Figure 23-34 shows a closed Gaussian surface in the shape of...

Figure 23-34 shows a closed Gaussian surface in the shape of a cube of edge length $2.00 \mathrm{~m}$. It lies in a region where the nonuniform electric field is given by $\vec{E}=(3.00 x+$ 4.00 $) \hat{\mathrm{i}}+6.00 \hat{\mathrm{j}}+7.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$, with $x$ in meters What is the net charge contained by the cube?

Key Concepts

Divergence Theorem

The divergence theorem, also known as Gauss's divergence theorem, connects the flux of a vector field through a closed surface to the volume integral of the divergence of that field. It serves as the mathematical foundation for Gauss's Law in differential form and is essential for calculating the net charge within a volume by integrating the divergence of the electric field over that volume.

Divergence of a Vector Field

The divergence of a vector field is a scalar quantity that measures the net rate at which the field 'spreads out' from a point. In the context of electromagnetism, the divergence of the electric field at any given point is directly proportional to the local charge density, as expressed in the differential form of Gauss’s Law.

Electric Field

The electric field is a vector field that represents the force per unit charge at each point in space. In problems with nonuniform fields, the field strength or direction varies with position, which can significantly affect the flux through surfaces.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the net electric flux through a closed surface to the total charge enclosed by that surface. It provides a powerful method to calculate the net charge or the field in highly symmetric charge distributions.

Transcript

00:01 Hi, here in this given problem first of all we draw a cartesian space having y axis horizontal towards right, z axis upward and x axis coming out of the plane of paper.

00:25 In this cartesian space there is a cube, these are the faces of this cube and it is starting from the origin but lying along negative x axis, it is clear from the figure like this.

01:00 Electric field in this region that is variable with the x axis but constant along y axis and z axis and that is 3 .00 x plus 4 .00 i cap plus 6 .00 j cap plus 7 .00 k cap newton per coulomb side of the cube that is a is equal to 2 .00 meter.

01:53 Now electric field along y and z axis is constant, it is clear from this expression that electric field is constant means uniform along y and z but it is varying along x axis.

02:31 Area of each face of the cube a that is square of side means square of 2 means this is 4 meter square.

02:48 Now as electric field is uniform along y axis, so whatever the field will be entering to the left face same will be leaving through the right face and whatever the electric field will be entering through the bottom of the cube it will be leaving through the top surface of the cube.

03:21 So, we can say electric flux which is given by the expression product of electric field with the area entering through the left face will be equal to that leaving through the right face and that entering through the bottom face that will be equal to leaving through the top face.

04:38 So, no net electric flux along y and z axis but along x axis the flux entering phi 1 as electric flux which enters into the face it is taken as negative...

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