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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2022 · 25 June · Morning Shift · Physics · Electrostatics

In the figure, a very large plane sheet of positive charge is...

In the figure, a very large plane sheet of positive charge is shown. P1{P}_{1} and P2{P}_{2} are two points at distance ll and 2l2l from the charge distribution. If σ\sigma is the surface charge density, then the magnitude of electric fields E1{E}_{1} and E2{E}_{2} at P1{P}_{1} and P2{P}_{2} respectively are

In the figure, a very large plane sheet of positive charge is shown. P _ 1 and P _ 2 are two points at distance l and

Options

  1. AE1=σεo,E2=σ2εo{E}_{1} = \dfrac{\sigma}{{ε}_{o}}, {E}_{2} = \dfrac{\sigma}{2{ε}_{o}}
  2. BE1=2σεo,E2=σεo{E}_{1} = \dfrac{2\sigma}{{ε}_{o}}, {E}_{2} = \dfrac{\sigma}{{ε}_{o}}
  3. CE1=E2=σ2εo{E}_{1} = {E}_{2} = \dfrac{\sigma}{2{ε}_{o}}
  4. DE1=E2=σεo{E}_{1} = {E}_{2} = \dfrac{\sigma}{{ε}_{o}}

Correct answer

C. E1=E2=σ2εo{E}_{1} = {E}_{2} = \dfrac{\sigma}{2{ε}_{o}}

Step-by-step solution

The electric field due to a large uniformly charged non-conducting sheet is given by, E=σ2εoE = \dfrac{\sigma}{2{ε}_{o}}. It is independent of the distance from the sheet. Therefore,

E1=E2=σ2εo{E}_{1} = {E}_{2} = \dfrac{\sigma}{2{ε}_{o}}

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