### Correct Option: A. \( \Phi = E \cdot A \cdot \cos(60^\circ) \) ### Detailed Explanation: **Understanding Electric Flux:** Electric flux (\( \Phi \)) through a surface in an electric field is a measure of the number of electric field lines passing through that surface. The formula for electric flux is given by: \[ \Phi = E \cdot A \cdot \cos(\theta) \] where: - \( \Phi \) is the electric flux, - \( E \) is the electric field strength, - \( A \) is the area of the surface, - \( \theta \) is the angle between the electric field lines and the normal (perpendicular) to the surface. **Identifying the Angle:** In this problem, we are given that the angle between the electric field lines and the normal to the surface is \( 60^\circ \). This means that \( \theta = 60^\circ \). **Calculating the Electric Flux:** Using the formula for electric flux, we substitute the values we have: \[ \Phi = E \cdot A \cdot \cos(60^\circ) \] Now, we know that \( \cos(60^\circ) = \frac{1}{2} \). Therefore, we can rewrite the expression for electric flux as: \[ \Phi = E \cdot A \cdot \frac{1}{2} \] This confirms that the expression for electric flux can also be represented as: \[ \Phi = \frac{E \cdot A}{2} \] However, the correct answer in the context of the options provided is option A, which correctly states the relationship using the cosine function. ### Why Other Options Are Incorrect: - **Option B: \( \Phi = E \cdot A \cdot \sin(60^\circ) \)** - This option incorrectly uses the sine function instead of the cosine function. The sine function is not applicable here because we need the component of the electric field that is perpendicular to the surface, which is given by the cosine of the angle. - **Option C: \( \Phi = \frac{E \cdot A}{2} \)** - While this expression is mathematically correct when substituting \( \cos(60^\circ) \), it does not directly reflect the relationship involving the angle. It is a derived form rather than the original expression that includes the angle, which is what the question asks for. - **Option D: \( \Phi = E \cdot A \cdot 60 \)** - This option is incorrect because it suggests multiplying the electric field strength and area by the angle in degrees, which is not a valid operation in the context of electric flux. The angle must be used in a trigonometric function (cosine or sine) to find the correct component of the electric field. ### Common Pitfalls: - Confusing the use of sine and cosine: Remember that electric flux depends on the component of the electric field that is perpendicular to the surface, which is why we use cosine. - Misinterpreting the angle: Ensure you understand whether the angle given is with respect to the normal or the surface itself. - Forgetting to convert angles: If angles are given in degrees, ensure you are using the correct trigonometric values for those angles. ### Revision Summary: - Electric flux is calculated using the formula \( \Phi = E \cdot A \cdot \cos(\theta) \). - The angle \( \theta \) is the angle between the electric field and the normal to the surface. - For \( \theta = 60^\circ \), \( \cos(60^\circ) = \frac{1}{2} \), leading to \( \Phi = E \cdot A \cdot \frac{1}{2} \). - Always use cosine for angles with respect to the normal when calculating electric flux.