JEE Main 2017 · 9 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

Four closed surfaces and corresponding charge distributions are shown below.

Let the respective electric fluxes through the surfaces be $\Phi_1$ , $\Phi_2$ , $\Phi_3$ and $\Phi_4$ . Then :

Four closed surfaces: S₁ encloses 2q; S₂ encloses q, −q, q and q; S₃ encloses q and q, with 5q just outside; S₄ encloses 8q, −2q and −4q, with 3q outside.

\Phi_1<\Phi_2=\Phi_3>\Phi_4

\Phi_1>\Phi_2>\Phi_3>\Phi_4

\Phi_1=\Phi_2=\Phi_3=\Phi_4

\Phi_1>\Phi_3;\ \Phi_2<\Phi_4

Answer and worked solution

Official answer: C

By Gauss's law only the enclosed charge counts: S₁ has 2q; S₂ has q − q + q + q = 2q; S₃ has q + q = 2q (5q is outside); S₄ has 8q − 2q − 4q = 2q (3q is outside). All four fluxes are $\frac{2q}{\epsilon_0}$ .

Topics:Electric flux and Gauss's law

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 14