In the given figure, there is no charge is placed inside the closed area. so, no flux will pass through the the cross section area.
Hence,
Flux through the closed surface = 0
Find the electric flux through the closed surface whose cross-section is shown below. -10×104c 4.0 ×...
Find the electric flux through the closed surface whose cross-section is shown below. Select the correct answer 4.25 x 105 N m2/C O 9.04 x 104 N m2/C O 1.89 x 105 N m2/C Your Answer O 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
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
Part A.) The electric flux through a spherical surface is 4.0\times 10^4~\text{N}\cdot\text{m}^2/\text{C}4.0×10 4 N⋅m 2 /C. What is the net charge enclosed by the surface? Part B.) The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the ball’s center and has magnitude 9.0*10^2N/C. How much charge is on the surface of the ball? Select the correct answer a.) 2.2*10^{-10} . b.)-1.25*10^{-10} . c.)-1.0* 10^{-9} . d.)-4.5* 10^{-10) ....
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?
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...
(a) The electric field in a certain region is E (4.0)) N/C. Determine the electric flux due to this freld through an area represented by the vector A (1.51-7.k) m 29 How is the flux defined in terms of the electric field vector, and the area vector? Review dot product rules. 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 -(1.51-7.3)...
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
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...
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...