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Electrostatics question

2021 · 31 Aug · Shift 2 · Q56
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Electrostatics question

2021 · 31 Aug · Shift 2 · Q56

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Choose the incorrect statement : (1) The electric lines of force entering into a Gaussian surface provide negative flux. (2) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same. (3) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (4) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below
  1. A
    (3) and (4) only
  2. B
    (2) and (4) only
  3. C
    (4) only
  4. D
    (1) and (3) only
View written solutionFree

Correct answer: C

  1. Use the definition of electric flux

Electric flux through a surface is

Φ=∫E⃗⋅dA⃗\Phi = \int \vec E \cdot d\vec AΦ=∫E⋅dA

where dA⃗d\vec AdA is the outward area vector.

  • If field lines leave the closed surface, E⃗⋅dA⃗>0\vec E \cdot d\vec A > 0E⋅dA>0, so flux is positive.
  • If field lines enter the closed surface, E⃗⋅dA⃗<0\vec E \cdot d\vec A < 0E⋅dA<0, so flux is negative.
  1. Check statement (1)

The electric lines of force entering into a Gaussian surface provide negative flux.

This is correct, because when field enters a closed surface, the field is opposite to the outward normal.

So, (1) is correct.

  1. Check statement (2)

A charge qqq is placed at the centre of a cube. The flux through all the faces will be the same.

By Gauss's law,

Φtotal=qε0\Phi_{\text{total}} = \frac{q}{\varepsilon_0}Φtotal​=ε0​q​

Since the charge is at the center of the cube, all 6 faces are symmetric with respect to the charge. Hence flux is equally distributed:

Φeach face=q6ε0\Phi_{\text{each face}} = \frac{q}{6\varepsilon_0}Φeach face​=6ε0​q​

So, (2) is correct.

  1. Check statement (3)

In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero.

By Gauss's law,

Φnet=qenclosedε0\Phi_{\text{net}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}Φnet​=ε0​qenclosed​​

If the closed surface contains no net charge, then

qenclosed=0  ⟹  Φnet=0q_{\text{enclosed}}=0 \implies \Phi_{\text{net}}=0qenclosed​=0⟹Φnet​=0

This is true even in a uniform electric field.

So, (3) is correct.

  1. Check statement (4)

When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux.

If electric field is parallel to the surface, then it is perpendicular to the area vector. Hence,

E⃗⋅dA⃗=EdAcos⁡90∘=0\vec E \cdot d\vec A = EdA\cos 90^\circ = 0E⋅dA=EdAcos90∘=0

Therefore flux is zero, not finite non-zero.

So, (4) is incorrect.

  1. Identify the incorrect statement(s)

Only statement (4) is incorrect.

Therefore the correct option is:

C: (4) only\boxed{\text{C: (4) only}}C: (4) only​
  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They match.

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