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JEE Advanced2019PhysicsElectrostaticsActual

A charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closed cylindrical surface of height h , radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? [ ∈ 0 is the permittivity of free spa

Options

  1. AIf h > 2 R and r > R then Φ = Q ∈ 0
  2. BIf h > 2 R and r = 3 R 5 then Φ = Q 5 ∈ 0
  3. CIf h < 8 R 5 and r = 3 R 5 then Φ = 0
  4. DIf h > 2 R and r = 4 R 5 then Φ = Q 5 ∈ 0

Correct answer

A. If h > 2 R and r > R then Φ = Q ∈ 0

Step-by-step solution

( A ) ⇒ h > 2 R r > R ϕ = Q ε 0 Clearly from Gauss' Law ⇒ Option A is correct. ⇒ for h = 2 r R = 4 R 5 s h a d e d charge = 2 π ( 1 - c o s 53 o ) × Q 4 π = Q 5 Where, 2 π ( 1 - c o s 53 ° ) = s o l i d a n g l e ∴ q e n c l o s e d = 2 Q 5 Q 5 a b o v e c e n t e r a n d Q 5 b e l o w c e n t e r ∴ ϕ = 2 Q 5 ε 0 ∴ f o r h > 2 R , r = 4 R 5 ∴ ϕ = 2 Q 5 ε 0 ⇒ Option D is incorrect. C ⇒ suppose h = 8 R 5 r = 3 R 5 ϕ = 0 (as no charge is enclosed inside the cylinder, as shown in diagram) So for h 8 R 5 , ϕ = 0 ⇒ Optio

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