Assume the magnitude of the electric field on each face of the cube of edge L = 1.10 m in the figure below is uniform and the directions of the fields on each face are as indicated. (Take E, = 33.0 N/C and E,24.9 N/C.)E220.0 N/C,20.0 N/CE120.0 N/C15.0 N/C(a) Find the net electric flux through the cube.N• m?/c(b) Find the net charge inside the cube.C(c) Could the net charge be a single point charge?O YesNo

(5) (a) ϕnet=ϕ1 +ϕ2 +ϕ3 + ϕ4 +ϕ5 + ϕ6we know that the flux is given by ϕ = E.Awhere E is…

Assume the magnitude or the the figure below is uniform and the directions each face are as indicated. (Take E1 U N/C and E2 E2 20.0 N/C 20.0 N/C E1 20.0 N/C 15.0 N/C (a) Find the net electric flux through the cube. N- m?/c (b) Find the net charge inside the cube. (c) Could the net charge be a single point charge? O Yes O No

Assume the magnitude or the the figure below is uniform and the directions each face are as indicated. (Take E1 U N/C and E2 E2 20.0 N/C 20.0 N/C E1 20.0 N/C 15.0 N/C (a) Find the net electric flux through the cube. N- m?/c (b) Find the net charge inside the cube. (c) Could the net charge be a single point charge? O Yes O No

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Gauss’s Law

Section: Chapter Questions

Problem 24.48AP: Consider a plane surface in a uniform electric field as in Figure P24.48, where d = 15.0 cm and =...

See similar textbooks

Similar questions

Find the electric flux through the plane surface shown in Figure P23.37 if = 60.0, E = 350 N/C, and d = 5.00 cm. The electric field is uniform over the entire area of the surface. Figure P23.37
Consider a plane surface in a uniform electric field as in Figure P24.48, where d = 15.0 cm and = 70.0. If the net flux through the surface is 6.00 N find the magnitude of the electric field. Figure P24.48
Find the electric flux through the surface in Figure 15.28. Assume all charges in the shaded area are inside the surface. (a) (3 C)/0 (b) (3 C)/0 (c) 0 (d) (6C)/0
The figure below shows a uniform electric field of magnitude E = 470 N/C making an angle of ? = 61.0° with a flat surface of area A = 3.10 m2. What is the magnitude of the electric flux through this surface (in N · m2/C)?
The surface shown is a square and has a side length of 11.0 cm. The electric field acting on this area has a magnitude of 108 N/C at an angle of 29.3°. Calculate the electric flux through the shown surface. -0.640 N m2/C 1.14 N m2/C 1.30×104 N m2/C 1.14×104 N m2/C
A uniform electric field of magnitude E = 38.5 N/C makes and angle of θ = 67.3° relative to a line normal to a surface with area A = 3.01 m2. What is the electric flux (Φ) that passes through this surface? Your answer should be in N·m2/C
There is an electric flux of 38.5 N⋅m^2/C through the square region shown in the figure. Each side of the square has a length of 0.100 m, n^ is the unit vector normal to the plane of the surface, and the electric field is at an angle of ?=141 relative to the plane of the surface. Assuming that the field is uniform over the region shown, what is the magnitude ? of the electric field?
What is the electric flux if a 3.2 m2 surface is inclined at 65O from the vertical to an 8.5 N/C horizontal uniform electric field? \ 25 Nm2/C 20 Nm2/C 11 Nm2/C 31 Nm2/C
) In Figure below, consider a plane surface in a uniform electric field, where d = 15 cm and θ = 70o . Find the magnitude of the electric field if the net flux through the surface is 6 N. m2 /C, (b) In the same figure, find the electric flux through the plane surface if θ = 60o , E = 350 N/C, and d = 5 cm. The electric field is uniform over the entire area of the surface
A flat square with sides 18 cm is placed in a uniform electric field of magnitude 5784 N/C. What is the electric flux (in N·m2/C) through the square when its face is perpendicular to the field lines?
A circular surface with a radius of 0.062 m is exposed to a uniform electric field of magnitude 1.58 × 104 N/C. The electric flux through the surface is 68 N·m2/C. What is the angle between the direction of the electric field and the normal to the surface?
We have an electric flux of magnitude 141 Nm2/ C passing through a flat horizontal surface with an area of 0.58 m2. The flux is due to a uniform electric field, which points 14.4 degrees above the horizontal. Find the magnitude of the electric field.