An infinitely long wire has uniform linear charge density λ = 2 nC/m. The net flux through a Gaussian cube of side length √3 cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm²C⁻¹, where x is:

⚡ Gauss’s Law – Flux Through a Cube | JEE Physics | Doubtify

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📌 Question:

An infinitely long wire has uniform linear charge density λ = 2 nC/m. The net flux through a Gaussian cube of side length √3 cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm²C⁻¹, where x is:

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🖼️ Question Image:

🧠 Solution Image:

🧮 Concept Explained:

🧾 What is Gauss’s Law?

Gauss’s Law states:

$$\phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$

Where:

• $\phi$ = net electric flux

• $q_{\text{enclosed}}$ = charge enclosed by the surface

• $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2$ (permittivity of free space)

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🚩 What’s Given:

• $\lambda =2\text{nC/m}=2\times {10}^{-9}\text{C/m}$

• Cube side = $\sqrt{3} \, \text{cm} = \sqrt{3} \times 10^{-2} \, \text{m}$
  

If the wire passes from one corner of the cube to the opposite corner (body diagonal), the length of the wire inside the cube is equal to the body diagonal, which is:

$$l=\sqrt{3}\cdot (\text{side})=\sqrt{3}\cdot \sqrt{3}\times {10}^{-2}=3\times {10}^{-2}\text{m<span class="base"></span>}$$

🧮 Total Charge Enclosed:

$$q = \lambda \cdot l = 2 \times 10^{-9} \cdot 3 \times 10^{-2} = 6 \times 10^{-11} \, \text{C}$$

🔍 Apply Gauss’s Law:

$$\phi = \frac{q}{\varepsilon_0} = \frac{6 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 6.78 \, \text{Nm}^2/\text{C}$$

✅ Final Answer:

$$\boxed{{\displaystyle x=\mathrm{6.78}}}$$

                      

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🎯 Why This Question is Important:

• It’s a classic Gauss’s Law application with non-standard geometry (wire through cube’s diagonal).

• Frequently appears in JEE Advanced and JEE Mains to test concept + visualization.

• Sharpens your understanding of charge distribution and flux calculation using symmetry.

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🧠 Pro Tip:

Even if the surface is complex, only the charge enclosed matters, not the path or shape of the surface — this is the beauty of Gauss’s Law!

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