Guided course 03:01Flux Through Spherical Shell due to Point ChargePatrick Ford4114views75rank6comments
06:46Physics - E&M: Ch 36.1 The Electric Field Understood (2 of 17) What is Electric Flux?Michel van Biezen332views
12:52Electric Flux, Gauss's Law & Electric Fields, Through a Cube, Sphere, & Disk, Physics ProblemsThe Organic Chemistry Tutor630views
Multiple ChoiceThe electric flux through each surface of a cube is given below. Which surfaces of the cube does the electric field run parallel to? Φ1 = 100 Nm2 /C Φ4 = 0 Nm2 /CΦ2 = 20 Nm2 /C Φ5 = −40 Nm2 /CΦ3 = 0 Nm2 /C Φ6 = −80 Nm2 /C1615views15rank3comments
Multiple ChoiceWhat is the total flux through the two surfaces depicted in the following figure? Note that surface 1 has an area of 50 cm2 and surface 2 has an area of 100 cm2 , and E = 500 N/C.1068views21rank5comments
Multiple ChoiceA uniform electric field points in the positive x direction and has a magnitude of 40N/C. What is the total flux through a rectangle with height 20cm and width 45cm? The rectangle lies in the y-z plane.417views
Multiple ChoiceA uniform electric field points in the positive x direction and has a magnitude of 40N/C. What is the net flux through a cube with sides 4.0cm? The cube has one corner on the origin and the positive x, y, and z axes form three of its edges.864views1comments
Multiple ChoiceA cube has sides of 4.0cm. Inside the cube is a 20nC and a −30nC charge. What is the net flux through the cube?320views
Multiple ChoiceA 4.0cm×5.0cm rectangle lies in the x-y plane; let the normal vector point in the z direction. What is the flux through the rectangle of a uniform electric field E⇀=(−7.8×105j^+2.3×105k^)N/C338views
Textbook QuestionA hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.1136views
Textbook QuestionA flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (c) For what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.391views1rank
Textbook QuestionA flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?1365views2rank
Textbook QuestionA 2.0 cm×3.0 cm rectangle lies in the 𝓍𝒵-plane with unit vector nˆ pointing in the +y -direction. What is the electric flux through the rectangle if the electric field is (b) E (→ above E) = (4000 î−2000 kˆ) N/C?357views
Textbook QuestionA 12 cm×12 cm rectangle lies in the first quadrant of the xy-plane with one corner at the origin. Unit vector nˆ points in the +𝒵 -direction. What is the electric flux through the rectangle if the electric field is E (→ above E) = (2000 m¯¹) x kˆ N/C? Hint: Divide the rectangle into narrow strips of width .546views1comments
Textbook QuestionA 10 nC charge is at the center of a 2.0 m x 2.0 m x 2.0 m cube. What is the electric flux through the top surface of the cube?232views
Textbook QuestionThe electric flux through the surface shown in FIGURE EX24.10 is 25 N m²/C . What is the electric field strength?309views
Textbook QuestionA spherically symmetric charge distribution produces the electric field E (→ above E) = (5000r²) rˆ N/C, where r is in m. (c) How much charge is inside this 40-cm-diameter spherical surface?120views
Textbook QuestionA spherically symmetric charge distribution produces the electric field E (→ above E) = (5000r²) rˆ N/C, where r is in m. (b) What is the electric flux through a 40-cm-diameter spherical surface that is concentric with the charge distribution?190views
Textbook QuestionFIGURE P31.38 shows the electric field inside a cylinder of radius R=3.0 mm. The field strength is increasing with time as E=1.0×10^8t^ 2 V/m, where t is in s. The electric field outside the cylinder is always zero, and the field inside the cylinder was zero for t<0. a. Find an expression for the electric flux Φₑ through the entire cylinder as a function of time.300views
Textbook QuestionAll examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net Φₑ = Qᵢₙ / ϵ₀ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. (b) Now integrate dΦ to find the total flux through this face.167views
Textbook Question(I) A uniform electric field of magnitude 6.4 x 10² N/C passes through a circle of radius 13 cm. What is the electric flux through the circle when its face is (b) at 45° to the field lines, and103views
05:43
