JEE Main 2020PhysicsElectrostaticsHardMCQ

JEE Main 2020 — Electrostatics Question with Solution

JEE Main 2020 (09 Jan Shift 1)

Question

Consider a sphere of radius R which carries a uniform charge density ρ . If a sphere of radius R2 is carved out of it, as shown, the ratio EA→EB→ of magnitude of electric field EA→ and EB→ , respectively, at points A and B due to the remaining portion is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B1834

Step-by-step explanation

E=ρr3ε0
EA=-ρR23ε0
EA=ρR6ε0
Electric filed at point B=EB=E1A+E2A
E1A= Electric Filed Due to solid sphere of radius R at point B=ρR3ε0
E2A= Electric Filed Due to solid sphere of radius R2 (which having charge density –ρ ) at point B
=-KQ'×49R2=-ρR54ε0
EB=E1A+E2A=ρR3ε0-ρR54ε0=17ρR54ε0
EAEB=917

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About this question

This is a previous-year question from JEE Main 2020, 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.