A cube is located with one corner situated at the origin of...

A cube is located with one corner situated at the origin of an x, y, z coordinate system. One of the cube’s faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the y axis. The magnitude of the field is 1500 N/C. (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss’ law predicts, since there is no net charge within the cube.

A cube is located with one corner situated at the origin of an x, y, z coordinate system. One of the cube’s faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the
cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the y axis. The magnitude of the field is 1500 N/C. (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss’ law predicts, since there is no net charge within the cube.

Key Concepts

Electric Flux

Electric flux is a measure of the number of electric field lines passing through a given surface. It is defined mathematically as the surface integral of the electric field over the area, taking the dot product with the area’s outward normal vector. This concept is crucial for quantifying the effect of an electric field on surfaces and is a foundational idea in the study of electromagnetism.

Gauss’ Law

Gauss’ Law is a fundamental principle in electromagnetism which states that the total electric flux through a closed surface is proportional to the net charge enclosed within that surface. Mathematically expressed as the integral of the electric field dotted with the area vectors equaling the enclosed charge divided by the permittivity of free space, this law provides a powerful tool for analyzing electric fields, especially in cases with high symmetry.

Uniform Electric Field

A uniform electric field is one where the field strength and direction remain constant at every point in the region of interest. This characteristic simplifies the analysis of electric flux because the field can be factored out of the surface integral, allowing for straightforward calculations over geometrically simple surfaces.

Surface Orientation and Normal Vectors

The orientation of surfaces in relation to the electric field is essential for determining the electric flux. Each surface has an associated outward normal vector, which dictates the direction against which the electric field’s contribution is measured. The dot product of the electric field vector and these normal vectors determines whether there is a positive, negative, or zero flux contribution through a specific face.

Closed Surface Integration

Integration over a closed surface, such as that of a cube, involves summing the electric flux through each individual face. In the context of Gauss’ Law, this total flux is directly related to the net charge enclosed within the surface. For regions with no enclosed charge, even though individual surfaces might have nonzero flux, their algebraic sum over the entire closed surface must be zero.

Transcript

00:01 In the given problem, the cube is located such that one of its corner is at the origin.

00:13 Suppose this is y -axis, this is x -axis, and this is x -axis.

00:22 Then one of the face of this cube is lying in the x -y plane, another face is lying in.

00:35 The x -z plane and one more phase which is lying in y -z plane and then we can complete this cube here like this so this is the complete cube which is lying overall in the first octant of the space here this is the origin then electric field is along y -axis positive y -axis so this is electric field lines these are electric field lines if we consider various faces the six faces of this cube the electric field is parallel to the four faces and these four faces are we can say this one electric field is parallel to this phase the electric field is parallel to this phase the electric field is parallel to this face also and this one and this is which is lying along x y plane so along these four faces the electric field is parallel and electric field is coming out of this phase let the flux be five two and electric field is entering here in this face which is lying along x -z plane.

02:23 So the flux, electric flux, let it be 5 -1.

02:27 We know the concept, electric flux, which enters into a closed surface.

02:35 It is taken as negative.

02:37 Electric flux entering into a closed surface is taken as negative.

03:00 And that leaving the closed surface is taken as positive.

03:21 Or rather we can also say if we have to find the first part of the problem, in the first part of the problem we have to find electric flux linked through all the six faces of this cube.

03:35 So the expression for the electric flux is this is given as the dot product of electric field.

03:43 With the area vector and area vector is always a normal outward through the surface.

04:03 So if you look for 5 -1, the electric field is entering, but this ds should be coming out.

04:14 So it becomes cosine of 180 degree.

04:18 Hence it becomes minus e .d .s or sorry, eds.

04:23 And when we look for phi 2, here electric field lines are coming out.

04:30 The area vector is also coming out.

04:33 Means both of them are in the same direction.

04:35 So this is cosine of 0 degree, which becomes 1.

04:39 So this is eds means positive.

04:42 So plugging in the known values, 5 1 here becomes electric field intensity is given as 1500 newton per coulame negative multiplied by area of the face as the side as the length of the edge is given as 0 .20 meter so this is the square of 0 .20 meter hence flux this flux comes out to be 30 minus 30 300 newton meter square per coulame similarly when we find 5 2 it will also be having the same magnitude but the positive sign...

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