Find the electric flux through the closed surface whose cross-sections are shown below. -2.0 x 10-6 C a
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Q: surface, as sha Calculate the electric flux through the surface.
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Q: flat surface of area 3.10 m² is rotated in a uniform electric field of magnitude E = 6.65 x 105 N/C.…
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- Use Gauss's law to find the electric field at the field point in the following case. The distance between the field point and the surface of the conductor is d. A semi-infinitely large conductor with surface charge density o. Field point d х ConductorIn the figure shown, the magnitude of the uniform electric field is ° E = 70 kN/C . Calculate the electric flux through: A. The vertical rectangular surface B. The slanted surface C. The entire surface of the box 320 cm 70Part A A closed surface encloses a net charge of 3.10 μC . What is the net electric flux through the surface? Express your answer in newton-meters squared per coulomb. %0 ΑΣΦ Submit Request Answer Part B Submit ? If the electric flux through a closed surface is determined to be 2.20 Nm²/C, how much charge is enclosed by the surface? Express your answer in coulombs. IVE ΑΣΦ Request Answer N.m²/C ? C Pearson
- A point charge q is situated in a "tent" with a cross section of an equilateral triangle of side length h. The length of the tent is I where I ≫ h. The charge is situated in the middle of the base face tent (as shown in the figure). L X A Find the electric flux through one of the side faces of the tent. Select one: ○ a. O b. ○ c. ΦΕ За 4Є0 ΦΕ 200 -- ΦΕ 4Є0 O P ○ e. ΦΕ = 3q 8€0 = 0 ΦΕ O f. 5q ΦΕ 850Please solve the given questionA long silver rod of radius 5 cm has a charge of -4 µC/cm on its surface. (a) Find the electric field (in N/C in the f direction) at a point 15 cm from the center of the rod (an outside point). X N/C î