Find the electric flux through the closed surface whose cross-section is shown below.
According to Gauss' law the net electric Flux linked with closed surface is always equals to the 1/(epsilon not) times the charge enclosed by the body.
In given figure all the charges are out side the surface, so charge enclosed is zero. Hence electric flux through the closed surface is zero.
Find the electric flux through the closed surface whose cross-section is shown below. Select the correct...
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
Find the net electric flux through the spherical closed surface shown in the figure below. The two charges on the right are inside the spherical surface. (Take 91 = +2.08 nC, 92 = +1.10 nC, and 93 = -2.73 nC.) N·m2/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...
If the electric flux through a closed surface is determined to be 2.30 N⋅m2/C , how much charge is enclosed by the surface?
Constants The three small spheres shown in the figure (Figure 1) charges q1 4.40 nC,92 --7.60 nC, and q3 2.25 n Figure 1 of 1 S3 042 Surface What it encloses S, S2 42 S, $ i and Part A Find the net electric flux through the closed surface S1 shown in cross section in the figure. N m2/C Submit Request Answer Part B Find the net electric flux through the closed surface shown in cross section in the figu...
4) 18. Find the net electric flux through (a) the closed spherical surface in a uniform electric field shown, and (b) the closed cylindrical surface shown. (c) What can you conclude about the charges, if any, inside the cylindrical surface? 2R
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
12 points OSUniPhys1 23 2 WA014 what is the net electric flux through a closed surface that encloses the following four charges: q1- 2.50 pC, o:" +7.65 μC q3--22.51 C ande.-+45.0c, N m2/c Additional Materials 華eBook
Part A If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero. True False Submit Request Answer
56. A surface completely surrounds a +2.0 × 10-6 C charge. Find the electric flux through this surface when the surface is (a) a sphere with a radius of 0.50 m, (b) a sphere with a radius of 0.25 m, and (c) a cube with edges that are 0.25 m long. 57. A circular surface with a radius of 0.057 m is exposed to a uniforrm external electric field of magnitude 1.44 × 104 N/C. The magnitude of the electric...