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Electrostatics question

2021 · 25 Feb · Shift 2 · Q57
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  5. /2021 · 25 Feb · Shift 2 · Q57

Electrostatics question

2021 · 25 Feb · Shift 2 · Q57

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E→\overrightarrow EE through the shaded area is : JEE Main 2021 (Online) 25th February Evening Shift Physics - Electrostatics Question 156 English
  1. A
    q24ε0{q \over {24{\varepsilon _0}}}24ε0​q​
  2. B
    q48ε0{q \over {48{\varepsilon _0}}}48ε0​q​
  3. C
    q4ε0{q \over {4{\varepsilon _0}}}4ε0​q​
  4. D
    q8ε0{q \over {8{\varepsilon _0}}}8ε0​q​
View written solutionFree

Correct answer: A

  1. Use Gauss's law idea with symmetry

A charge qqq is placed at a corner of a cube. To apply symmetry conveniently, imagine 8 identical cubes arranged around that corner so that the charge comes at the center of a bigger cube.

Then, by Gauss's law, the total electric flux through the surface of the bigger cube is

Φbig cube=qε0.\Phi_{\text{big cube}} = \frac{q}{\varepsilon_0}.Φbig cube​=ε0​q​.
  1. Flux through one face of the big cube

Since the charge is at the center of the big cube, symmetry gives equal flux through all 6 faces:

Φone face of big cube=q6ε0.\Phi_{\text{one face of big cube}} = \frac{q}{6\varepsilon_0}.Φone face of big cube​=6ε0​q​.
  1. Relation to the shaded area

Each face of the big cube is made up of 4 small square faces (from the 8 small cubes). So flux through one such small square area is

Φshaded square=14⋅q6ε0=q24ε0.\Phi_{\text{shaded square}} = \frac{1}{4}\cdot \frac{q}{6\varepsilon_0} = \frac{q}{24\varepsilon_0}.Φshaded square​=41​⋅6ε0​q​=24ε0​q​.
  1. Final answer

Thus the flux through the shaded area is

q24ε0.\boxed{\frac{q}{24\varepsilon_0}}.24ε0​q​​.

So, the correct option is A.

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