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QJEEAdv19Ph2c04: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph2c04 Solution: A charged shell of radius R carries a total charge Q. Given as…

A charged shell of radius R carries a total charge Q. Given ϕ\phi 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?

[ ∈\in 0 is the permittivity of free space]

Multiple Choice Options

  • A If h > 2R and r = 4R / 5 then ϕ\phi = Q / 5 ∈\in 0
  • B If h > 2R and r = 3R / 5 then ϕ\phi = Q / 5 ∈\in 0
  • C If h > 8R /5 and r = 3R / 5 then ϕ\phi = 0
  • D If h > 2R and r = R then ϕ\phi = Q / ∈\in 0

Step-by-Step Verified Solution

(a) h > 2R and r > R

JEE Advanced 2019 Paper 1 Offline Physics - Electrostatics Question 33 English Explanation 1

ϕ=Qε0\phi = {Q \over {{\varepsilon _0}}} , clearly from Gauss' Law

(b) suppose h=8R5h = {{8R} \over 5} and r=3R5r = {{3R} \over 5}

JEE Advanced 2019 Paper 1 Offline Physics - Electrostatics Question 33 English Explanation 2

∴\therefore ϕ\phi = 0

so, for h < {{8R} \over 5} then ϕ\phi = 0.

(c) For h = 3R and r = 4R5{{4R} \over 5}

JEE Advanced 2019 Paper 1 Offline Physics - Electrostatics Question 33 English Explanation 3

Shaded charge = 2π(1−cos⁡ 53∘)×Q4π=Q52\pi (1 - \cos \,53^\circ ) \times {Q \over {4\pi }} = {Q \over 5}

∴\therefore qenclosed=2Q5{q_{enclosed}} = {{2Q} \over 5}

∴\therefore ϕ=2Q5ε0\phi = {{2Q} \over {5{\varepsilon _0}}}

∴\therefore for h > 2R and r=4R5r = {{4R} \over 5}

∴\therefore ϕ=2Q5ε0\phi = {{2Q} \over {5{\varepsilon _0}}}

(d) like option c for h = 2R and r=3R5r = {{3R} \over 5}

qenclosed=2×2π(1−cos⁡ 37∘)Q4π=Q5{q_{enclosed}} = 2 \times 2\pi (1 - \cos \,37^\circ ){Q \over {4\pi }} = {Q \over 5}

∴\therefore Electric flux, ϕ=Q5ε0\phi = {Q \over {5{\varepsilon _0}}}

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