Question

11:45イ adpoly.ankabut.ac.ae A flat disk 1.0 m in diameter is oriented so that the plane of the disk make an angle of s radians with a uniform electric field. If the field strength is BONC find the electric flux through the surface. A) 160m Nnm2/C #2500" Nm 2/C O 미 90m Nm2/C Question 3 (1 point) 0-50 +-40. E-2x10 N/C Electric field lines pass through a cube as shown in the diagram above. What is the electric flux through sides A and B? C) 5.1 × 10-7 Nm2C Question 4 (1 point What is the totall flux through the entire cube? #2s3 E cos(45.) Question 5 (1 pont The electric field at point P due to a point charge Q a distance R away ha magnitude E. In order to double the magnitude of the field at P. you could O A reduce the distance to R2 O s double the distance to 2R O ci double the charge to 20 and at the same time reduce the distance to R2 O oi double the charge to 20 O Ee reduce the distance to R

Answer 1

Area (A) = pi d^2/4 = pi times 1^2/4 = pi/4 m^2 plane of disk makes angle pi/6 with electric field so angle b/w area vector ad electric field = pi/2 - pi/6 = pi/3 so phi = E A cos theta therefore 719 times pi/4 cos (pi/3) phi = 90 pi N-m^2/c phi_a = -E. A. cos (90 - theta) = -2 times 10^5 times 4 times 10^-12 times cos 40 = -5.14 times 10^-7 N-m^2/c phi_A = -6.128 times 10^-7 N-m^2/c phi_B = E A. co (90 - phi) = 2 times 10^5 times 4 times 0^-21 cos (50) = 5.14 times 10^-7 N-m^2/c phi_A b = phi_a + phi_B = (- 6.128 + 5.14) times 10^-7 N_m^2 = -0.988 times 10^-7 phi_A B = 9.88 times 10^-8 N-m^2