Part A A cube has sides of length L = 0.400 m. It is placed...

Part A A cube has sides of length L = 0.400 m. It is placed with one corner at the origin as shown in the figure (Figure 1). The electric field is not uniform but is given by \(\vec{E} = (-4.11 \text{ N/(C\cdot m)})\hat{z}i + (3.30 \text{ N/(C\cdot m)})\hat{z}k\). Find the electric flux through each of the six cube faces S1, S2, S3, S4, S5, and S6. Enter your answers numerically separated by commas. Figure S1 (left side) L S5 (front) S2 (top) S6 (back) \(\Phi_1, \Phi_2, \Phi_3, \Phi_4, \Phi_5, \Phi_6 = \) (N/C)\cdot m^2 Submit Previous Answers Request Answer Incorrect; Try Again; 8 attempts remaining Part B Find the total electric charge inside the cube. 1 of 1 q= C Submit Previous Answers Request Answer Incorrect; Try Again; 2 attempts remaining S3 (right side) < Return to Assignment Provide Feedback S4 (bottom)

Transcript

00:01 Hi, here in this given problem there is a cube in cartesian space.

00:08 So to show the cube, first of all we draw coordinate axis, the vertical is z -axis, horizontal this is y -axis in the plane of the paper and coming out of the plane of paper perpendicularly, this is x -axis.

00:30 Now, we draw the cube here.

00:38 First of all, this is the face of the cube in zy plane, then this is the face of the cube parallel to zy plane, then the face in x, z, plane, and a face parallel to x, z plane.

01:04 These faces are named as, first of all, this is s1, the face in x -z plane, s -2 is the topmost plane, s -3 is the rightmost plane, means face, this is s -3, s -4, that is the bottom -most plane, is the bottom most face means face in x y plane this is s 4 s 5 that is parallel to y z plane and then s 6 this face is in y z plane side of the cube that is given as l is equal to 0 .370 meter.

02:23 Electric field in vector form that is given as minus 5 .38 newton per column per meter into xi cap means this electric field is along x -axis means along negative x -axis because of this minus and it is varying with the x coordinate.

02:53 Means this component of the electric field will be 0 in yz plane.

02:59 Means in the face s6, this first component of the electric field will be 0.

03:06 And then here it is the second component plus 2 .23 newton per coulum per meter z.

03:19 K cap.

03:22 It is along z -axis and it is varying with the z coordinate.

03:29 So its value will be 0 in x -y plane means for the plane s -4 it will be 0.

03:37 So we can say electric field along x -axis is minus 5 .38 x culum sorry this is newton per coulame means electric field along y axis 0 there is no component along y axis and electric field along x axis that is 2 .23 z newton per coulum meter in the first part of the problem we have to find electric flux linked with each face of this cube for which first of all we find area of each face of the cube of the cube cube and that will be a is equal to square of the side.

04:30 This is 0 .370 meter square.

04:35 Means this is 0 .137 meter square.

04:41 Now as the face is s1 and s3, these are parallel to ex and ez, so no flux will be linked through them.

04:56 So, electric flux through s1 and s3 is 0 as ex and ez.

05:16 Both the components are parallel to these faces.

05:29 Now we find electric flux passing through surface s 2 and that will be equal to electric field e z into area into angle cosine of 0 degree angle between electric field vector and area vector so this is 2 .23 into z and for z that will be 0 .370 into area 0 .137 and for cos 0 .137 and for cos 0 .0 degree it will be 1.

06:06 So, phi 2 comes out to be equal to 0 .1 ,13 newton meter square per coulum.

06:16 Then we find flux linked through surface s4 and that is phi 4.

06:27 It will be ez again, a again, but cosine of 180 degree.

06:38 And here, if you look into the figure, this is the surface s4 and for this surface, ez will become zero because for x, y, plane, the value of z is zero...

Question

Please give Ace some feedback