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=1.0×10^8t^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 t<0.a. Find an expression for the electric flux Φₑ through the entire cylinder as a function of time.

A 10 nC charge is at the center of a 2.0 m x 2.0 m x 2.0 m cube. What is the electric flux through the top surface of the cube?

The electric flux through the surface shown in FIGURE EX24.10 is 25 N m²/C . What is the electric field strength?

Find the electric fluxes ΦA to ΦE through surfaces A to E in FIGURE P24.29.

Charges q₁ = -4Q and q2 = +2Q are located at 𝓍 = -a and 𝓍 = + a, respectively. What is the net electric flux through a sphere of radius 2a centered (a) at the origin and (b) at 𝓍 = 2a?

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. (a) Consider the face parallel to the yz-plane. Define area dA (→ above 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.