Electric Flux Through Cube with Line Charge

❓ Question

An infinitely long wire has uniform linear charge density:

$$\lambda = 2\times10^{-6}\ \text{C/m}$$

The net flux through a Gaussian cube of side length:

$$\sqrt{3}\text{ m}$$

if the wire passes through any two opposite vertices of the cube, is:

(Neglect edge effects and use:

$$\varepsilon_0=9\times10^9$$

SI units.)

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Gauss’s Law
👉 Electric flux
👉 Charge enclosed by Gaussian surface.

Main idea:

$$\Phi=\frac{Q_{enc}}{\varepsilon_0}$$

So first find charge enclosed inside cube.

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🔷 Step 1 — Length of Wire Inside Cube 💯

Wire passes through opposite vertices of cube.

Hence wire lies along body diagonal.

Body diagonal of cube:

$$d=\sqrt{3}\,a$$

Given:

$$a=\sqrt3\text{ m}$$

So:

$$d=\sqrt3\times\sqrt3$$
$$d=3\text{ m}$$

Thus length of wire inside cube:

$$L=3\text{ m}$$

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🔷 Step 2 — Find Charge Enclosed

Linear charge density:

$$\lambda=2\times10^{-6}\text{ C/m}$$

Charge enclosed:

$$Q=\lambda L$$
$$=(2\times10^{-6})(3)$$
$$=6\times10^{-6}\text{ C}$$

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🔷 Step 3 — Apply Gauss Law

Using:

$$\Phi=\frac{Q}{\varepsilon_0}$$

Also:

$$\frac1{4\pi\varepsilon_0}=9\times10^9$$

Hence:

$$\frac1{\varepsilon_0}=36\pi\times10^9$$

Therefore:

$$\Phi = 6\times10^{-6}\times36\pi\times10^9$$
$$=216\pi\times10^3$$
$$=2.16\pi\times10^5$$

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🔷 Step 4 — Final Answer

$$\boxed{2.16\pi\times10^5}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Side length ko directly wire length maan lena

❌ Face diagonal use kar lena instead of body diagonal

❌ Gauss law me enclosed charge galat lena

Remember:

$$\boxed{ \text{Body diagonal of cube}=\sqrt3\,a }$$

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✅ Final Answer

$$\boxed{2.16\pi\times10^5}$$

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📚 Related Topics

• Potential Energy Change on Circular Path
• Find Dielectric Constant Using Induced Charge Method
• Effective Capacitance with Parallel Dielectrics and Capacitor