MHT CET202314 May 2023Evening ShiftPhysicsElectrostaticsActual
A charge 17.7 10⁻⁴ C is distributed uniformly over a large sheet of area 200 ~m ^2 . The electric field intensity at a distance 20 ~cm from it in air will be [ ₀=8.85 10⁻¹² C ^2 / Nm ^2 ]
Options
- A5 10^5 ~N / C
- B6 10^5 ~N / C
- C7 10^5 ~N / C
- D8 10^5 ~N / C
Correct answer
A. 5 10^5 ~N / C
Step-by-step solution
The surface charge density is given by, = q A = 17.7 10⁻⁴ 200 =8.85 10⁻⁶ C _ m m ^2 The electric field intensity at a distance of 20 ~cm in air is, E = 2 ₀ = 8.85 10⁻⁶ 2 8.85 10⁻¹² =5 10^5 ~N / C Hence, option (A).