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.

To find the electric flux through the sheet, we can use the equation for electric flux, which is defined as: \[ \Phi_E = E \cdot A \cdot \cos(\theta) \] where \( \Phi_E \) is the electric flux, \( E \) is the magnitude of the electric field, \( A \) is the area of the sheet, and \( \theta \) is the angle between the normal to the surface and the electric field. ### (a) Given: - \( E = 14 \, \text{N/C} \) - \( A = 0.250 \, \text{m}^2 \) - \( \theta = 60^\circ \) We can substitute these values into the equation to find the electric flux: \[ \Phi_E = 14 \, \text{N/C} \times 0.250 \, \text{m}^2 \times \cos(60^\circ) = 14 \times 0.250 \times 0.5 = 1.75 \, \text{N m}^2/\text{C} \] So, the magnitude of the electric flux through the sheet is \( 1.75 \, \text{N m}^2/\text{C} \). ### (b) The answer to part (a) does not depend on the shape of the sheet. Electric flux depends on the area and the angle of orientation with respect to the electric field, not on the specific dimensions of the shape. As long as the area remains constant and the angle with the electric field remains unchanged, the resulting flux will be the same, demonstrating that electric flux is a property solely influenced by area and orientation. ### (c) The magnitude of the flux through the sheet is largest when the angle \( \phi \) between the normal to the sheet and the electric field is \( 0^\circ \), as this maximizes \( \cos(\phi) \) (which is 1). Conversely, the magnitude of the flux is smallest when \( \phi \) is \( 90^\circ \), since \( \cos(90^\circ) = 0\), meaning no electric field lines penetrate the surface. Therefore, the electric flux is maximized with perfect alignment and vanishes completely when perpendicular to the field lines.