Electric Flux Trick Question Explained in 1 Minute ⚡ | JEE Main Physics

❓ 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 Solution

We’ll use the electric flux formula:

$$\mathrm{\Phi }=\vec{E}\cdot \vec{A}=EA\cos \theta$$

Where:

• $\Phi$ = Electric flux

• $\vec{E}$ = Electric field

• $\vec{A}$ = Area vector (perpendicular to the surface)

• $\theta$ = angle between $\vec{E}$ and $\vec{A}$

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🔹 Step 1 — Determine direction of the area vector

The surface is parallel to the x–z plane,
⇒ its normal vector is along the y-axis (positive or negative ĵ).

Thus,

$$\vec{A}=A\hat{j}$$

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🔹 Step 2 — Find component of electric field perpendicular to surface

Electric field:

$$\vec{E}=(2\widehat{i}+4\widehat{j}+6\widehat{k})\times {10}^3$$

Only the ĵ-component (i.e., $E_y$) contributes to the flux through the x–z surface.

So,

$$E_y=4\times {10}^3\text{N/C}$$

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🔹 Step 3 — Use electric flux formula

$$\mathrm{\Phi }=E_y\times A$$

Given,

$$\mathrm{\Phi }=6.0{\text{N}\text{<span style="color: black;"></span>}\text{m}}^2{\text{C}}^{-1}$$

Substitute:

$$6.0=4\times {10}^3\times A$$

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🔹 Step 4 — Solve for area (A)

$$A = \frac{6.0}{4 \times 10^3}$$ $$A=1.5\times {10}^{-3}{\text{m}}^2$$

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🧮 Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{A = 1.5 \times 10^{-3} \, \text{m}^2}$$​