Guided course 03:01Flux Through Spherical Shell due to Point ChargePatrick Ford3003views64rank6comments
06:46Physics - E&M: Ch 36.1 The Electric Field Understood (2 of 17) What is Electric Flux?Michel van Biezen174views
12:52Electric Flux, Gauss's Law & Electric Fields, Through a Cube, Sphere, & Disk, Physics ProblemsThe Organic Chemistry Tutor395views
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 /C1234views12rank3commentsHas a video solution.
Multiple ChoiceWhere does the normal vector point for a spherical shell?891views10rank2commentsHas a video solution.
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.825views18rank3commentsHas a video solution.
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.240views
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.594views
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?166views
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/C180views
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.683viewsHas a video solution.
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.197views1rankHas a video solution.
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?1006views1rankHas a video solution.
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?143viewsHas a video solution.
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 .169views1commentsHas a video solution.
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?61viewsHas a video solution.
Textbook QuestionThe electric flux through the surface shown in FIGURE EX24.10 is 25 N m²/C . What is the electric field strength?29viewsHas a video solution.
Textbook QuestionFind the electric fluxes ΦA to ΦE through surfaces A to E in FIGURE P24.29.106viewsHas a video solution.
Textbook QuestionCharges q₁ = -4Q and q2 = +2Q are located at 𝓍 = -a and 𝓍 = + a, respectively. What is the net electric flux through a sphere of radius 2a centered (a) at the origin and (b) at 𝓍 = 2a?64viewsHas a video solution.
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.87viewsHas a video solution.