(a) electric flux = E.A
= (-4j) . (1.5i - 7.3k)
= 0
(b) E.A
= (-4j).(1.5i - 7.3j)
= (-4)(-7.3) = 29.2 Nm^2/C
(a) The electric field in a certain region is E (4.0)) N/C. Determine the electric flux...
(a) The electric field in a certain region is E = (-4.0k) N/C. Determine the electric flux due to this field through an area represented by the vector A = (3.41 – 5.4K) m2. N- m²/c (b) Determine the flux due to the same electric field when the surface orientation has changed such that the area is now represented by the vector A = (3.4 - 5.49) m2. Nm2/C
(a) The electric field in a certain region is E = (-4,0 ) NC Determine the electric flux due to this field through an area represented by the vector A = 4.8-5.0k) m2 (b) Determine the flux due to the same electric field when the surface orientation has changed such that the area is now represented by the vector A = (4.8↑-5.0j) m2. N m2/c
An electric field of intensity 2.50 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane. 612.5 、N-m2/C (b) The plane is parallel to the xy-plane. (c) The plane contains the y-axis, and its normal makes an angle of 32.5° with the x-axis. 737.96X Use the area, the electric field, and the relative orientation of...
PRACTICE: Find the flux Og of the electric field E = (24î + 30ĵ+ 168) N/C through a 2.0 mp portion of the xy plane (k is the normal). [A] 32 N·m? [B] 34 N • m2 [C] 42 N·m? [D] 48 N•m? [E] 60 N·m2
Consider the uniform electric field } = (2.0ġ + 7. Ok) 103 N/C. (a) What is its electric flux (in N.m2/C) through a circular area of radius 1.6 m that lies in the yz-plane? (Enter the magnitude.) 0 N·m2/C (b) What is its electric flux (in N·m-/C) through a circular area of radius 1.6 m that lies 45.0° above the xy-plane? (Assume that the unit normal vector is n = (î+k).) 51200 X N·m2/C
An electric flux of 157 N·m2/C passes through a flat horizontal surface that has an area of 0.72 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15° above the horizontal? Submit Answer Tries 0/5
Consider the uniform electric field E =(4.0 j^+3.0 k^)×103^ N/C. What is its electric flux through a circular area of radius 1.34 m that lies in the xy-plane? Answer in Nxm^2/C
A flat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E-5.90 x 105 N/C. (a) Determine the electric flux through this area when the electric field is perpendicular to the surface. The correct answer is not zero. N m2/C (b) Determine the electric flux through this area when the electric field is parallel to the surface. 15 34 Your response differs significantly from the correct answer. Rework your solution from the beginning and check...
Find the electric flux through the closed surface whose cross-section is shown below. -10×104c 4.0 × 10-6 C Select the correct answer ○ 1.89 × 105 N . m2/ Your Answer O 9.04 × 104 N·m2/C ○ 4.25 × 105 N·m2/C 2.22 × 106 N·m2/C
The electric flux from a uniform field through a flat surface is 7.80 N·m2/C when the field lines make an angle of 13.0º with the plane of the surface. What is the electric flux when the angle is changed to 77.0º? (Enter units as Nm^2/C or Vm