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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