A uniform electric field is given as $\vec{E} = 4 \times 10^4\,\hat{j}\ \text{N/C}$. The flux of this field through a square of side $10\ \text{cm}$ whose plane is parallel to the Y-Z plane is

The electric field is given as $\vec{E} = 4 \times 10^4\,\hat{j}\ \text{N/C}$.

The area of the square is $A = (10\ \text{cm})^2 = 0.01\ \text{m}^2$.

Since the plane of the square is parallel to the Y-Z plane, its area vector is perpendicular to the Y-Z plane, which means it is directed along the X-axis.

$\vec{A} = 0.01\,\hat{i}\ \text{m}^2$

The electric flux $\Phi$ through the square is given by the dot product of the electric field and the area vector:

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

$\Phi = (4 \times 10^4\,\hat{j}) \cdot (0.01\,\hat{i})$

Since $\hat{j} \cdot \hat{i} = 0$, we get:

$\Phi = 0$

Answer: zero