Electric Flux Through Surface Parallel to xz Plane

❓ Question

The electric field in a region is given by

$$\mathbf{E}=(2\widehat{i}+4\widehat{j}+6\widehat{k})\times {10}^3\text{N/C}\mathrm{.}$$

The flux of the field through a rectangular surface parallel to the x–z plane is 6.0 N·m²·C⁻¹.
Find the area of the surface.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Electric flux
👉 Vector components
👉 Area vector concept.

Main idea:

Electric flux:

$$\boxed{ \Phi=\vec E\cdot\vec A }$$

For a surface parallel to $x-z$ plane:

✔ Area vector is along $y$-axis.

Thus only $E_y$ contributes.

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🔷 Step 1 — Identify Relevant Electric Field Component 💯

Given:

$$\vec E=(2\hat i+4\hat j+6\hat k)\times10^3$$

Surface is parallel to $x-z$ plane.

Hence normal direction:

$$\boxed{ \hat j }$$

Thus effective electric field:

$$E_y=4\times10^3\ \text{N/C}$$

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🔷 Step 2 — Apply Flux Formula

Flux:

$$\Phi=EA$$

Given:

$$\Phi=6$$

So:

$$6=(4\times10^3)A$$
$$A=\frac6{4\times10^3}$$
$$=1.5\times10^{-3}\text{ m}^2$$

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🔷 Step 3 — Convert into ${\text{cm}}^2$

$$1\text{ m}^2=10^4\text{ cm}^2$$

Thus:

$$A=1.5\times10^{-3}\times10^4$$
$$=15\text{ cm}^2$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Full electric field magnitude use kar lena

❌ Surface parallel condition ignore kar dena

Remember:

Surface parallel to $x-z$ plane

$$\Rightarrow \text{Normal along } y\text{-axis}$$

So only $E_y$ matters.

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✅ Final Answer

$$\boxed{{\displaystyle 15{\text{ cm}}^{\mathrm{2}}}}$$

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