An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge + Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge -Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. A uranium atom has Z = 92 and 𝑅 = 0.10nm . What is the electric field strength at r = ½𝑅?

An infinite slab of charge is centered in the xy-plane. It has charge density $\rho ={\rho }_0{e}^{-\mid z\mid /z_e}$, where ρ₀ and z₀ are constants. This is a charge density that decreases exponentially as you move away from z = 0 in either the positive or negative direction. Find the electric field strength at distance z from the center of the slab.

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.

FIGURE P24.46 shows an infinitely wide conductor parallel to and distance d from an infinitely wide plane of charge with surface charge density η. What are the electric fields $\overrightarrow{E_A}$ to $\overrightarrow{E_D}$ in regions A to D?

An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m³) is a constant. Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

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.

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 ${\Phi }_e=Q_{in}/{ϵ}_0$.