### Correct Option: B. \( \Phi_E = E \cdot A \cdot \cos(\theta) \) ### Detailed Explanation: **Understanding Electric Flux:** Electric flux (\( \Phi_E \)) is a measure of the electric field (\( \vec{E} \)) passing through a given surface area (\( \vec{A} \)). It quantifies how much electric field lines penetrate a surface. The concept is crucial in understanding Gauss's Law and the behavior of electric fields in various configurations. **Formula for Electric Flux:** The electric flux through a surface is defined mathematically as: \[ \Phi_E = \vec{E} \cdot \vec{A} \] This dot product can also be expressed in terms of the angle (\( \theta \)) between the electric field vector and the area vector: \[ \Phi_E = E \cdot A \cdot \cos(\theta) \] Where: - \( E \) is the magnitude of the electric field, - \( A \) is the magnitude of the area vector, - \( \theta \) is the angle between the electric field vector and the area vector. **Why Option B is Correct:** In the scenario described, the surface area vector (\( \vec{A} \)) is perpendicular to the electric field (\( \vec{E} \)). When two vectors are perpendicular, the angle \( \theta \) between them is \( 0^\circ \). The cosine of \( 0^\circ \) is \( 1 \), so the formula simplifies to: \[ \Phi_E = E \cdot A \cdot \cos(0^\circ) = E \cdot A \cdot 1 = E \cdot A \] Thus, option B correctly represents the general case of electric flux, which can be simplified when the vectors are perpendicular. ### Why Other Options are Incorrect: - **Option A: \( \Phi_E = E \cdot A \)** - While this expression is true when the electric field is perpendicular to the surface, it does not account for the general case where the angle \( \theta \) could be anything other than \( 0^\circ \). Therefore, it is a special case rather than a general definition. - **Option C: \( \Phi_E = E \cdot A \cdot \sin(\theta) \)** - This option is incorrect because the sine function is not used in the context of electric flux. The sine function relates to the component of a vector that is perpendicular to another vector, which is not relevant here. Electric flux is concerned with the component of the electric field that is parallel to the area vector, which is captured by the cosine function. - **Option D: \( \Phi_E = 0 \)** - This option would only be true if the electric field is parallel to the surface (i.e., \( \theta = 90^\circ \)). In that case, no electric field lines pass through the surface, resulting in zero flux. However, since the problem states that the area vector is perpendicular to the electric field, this option is not applicable. ### Summary of Key Points: - Electric flux (\( \Phi_E \)) is defined as \( \Phi_E = E \cdot A \cdot \cos(\theta) \). - When the area vector is perpendicular to the electric field, \( \theta = 0^\circ \), leading to \( \Phi_E = E \cdot A \). - The cosine function is used to determine the effective component of the electric field that passes through the surface. - Understanding the relationship between the electric field, area vector, and angle is crucial for applying Gauss's Law and solving problems in electrostatics.