The electric field in a region is given by `E=ahati+bhatj`. Hence as and b are constants. Find the net flux passing through a square area of side `I` parallel to y-z plane.

Explanation

To find the net electric flux passing through a square area of side length \( l \) that is parallel to the y-z plane, we can follow these steps:

### Step 1: Understand the Electric Field and Area Vector
The electric field is given by:
\[ \mathbf{E} = a \hat{i} + b \hat{j} \]
Since the square area is parallel to the y-z plane, the area vector \( \mathbf{A} \) will be directed along the x-axis. The magnitude of the area vector is given by the area of the square, which is \( l^2 \), and its direction is along the x-axis:
\[ \mathbf{A} = l^2 \hat{i} \]

### Step 2: Calculate the Electric Flux
The electric flux \( \Phi \) through a surface is given by the dot product of the electric field vector \( \mathbf{E} \) and the area vector \( \mathbf{A} \):
\[ \Phi = \mathbf{E} \cdot \mathbf{A} \]
Substituting the values we have:
\[ \Phi = (a \hat{i} + b \hat{j}) \cdot (l^2 \hat{i}) \]

### Step 3: Perform the Dot Product
Using the properties of the dot product, we can calculate:
\[ \Phi = (a \hat{i} \cdot l^2 \hat{i}) + (b \hat{j} \cdot l^2 \hat{i}) \]
Since \( \hat{j} \cdot \hat{i} = 0 \) (the unit vectors are orthogonal), the second term becomes zero:
\[ \Phi = a l^2 \]

### Final Answer
Thus, the net electric flux passing through the square area is:
\[ \Phi = a l^2 \]