22.1 A flat sheet of paper of area \( 0.250 \mathrm{~m}^{2} \) is oriented so that the normal to the sheet is at an angle of \( 60^{\circ} \) to a uniform electric field of magnitude \( 14 \mathrm{~N} / \mathrm{C} \). (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \( \phi \) between the nor- mal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

Let's tackle each part of the problem step by step. --- ### **Given:** - **Area of the sheet,** \( A = 0.250 \, \text{m}^2 \) - **Electric field magnitude,** \( E = 14 \, \text{N/C} \) - **Angle between the normal to the sheet and the electric field,** \( \theta = 60^\circ \) --- ### **Part (a): Calculate the Electric Flux Through the Sheet** **Electric Flux (\( \Phi_E \))** through a surface is given by the formula: \[ \Phi_E = E \cdot A \cdot \cos(\theta) \] **Where:** - \( E \) = Electric field magnitude - \( A \) = Area through which the field lines pass - \( \theta \) = Angle between the electric field and the normal (perpendicular) to the surface **Plugging in the given values:** \[ \Phi_E = 14 \, \text{N/C} \times 0.250 \, \text{m}^2 \times \cos(60^\circ) \] \[ \cos(60^\circ) = 0.5 \] \[ \Phi_E = 14 \times 0.250 \times 0.5 = 14 \times 0.125 = 1.75 \, \text{N·m}^2/\text{C} \] **Answer:** \[ \Phi_E = 1.75 \, \text{N·m}^2/\text{C} \] --- ### **Part (b): Does the Electric Flux Depend on the Shape of the Sheet?** **Answer:** No, the electric flux through the sheet does not depend on its shape. **Explanation:** Electric flux is determined by the component of the electric field passing perpendicular to the surface and the area of that perpendicular projection. As long as the area \( A \) and the angle \( \theta \) between the electric field and the normal to the surface are the same, the shape of the sheet does not affect the flux. Whether the sheet is square, circular, or of any other shape, as long as its area and orientation relative to the electric field remain constant, the electric flux will be identical. --- ### **Part (c): Angle \( \phi \) for Maximum and Minimum Electric Flux** **Objective:** Determine the angles \( \phi \) between the normal to the sheet and the electric field at which the magnitude of the electric flux is: 1. **Largest** 2. **Smallest** **Understanding the Relationship:** Electric flux is given by: \[ \Phi_E = E \cdot A \cdot \cos(\phi) \] The magnitude of the flux depends on the value of \( \cos(\phi) \), which varies between -1 and 1 as \( \phi \) varies from \( 0^\circ \) to \( 180^\circ \). 1. **Maximum Electric Flux:** - Occurs when \( \cos(\phi) \) is at its maximum value. - \( \cos(\phi) \) is maximum when \( \phi = 0^\circ \). - **Interpretation:** The electric field is parallel to the normal (perpendicular) to the sheet, meaning all electric field lines pass directly through the sheet. 2. **Minimum Electric Flux:** - Occurs when \( \cos(\phi) \) is at its minimum value. - \( \cos(\phi) \) is minimum when \( \phi = 90^\circ \). - **Interpretation:** The electric field is parallel to the surface of the sheet. In this orientation, no electric field lines pass through the sheet, resulting in zero flux. **Answers:** 1. **Largest Flux:** \( \phi = 0^\circ \) 2. **Smallest Flux:** \( \phi = 90^\circ \) **Summary:** - **Maximum Flux** occurs when the electric field is perpendicular to the sheet (\( \phi = 0^\circ \)). - **Minimum Flux** occurs when the electric field is parallel to the sheet (\( \phi = 90^\circ \)). --- ### **Final Summary:** - **(a)** The electric flux through the sheet is **1.75 N·m²/C**. - **(b)** The flux does **not** depend on the shape of the sheet, only on its area and orientation. - **(c)** The flux is **largest** when \( \phi = 0^\circ \) and **smallest** when \( \phi = 90^\circ \).