An electric field given by E⃗=4.0 i-3.0(y^2+2.0) ĵ pierces a Gaussian cube of edge length 2.0

An electric field given by E⃗=4.0 i-3.0(y^2+2.0) ĵ pierces...

An electric field given by $\vec{E}=4.0 \mathrm{i}-3.0\left(y^{2}+2.0\right) \hat{\mathrm{j}} \quad$ pierces a Gaussian cube of edge length $2.0 \mathrm{~m}$ and positioned as shown in Fig. 23-7. (The magnitude $E$ is in newtons per coulomb and the position $x$ is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube?

Key Concepts

Non-uniform Field Considerations

When dealing with non-uniform electric fields, such as fields that vary with spatial coordinates, the integration must account for these variations. This means that the field’s magnitude and direction can change over the surface, and the simple product of constant field and area does not apply. Instead, the local variation must be integrated, making the calculation more complex and highlighting the importance of properly setting the limits and expressions for the integration over each face.

Orientation of Surface Normals

The orientation of surface normals is crucial when calculating electric flux since the differential area vector, by convention, points outward from the enclosed volume in a closed surface. The sign of the flux contribution from each face depends on the relative directions of the electric field and the area’s normal vector. Correctly determining these orientations for each face of a body like a cube is essential for accurately computing both individual face contributions and the net flux.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in the application of Gauss’s law to simplify the calculation of electric fields. By considering the flux through such a closed surface, one can relate the total electric flux to the charge enclosed. Although the problem may not require the use of Gauss’s law directly, the concept of a Gaussian surface underlies many procedures in electrostatics for organizing and simplifying flux calculations.

Surface Integration

Surface integration involves computing the total effect of a field over a surface by integrating the field’s component along the direction of the unit normal to the surface at each point. In the context of electric flux, this requires setting up the dot product of the electric field and the area element, then integrating over the entire surface, taking into account any variations in the field or the orientation of the surface.

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 dot product with the differential area vector. This concept allows one to quantify the influence of an electric field in a region of space by summing its effect over a specific surface.

Transcript

00:01 For this problem on the topic of kalses law, we are given the electric field of 4 i -hat minus 3 into y squared plus 2 j hat that is passing through a gaussian cube, which has an age length of 2 meters and is positioned as shown in the figure.

00:14 We want to find the electric flux through the top face, bottom face, left face, and back face, and then find the net electric flux through the cube.

00:23 Now, on the top face of the cube, we have y equal to two meters, and area element da equal to daj hat.

00:44 And so therefore, we have the electric field vector e to be 4i hat minus 3 into 2 squared plus 2, j hat, which simplifies to 4 i .hat minus 18 j hat in s i units.

01:13 And so therefore the flux to the top phase is the integral over the top face of e .da, which is the integral of 4 i .hat minus 18 jhat dot d d .m.

01:40 J hat and so this gives us minus 18 times the integral of da over the top surface.

01:51 So this is minus 18 times the area of the top surface which is 2 meters squared which gives us the flux of the top surface to be minus 72 newton meter squared per cool on the bottom face of the cube we have y equal to 0.

02:19 And we have the area element da equal to da times minus j -hat, which therefore means that the electric field vector is 4 -i -hat minus 3 into 0 squared plus 2 j -hat, which is 4 -i -hat minus 6 j -hat, newton's per k -cule.

02:53 Which means that the flux is the integral over the bottom surface of e .da, which is the integral over the surface of 4 i .hat minus 6 jhat dotted with da...

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