_0}$ .
So the answer is (c).

Report a mistake

Electric Flux & Gauss's Law question from JEE Main 2022 (25 Jun, Shift 1)

More questions on Electric Flux & Gauss's Law