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Electrostatics question

2017 · 9 Apr · Shift 1 · Q72

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Electrostatics
2017 · 9 Apr · Shift 1 · Q72

Electrostatics question

2017 · 9 Apr · Shift 1 · Q72

JEE MainPhysicsElectrostaticsMCQ+4 / −1

Four closed surfaces and corresponding charge distributions are shown below. JEE Main 2017 (Online) 9th April Morning Slot Physics - Electrostatics Question 210 English

JEE Main 2017 (Online) 9th April Morning Slot Physics - Electrostatics Question 210 English

Let the respective electric fluxes through the surfaces be

{\Phi _1},{\Phi _2},{\Phi _3}

and

{\Phi _4}

. Then :

A
${\Phi _1}$ < ${\Phi _2}$ = ${\Phi _3}$ > ${\Phi _4}$
B
${\Phi _1}$ > ${\Phi _2}$ > ${\Phi _3}$ > ${\Phi _4}$
C
${\Phi _1}$ = ${\Phi _2}$ = ${\Phi _3}$ = ${\Phi _4}$
D
${\Phi _1}$ > ${\Phi _3}$ ; ${\Phi _2}$ < ${\Phi _4}$

View written solutionFree

Correct answer: C

Use Gauss's law

For any closed surface,

\Phi = \oint \vec E \cdot d\vec A = \frac{Q_{\text{enclosed}}}{\varepsilon_0}

So, the electric flux through a closed surface depends only on the net charge enclosed by the surface, and not on:

shape of the surface,
size of the surface,
position of the charge inside the surface,
charge distribution outside the surface.

Key idea for all four figures

In each of the four given closed surfaces, the net enclosed charge is the same.

Hence, by Gauss's law,

\Phi_1 = \frac{Q}{\varepsilon_0},\quad \Phi_2 = \frac{Q}{\varepsilon_0},\quad \Phi_3 = \frac{Q}{\varepsilon_0},\quad \Phi_4 = \frac{Q}{\varepsilon_0}

Therefore,

\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4

Check options

A: Claims inequality between $\Phi_1$ and $\Phi_4$ → false
B: $\Phi_1 > \Phi_2 > \Phi_3 > \Phi_4$ → false
C: $\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4$ → true
D: Mixed inequalities → false

Final answer

The correct option is:

\boxed{\text{C}}

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