An electric field given by E = 4 xi- 7yj pierces through a cubic Gaussian surface of edge length 2.0 m and positioned as shown in the figure (E is in N/C and the positions x, y are in meters). Calculate the electric flux of through the top face (in N.m2/C) Gaussian surface

An electric field given by E = 4 xi- 7yj pierces through a cubic Gaussian surface of edge length 2.0 m and positioned as shown in the figure (E is in N/C and the positions x, y are in meters). Calculate the electric flux of through the top face (in N.m2/C) Gaussian surface

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A closed surface with dimensions a=b%3D0.40 m and c30.60 m is located as in the figure below. The left edge of the closed surface is located at position x= a. The electric field in the region is non-uniform and is given by E=(3.0+ 46x² ) i N/C, where x is in meters. Calculate the net electric flux leaving the closed surface? x= a Select one: OA 15.38 OB. 4.42 OC 6.18 OD. 147 OE 10.78 9:03 PM 4/30/2021
please help me
An electric field given by E = 7.80 i - 7.40(y² + 8.00) j pierces the Gaussian cube of edge length 58.0 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 Nm²/C through the cube? Gaussian surface
A uniform horizontal electric field of E = (136 Î +71 Ĵ) N/C passes through an inclined plane that is 52 cm wide and 88 cm long. If the inclined plane makes an angle of 45° with the horizontal, determine the electric flux in the plane in N.m^2/C.
Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed surface. y E L L We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux: Фnet Ф1 + Ф, +Ф; + Ф. ху + 0xz+@yz The flux through surface 1 is: 01=0. The flux through surface 2 is: 02= The flux through surface 3 is: 03 = The flux through the surface on the xy plane is: Oxy = The flux through the surface on the xz plane is: Oxz= The flux through the surface on the yz plane is: O yz Taking the algebraic sum, the net electric flux through the cube is: Onet = 1111
A very long wire carrying charge density λ is surrounded by a very long hollow cylindrical layer with an inner radius a outer radius b on the same axis as the wire. The volume charge density in the cylinder layer is p. a) Obtain qic in terms of charge density and r distance, which you will use in the equation you will create with Gauss's law to find an electric field at a point perpendicular to the wire and at a distance r, inside the cylinder layer. b) Find the size of the electric field generated at the point mentioned in part a, where λ = 3x10 ^ -9 C / m p = 9x10 ^ -9 C / m ^ 3 a = 0.3 m b = 0.7 m r = 0.5 m.
Find the electric field due to infinitely long cylindrical volume charge density (p,) and a cylindrical surface charge density (p,) for p
A disk of radius R = 3 m, lies in a region of uniform electric field E = 100 N / C. The electric field passes through the plane of the disk at an angle of 30. (a) Determine the electric flux through the disk.
The rectangular surface in the figure below is in a uniform electricfield region with strength of 45.4 N/C. If the surface has a dimensionof 5.00 m × 9.00 m, and the electric flux through the surface is 1040N · m2/C^2, the angle is?
Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed surface. E L L We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux: Фnet — Ф1 + Ф2 + Фз + Фху + Фxz + Фуz The flux through surface 1 is: 01 =0. The flux through surface 2 is: 02= The flux through surface 3 is: 3 The flux through the surface on the xy plane is: The flux through the surface on the xz plane is: 0 The flux through the surface on the yz plane is: Oyz Taking the algebraic sum, the net electric flux through the cube is: Pnet
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 90.0 N>C that is directed at 20° from the plane of the sheet. Find the magnitude of the electric flux through the sheet.
A right conical frustrum is placed in a uniform electric field that is parallel to the y axis, as shown. The magnitude of the electric field is 43 N/C. The electric flux measured through the top surface is 2240 N-m?/C, and the electric flux measured through the bottom surface is 4480 N-m²/C. What is the electric flux through the side of the frustrum?