Now let's use E = EA cos o to calculate the electric flux through a diskwith radius 0.10 m. The disk is oriented with its axis (the line through thecenter, perpendicular to the disk's surface) at an angle of 30° to a uniformelectric field E with magnitude 2.0 x 10³ N/C. (a) What is the totalelectric fluxthrough the disk?(b) What is thetotal flux throughthe disk if it isturned so that itsplane is parallelto E? (c) What isthe total fluxthrough the disk ifit is turned so thatits axis (markedby the dashedline perpendicularto the disk in thefigure) is parallelto E?0.10 mSOLUTIONSET UP AND SOLVE Part (a): The area isA = 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/CPart (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/CPart A Practice Problem:EREFLECT The flux through the disk is greatest when its axis is parallel toE, and it is zero when E lies in the plane of the disk. That is, it is greatestwhen the most electric field lines pass through the disk, and it is zero whenno lines pass through it.CE° with

The electric field is given by E=2∗103 N/CThe radius = 0.1 mThe area of the disc A=π(0.1)2=0.0314…

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