JEE Main 2014PhysicsElectrostaticsMediumMCQ

JEE Main 2014 — Electrostatics Question with Solution

JEE Main 2014 (09 Apr Online)

Question

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given : ∈ O = 8.85 × 1 0 - 1 2   C 2 / N-m 2 ,   R E = 6.37 × 1 0 6 m ]

Choose an option

Show full solutionCorrect option: A
Correct answer
A-680 kC

Step-by-step explanation

E=150 N/C

We know that  ϕ = E → · A →

                          = E × 4 π r 2

                          = 1 5 0 × 4 π × R e 2

Now, ϕ = Q in ∈ o



Q in = ϕ ∈ o = 1 5 0 × 4 π R e 2 × 8.85 × 1 0 - 1 2  C

      = 1 5 0 × 4 π 6.37 × 1 0 6 2 × 8.85 × 1 0 - 1 2

      = 6 0 0 π × 6.37 × 6.37 × 8.85 C

      =  6 8 0  kC

As Electric field is inward, Q will be negative

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About this question

This is a previous-year question from JEE Main 2014, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.