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Now let's use PE = EA cos to calculate the electric flux through a disk with radius 0.10 m. The disk is oriented with its axis (the line through the center, perpendicular to the disk's surface) at an angle of 30° to a uniform electric field E with magnitude 2.0 × 10³ N/C. (a) What is the total electric flux

Now let's use PE = EA cos to calculate the electric flux through a disk with radius 0.10 m. The disk is oriented with its axis (the line through the center, perpendicular to the disk's surface) at an angle of 30° to a uniform electric field E with magnitude 2.0 × 10³ N/C. (a) What is the total electric flux

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Now let's use E = EA cos o to calculate the electric flux through a disk with radius 0.10 m. The disk is oriented with its axis (the line through the center, perpendicular to the disk's surface) at an angle of 30° to a uniform electric field E with magnitude 2.0 x 10³ N/C. (a) What is the total electric flux through the disk? (b) What is the total flux through the disk if it is turned so that its plane is parallel to E? (c) What is the total flux through the disk if it is turned so that its axis (marked by the dashed line perpendicular to the disk in the figure) is parallel to E? 0.10 m SOLUTION SET UP AND SOLVE Part (a): The area is A = T(0.10 m)² = 0.0314 m². From E = EA cos , ΦΕ = √30 ▼ 30° (2.0 x 10³ N/C) (0.0314 m²) (cos 30°) 54 N·m²/C Part (b): The axis of the disk is now perpendicular to E, so = 90°, cos = 0, and PE = 0. = Part (c): The axis of the disk is parallel to E, so = 0, cos = 1, and, from EEA cos , ΦΕ (2.0× 10³ N/C) (0.0314 m²)(1) = 63 N m²/C Part A Practice Problem: E REFLECT The flux through the disk is greatest when its axis is parallel to E, and it is zero when E lies in the plane of the disk. That is, it is greatest when the most electric field lines pass through the disk, and it is zero when no lines pass through it. CE° with
Transcribed Image Text:Now let's use E = EA cos o to calculate the electric flux through a disk with radius 0.10 m. The disk is oriented with its axis (the line through the center, perpendicular to the disk's surface) at an angle of 30° to a uniform electric field E with magnitude 2.0 x 10³ N/C. (a) What is the total electric flux through the disk? (b) What is the total flux through the disk if it is turned so that its plane is parallel to E? (c) What is the total flux through the disk if it is turned so that its axis (marked by the dashed line perpendicular to the disk in the figure) is parallel to E? 0.10 m SOLUTION SET UP AND SOLVE Part (a): The area is A = T(0.10 m)² = 0.0314 m². From E = EA cos , ΦΕ = √30 ▼ 30° (2.0 x 10³ N/C) (0.0314 m²) (cos 30°) 54 N·m²/C Part (b): The axis of the disk is now perpendicular to E, so = 90°, cos = 0, and PE = 0. = Part (c): The axis of the disk is parallel to E, so = 0, cos = 1, and, from EEA cos , ΦΕ (2.0× 10³ N/C) (0.0314 m²)(1) = 63 N m²/C Part A Practice Problem: E REFLECT The flux through the disk is greatest when its axis is parallel to E, and it is zero when E lies in the plane of the disk. That is, it is greatest when the most electric field lines pass through the disk, and it is zero when no lines pass through it. CE° with
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