JEE Physics 2025 Question with Solution

Given: $\vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \, \text{N/C}$ and electric flux $\Phi = 6.0 \, \text{N m}^2\text{C}^{-1}$.

Find: The area of the rectangular surface in $\text{cm}^2$.

The surface is parallel to the x-z plane, so the area vector is along the y-axis:

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

Only the y-component of the electric field contributes to the flux. Therefore,

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

Here,

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

So,

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

Hence,

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

Now convert to $\text{cm}^2$:

$$A = 1.5 \times 10^{-3} \times 10^4 = 15 \, \text{cm}^2$$

Therefore, the area of the surface is $15 \, \text{cm}^2$.

Given: $\vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \, \text{N/C}$ and $\Phi = 6.0 \, \text{N m}^2\text{C}^{-1}$.

Find: The area $A$ of the surface.

For a surface parallel to the x-z plane, the normal is along $\hat{j}$, so

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

Using electric flux,

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

Substitute the vectors:

$$6.0 = \left[(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3\right] \cdot (A\hat{j})$$

Expand the dot product:

$$6.0 = (2 \times 10^3 \, \hat{i} \cdot A\hat{j}) + (4 \times 10^3 \, \hat{j} \cdot A\hat{j}) + (6 \times 10^3 \, \hat{k} \cdot A\hat{j})$$

Using $\hat{i} \cdot \hat{j} = 0$, $\hat{j} \cdot \hat{j} = 1$, and $\hat{k} \cdot \hat{j} = 0$,

$$6.0 = 0 + 4 \times 10^3 A + 0$$

Thus,

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

Now,

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

Therefore,

$$A = 1.5 \times 10^{-3} \times 10^4 = 15 \, \text{cm}^2$$

Therefore, the required numerical value is 15.