Question

10. A cube has one corner at the origin and the opposite corner at the point (2, 2, 2). The dimensions provided are in meters. The sides of the cube are parallel to the coordinate planes as shown in the figure below. The electric field in and around the cube depends on the coordinates x and y and is given by E--5ri + 2yj where x and y are in meters. What is the value of the net charge enclosed by the cube? a.-246 b.-16є0 c. 0 d. 160 e. 240 (2,2,2) (0,0,0)

The correct answer is A but the answer key does not show how to get the answer do it doesn't help me. I don't know how to do this problem. How does an E vector without a k component pass through a cube? I would think i and j only means it runs along the back surface? Please show how to do problem and sketch would be helpful. Thanks

Answer 1

E = -5 times j + 2yj + 0k we know: Differential form of Gauss law rightsinglearrow nabla rightsinglearrow E = rho/epsilon_0 [rho = volume charge density] (partial differential/ partial differential x i + partial differential/partial differential y j + partial differential/partial differential z k) (- 5xj + 2yj) rho/epsilon_0 rightdoublearrow partial differential/partial differential x (- 5x) + partial differential/partial differential y (2y) = rho/epsilon_0 - 5 + 2 = rho/epsilon_0 rho - 3 epsilon_0 q/volume of cube = rho = -3 epsilon_0 q/a^3 = -3 epsilon_0 q = -3 a^3 epsilon_0 side of cube (from figure) a = 2 therefore q = -3(2)^3 epsilon_0 = -3 times 8 epsilon_0 q = -24 epsilon_0