A surface element `vec(dS) = 5 hat(i)` is placed in an electric field `vec(E) = 4 hat(i) + 4 hat(j) + 4 hat(k)`. What is the electric flux emanating from the surface ?

Options

A. `20 units`
B. `25 units`
C. `10 units`
D. `15 units`

Explanation

To find the electric flux emanating from the surface, we will use the formula for electric flux, which is given by:

\[ \Phi_E = \vec{E} \cdot \vec{dS} \]

where \(\Phi_E\) is the electric flux, \(\vec{E}\) is the electric field vector, and \(\vec{dS}\) is the surface element vector.

### Step-by-Step Solution:

1. **Identify the Electric Field Vector**:
The electric field vector is given as:
\[ \vec{E} = 4 \hat{i} + 4 \hat{j} + 4 \hat{k} \]

2. **Identify the Surface Element Vector**:
The surface element vector is given as:
\[ \vec{dS} = 5 \hat{i} \]

3. **Calculate the Dot Product**:
The electric flux is calculated using the dot product of \(\vec{E}\) and \(\vec{dS}\):
\[ \Phi_E = \vec{E} \cdot \vec{dS} = (4 \hat{i} + 4 \hat{j} + 4 \hat{k}) \cdot (5 \hat{i}) \]

To compute the dot product, we use the property of dot products:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]

Here, we have:
- \(A_x = 4\), \(A_y = 4\), \(A_z = 4\)
- \(B_x = 5\), \(B_y = 0\), \(B_z = 0\)

Thus, the dot product becomes:
\[ \Phi_E = (4)(5) + (4)(0) + (4)(0) = 20 + 0 + 0 = 20 \]

4. **Conclusion**:
Therefore, the electric flux emanating from the surface is:
\[ \Phi_E = 20 \, \text{units} \]