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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2019 · Paper 1 · Physics · Electrostatics

A charged shell of radius R carries a total charge Q . Given Φ as...

A charged shell of radius RR carries a total charge Q.Q . Given ΦΦ as the flux of electric field through a closed cylindrical surface of height h,h, radius rr 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{\in}_{0} is the permittivity of free space]A charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closedA charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closedA charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closed

Options

  1. AIf h>2Rh > 2R and r>Rr > R then Φ=Q∈0Φ = \dfrac{Q}{{\in}_{0}}
  2. BIf h>2Rh > 2R and r=3R5r = \dfrac{3R}{5} then Φ=Q5∈0Φ = \dfrac{Q}{5{\in}_{0}}
  3. CIf h<8R5h < \dfrac{8R}{5} and r=3R5r = \dfrac{3R}{5} then Φ=0Φ = 0
  4. DIf h>2Rh > 2R and r=4R5r = \dfrac{4R}{5} then Φ=Q5∈0Φ = \dfrac{Q}{5{\in}_{0}}

Correct answer

A. If h>2Rh > 2R and r>Rr > R then Φ=Q∈0Φ = \dfrac{Q}{{\in}_{0}}

Step-by-step solution

( A ) ⇒ h > 2Rr > R<br><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/sqwcjchmnxveiygmzbocgidutnyeuaapvkzvxuqkauqtdvacfh.png"/><br><br > <img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/sqwcjchmnxveiygmzbocgidutnyeuaapvkzvxuqkauqtdvacfh.png"/> <br >ϕ = \dfrac{Q}{{ε}_{0}}<br>ClearlyfromGauss′Law<br >Clearly from Gauss' Law⇒Option(A)Option \left(A\right) is correct.
⇒⇒ for h=2rR=4R5h = 2rR = \dfrac{4R}{5}
A charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closed
shadedcharge=2π(1−cos53o)×Q4π=Q5shaded\text{charge} = 2\pi ( 1 - cos{53}^{o} ) \times \dfrac{Q}{4\pi} = \dfrac{Q}{5}
Where, 2π(1−cos⁡53°)=solidangle2\pi ( 1 - \cos 53 ° ) = solidangle
∴qenclosed=2Q5(Q5abovecenterandQ5belowcenter)∴ {q}_{enclosed} = \dfrac{2Q}{5}\left(\dfrac{Q}{5}abovecenterand\dfrac{Q}{5}belowcenter\right)
∴ϕ=2Q5ε0∴ ϕ = \dfrac{2Q}{5{ε}_{0}}
∴forh>2R,r=4R5∴ forh > 2R, r = \dfrac{4R}{5}
∴ϕ=2Q5ε0⇒∴ ϕ = \dfrac{2Q}{5{ε}_{0}} ⇒ Option \left(D\right)isincorrect.<br>(C)⇒is incorrect.<br > \left(C\right) ⇒ suppose h=8R5r=3R5h = \dfrac{8R}{5}r = \dfrac{3R}{5}
A charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a closed
ϕ=0ϕ = 0 (as no charge is enclosed inside the cylinder, as shown in diagram)
So for h<8R5,ϕ=0⇒h < \dfrac{8R}{5}, ϕ = 0 ⇒ Option (C)( C ) is correct.
\left(B\right) ⇒likeoption(D)like option \left(D\right) for h=2R,r=3R5h = 2R, r = \dfrac{3R}{5}
qenclosed=2×2π(1−cos37o)Q4π=Q5{q}_{enclosed} = 2 \times 2\pi\left(1 - cos{37}^{o}\right)\dfrac{Q}{4\pi} = \dfrac{Q}{5}
∴ϕ=Q5ε0⇒∴ ϕ = \dfrac{Q}{5{ε}_{0}} ⇒ Option (B)( B ) is correct.
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