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Electrostatics question

2025 · 8 Apr · Shift 2 · Q67
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Electrostatics question

2025 · 8 Apr · Shift 2 · Q67

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An infinitely long wire has uniform linear charge density λ=2 nC/m\lambda = 2 \text{ nC/m}λ=2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt{3}3​ cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm2C−1x \text{ Nm}^2\text{C}^{-1}x Nm2C−1, where xxx is: [Neglect any edge effects and use 14πϵ0=9×109\frac{1}{4\pi \epsilon_0} = 9 \times 10^94πϵ0​1​=9×109 SI units]
  1. A
    6.48π6.48 \pi6.48π
  2. B
    0.72π0.72 \pi0.72π
  3. C
    1.44π1.44 \pi1.44π
  4. D
    2.16π2.16 \pi2.16π
View written solutionFree

Correct answer: D

  1. Use Gauss's law

For any closed surface,

Φ=qenclosedε0\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}Φ=ε0​qenclosed​​

So we need the charge enclosed by the cube.

  1. Geometry of the wire inside the cube

The wire passes through two corners of the cube that are maximally displaced, i.e. opposite corners of the cube. Hence the wire lies along the body diagonal of the cube.

If side of cube is

a=3 cma = \sqrt{3}\text{ cm}a=3​ cm

then the body diagonal is

d=a3=3⋅3 cm=3 cmd = a\sqrt{3} = \sqrt{3}\cdot \sqrt{3}\text{ cm} = 3\text{ cm}d=a3​=3​⋅3​ cm=3 cm

So the length of wire inside the cube is

L=3 cm=0.03 mL = 3\text{ cm} = 0.03\text{ m}L=3 cm=0.03 m
  1. Charge enclosed

Given linear charge density,

λ=2 nC/m=2×10−9 C/m\lambda = 2\text{ nC/m} = 2\times 10^{-9}\text{ C/m}λ=2 nC/m=2×10−9 C/m

Therefore enclosed charge is

qenclosed=λL=2×10−9×0.03q_{\text{enclosed}} = \lambda L = 2\times 10^{-9}\times 0.03qenclosed​=λL=2×10−9×0.03 qenclosed=6×10−11 Cq_{\text{enclosed}} = 6\times 10^{-11}\text{ C}qenclosed​=6×10−11 C
  1. Compute flux

Using

Φ=qenclosedε0\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}Φ=ε0​qenclosed​​

and

14πε0=9×109⇒1ε0=4π×9×109=36π×109\frac{1}{4\pi\varepsilon_0} = 9\times 10^9 \Rightarrow \frac{1}{\varepsilon_0} = 4\pi \times 9\times 10^9 = 36\pi \times 10^94πε0​1​=9×109⇒ε0​1​=4π×9×109=36π×109

Hence,

Φ=6×10−11×36π×109\Phi = 6\times 10^{-11}\times 36\pi\times 10^9Φ=6×10−11×36π×109 Φ=216π×10−2\Phi = 216\pi\times 10^{-2}Φ=216π×10−2 Φ=2.16π N m2/C\Phi = 2.16\pi\ \text{N m}^2\text{/C}Φ=2.16π N m2/C
  1. Match with options
Φ=2.16π\Phi = 2.16\piΦ=2.16π

So the correct option is:

D) 2.16π2.16\pi2.16π

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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