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JEE Advanced · Physics · 17. Electrostatics

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 space]

  1. A If h>2R and r>R then Φ=Q0
  2. B If h>2R and r=3R5 then Φ=Q50
  3. C If h<8R5 and r=3R5 then Φ=0
  4. D If h>2R and r=4R5 then Φ=Q50
Verified Solution

Answer & Solution

Correct Answer

(A) If h>2R and r>R then Φ=Q0

Step-by-step Solution

Detailed explanation

(A) h>2R r>R

ϕ=Qε0
Clearly from Gauss' Law Option A is correct.
for h=2r R=4R5

shaded charge=2π(1-cos53o)×Q4π=Q5
Where, 2π(1-cos53°)= solid angle
\(\therefore q_{\text {enclosed }}=\frac{2 Q}{5}\left(\frac{Q}{5}\right.\) above center and \(\frac{Q}{5}\) below center \()\)
ϕ=2Q5ε0
for h>2R, r=4R5
ϕ=2Q5ε0 Option D is incorrect.
C suppose h=8R5 r=3R5

ϕ=0 (as no charge is enclosed inside the cylinder, as shown in diagram)
So for h<8R5, ϕ=0 Option (C) is correct.
B like option D for h=2R, r=3R5
qenclosed=2×2π1-cos37oQ4π=Q5
ϕ=Q5ε0 Option (B) is correct.
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