Verified Solution • 2025–26 Session moderate

QJEEM17Phc30e: JEE Main 2017 Physics - Class 12 JEEM17Phc30e Solution: Four closed surfaces and corresponding charge distributions are …

Four closed surfaces and corresponding charge distributions are shown below.

JEE Main 2017 (Online) 9th April Morning Slot Physics - Electrostatics Question 191 English

Let the respective electric fluxes through the surfaces be Φ1,{\Phi _1}, Φ2,{\Phi _2}, Φ3{\Phi _3} and Φ4{\Phi _4} . Then :

Multiple Choice Options

  • A Φ1{\Phi _1} < Φ2{\Phi _2} = Φ3{\Phi _3} > Φ4{\Phi _4}
  • B Φ1{\Phi _1} > Φ2{\Phi _2} > Φ3{\Phi _3} > Φ4{\Phi _4}
  • C Φ1{\Phi _1} = Φ2{\Phi _2} = Φ3{\Phi _3} = Φ4{\Phi _4}
  • D Φ1{\Phi _1} > Φ3{\Phi _3} ; Φ2{\Phi _2} < Φ4{\Phi _4}

Step-by-Step Verified Solution

Net flux through a closed surface,

ϕ\phi = qencloseε0{{{q_{enclose}}} \over {{\varepsilon _0}}}

qenclosed = charge enclosed by closed surface.

For surface S1,

ϕ\phi 1 = 1ε0{1 \over {{\varepsilon _0}}} (2q)

For surface S2,

ϕ\phi 2 = 1ε0{1 \over {{\varepsilon _0}}} (q + q + q −- q) = 1ε0{1 \over {{\varepsilon _0}}} (2q)

For Surface S3,

ϕ\phi 3 = 1ε0{1 \over {{\varepsilon _0}}} (q + q) = 1ε0{1 \over {{\varepsilon _0}}} (2q)

For surface S4,

ϕ\phi 4 = 1ε0{1 \over {{\varepsilon _0}}} (8q −- 2q −- 4q) = 1ε0{1 \over {{\varepsilon _0}}} (2q)

∴   \therefore\,\,\, ϕ\phi 1 = ϕ\phi 2 = ϕ\phi 3 = ϕ\phi 4

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