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- Compute for the volume charge density at (4.5 m, 100 degrees, 200 degrees) given the electric flux density is D = r2 sin2(0) sin(0) ao C/m2. O -15.4296 C/m3 O 4.1644 C/m3 O -11.9204 C/m3 O The answer cannot be found on the other choices.Find the flux in N.m2.C-1 of a constant electric field E = (5.85x10^3) i + (2.712x10^3) j + (-7.4910x10^3) k N/C, passing through an area defined by the area vector A = (3.633x10^0) i + (1.5050x10^0) j + (-1.9920x10^0) k m2.Find the flux passing through the surface y = 3, -1 < x < 2, 1 < z < 4 given the electric field everywhere in the region is E = 8x ax + 3(y^2)z ay + y^3 az volts per meter. All coordinates are in meters. Assume a positive direction for the surface vector. 57.1631 nanocoulombs 5.3788 nanocoulombs; 223.3473 nanocoulombs; 607.5 nanocoulombs
- The nonuniform charge density of a solid insulating sphere of radius R is given by = cr2 (r R), where c is a positive constant and r is the radial distance from the center of the sphere. For a spherical shell of radius r and thickness dr, the volume element dV = 4r2dr. a. What is the magnitude of the electric field outside the sphere (r R)? b. What is the magnitude of the electric field inside the sphere (r R)?In free space, let D = 8xyz*ax + 6x²z*ãy + 16x?yz'a, pC/m?. a. Find the total electric flux passing through the rectangular surface z = 2, 0An electric field given by E = 2.6 i – 6.1(y² + 6.2) j pierces the Gaussian cube of edge length 19.0 cm and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the net electric flux in Nm?/C through the cube? 3. Gaussian surface (a) -0.0076 (b) -0.0079 (c) -0.0082 (d) -0.0085 (e) None of the aboveQ.1. Given a uniform electric field É = 5 × 10³ î N/C. Find the flux of this field through a square of side 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis ?Two flat conductors are placed with their in- ner faces separated by 16 mm. If the surface charge density on inner face A is 105 pC/m and on inner face B is -105 pC/m², calculate the electric po- tential difference AV VA VB- Use €0 = 8.85419 x 10¬12 c² /Nm². Answer in units of V. %3DAn electric field of (9.242x10^3) i + (2.441x10^3) j + (4.8590x10^3) k N/C passes outwards through an area (7.04x10^0) m² in the xy plane. Calculate the flux in N.m2/C. y AThe flux through a closed surface is Φ = (-9.00x10^3) N.m2/C. Calculate the charge in Coulombs enclosed by the surface.Calculate the absolute value of the electric flux for the following situations (In all case provide your answer in N m2/C): a. A constant electric field of magnitude 300 N/C at a 30 degrees angle with respect to the flat rectangular surface shown in the Figure above. b. A uniform electric field E = (70 i + 90 k) N/C through a 4 cm ×5 cm in the x-y plane. c. A uniform electric field E = (−350 i + 350 j + 350 k) N/C through a disk of radius 3 cm in the x-z plane.1. Figure shows a non-conducting hollow sphere with inner radius a and outer radius b. It has non-uniform positive volume charge density p P. d that is a function of radial distance r from its center p = ar where a is a constant with the units of C/m*. a. Find the electrical flux through the spherical Gaussian surface with radius r for a b) element of a sphere Express your answers in term of a, b, d,r, a, e, and n.SEE MORE QUESTIONS