Question

The electric field of intensity \( \vec{E} = (3 \hat{i} + 4 \hat{j}) \, \text{N/C} \) passes through the plane of area \( \vec{A} = (10 \hat{i}) \, \text{m}^2 \). Find the electric flux.

Hint:

The electric flux is calculated as the dot product of the electric field vector and the area vector.

Answer 1

Step 1: Electric flux formula.
The electric flux \( \Phi_E \) is given by the dot product of the electric field \( \vec{E} \) and the area vector \( \vec{A} \): \[ \Phi_E = \vec{E} \cdot \vec{A} \]
Step 2: Substituting the values.
We are given: \[ \vec{E} = 3 \hat{i} + 4 \hat{j} \, \text{N/C}, \quad \vec{A} = 10 \hat{i} \, \text{m}^2 \] Now, calculate the dot product: \[ \Phi_E = (3 \hat{i} + 4 \hat{j}) \cdot (10 \hat{i}) \] Since the \( \hat{j} \) component of the area vector is 0, we only consider the \( \hat{i} \) component: \[ \Phi_E = 3 \times 10 = 30 \, \text{Nm}^2/\text{C} \]
Step 3: Conclusion.
Therefore, the electric flux is \( 30 \, \text{Nm}^2/\text{C} \).