All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net ${\Phi }_e=Q_{in}/{ϵ}_0$ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Consider the face parallel to the yz-plane. Define area $d\overrightarrow{A}$ as a strip of width dy and height L with the vector pointing in the 𝓍-direction. One such strip is located at position y. Use the known electric field of a wire to calculate the electric flux dΦ through this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

A spherically symmetric charge distribution produces the electric field $\overrightarrow{E}=\left(5000r^2\right)\hat{r}$ N/C, where r is in m. How much charge is inside this 40-cm-diameter spherical surface?

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net ${\Phi }_e=Q_{in}/{ϵ}_0$ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Show that the net flux through the cube is ${\Phi }_e=Q_{in}/{ϵ}_0$.

FIGURE P31.38 shows the electric field inside a cylinder of radius $R=3.0$ mm. The field strength is increasing with time as $E=1.0\times {10}^8{t}^2$ V/m, where t is in s. The electric field outside the cylinder is always zero, and the field inside the cylinder was zero for . Find an expression for the electric flux ${\Phi }_e$ through the entire cylinder as a function of time.

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net Φₑ = Qᵢₙ / ϵ₀ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Now integrate dΦ to find the total flux through this face.

(I) A uniform electric field of magnitude 6.4 x 10² N/C passes through a circle of radius 13 cm. What is the electric flux through the circle when its face is at 45° to the field lines?

The electric flux through each surface of a cube is given below. Which surfaces of the cube does the electric field run parallel to? 

Φ₁ = 100 Nm² /C      Φ₄ = 0 Nm² /C

Φ₂ = 20 Nm² /C       Φ₅ = −40 Nm² /C

Φ₃ = 0 Nm² /C        Φ₆ = −80 Nm² /C

Where does the normal vector point for a spherical shell?