A suface has the area vector `vecA(2hati + 3hatj)` what is the flux erof a uniform electric though the area if the field is a`vecE=4hatiN//c and vecE = 4 hatkN//C` ?

Answer

To find the electric flux through a surface given the area vector and the electric field, we can use the formula for electric flux:

\[ \Phi = \vec{E} \cdot \vec{A} \]

where \(\Phi\) is the electric flux, \(\vec{E}\) is the electric field vector, and \(\vec{A}\) is the area vector.

### Step 1: Identify the Area Vector
The area vector given in the problem is:

\[ \vec{A} = 2\hat{i} + 3\hat{j} \]

### Step 2: Calculate the Flux for the First Electric Field
The first electric field vector is:

\[ \vec{E_1} = 4\hat{i} \, \text{N/C} \]

Now, we calculate the flux using the dot product:

\[ \Phi_1 = \vec{E_1} \cdot \vec{A} = (4\hat{i}) \cdot (2\hat{i} + 3\hat{j}) \]

Calculating the dot product:

\[ \Phi_1 = 4 \cdot 2 + 4 \cdot 0 = 8 + 0 = 8 \, \text{N m}^2/\text{C} \]

### Step 3: Calculate the Flux for the Second Electric Field
The second electric field vector is:

\[ \vec{E_2} = 4\hat{k} \, \text{N/C} \]

Now, we calculate the flux using the dot product again:

\[ \Phi_2 = \vec{E_2} \cdot \vec{A} = (4\hat{k}) \cdot (2\hat{i} + 3\hat{j}) \]

Calculating the dot product:

\[ \Phi_2 = 4 \cdot 0 + 4 \cdot 0 = 0 + 0 = 0 \, \text{N m}^2/\text{C} \]

### Final Results
- The electric flux for the first electric field is \(\Phi_1 = 8 \, \text{N m}^2/\text{C}\).
- The electric flux for the second electric field is \(\Phi_2 = 0 \, \text{N m}^2/\text{C}\).