An infinite cylinder of radius R has a linear charge density λ. The volume charge density (C/m³) within the cylinder (r ≤ R) is $p\left(r\right)=r{p}_0/R$, where p₀ is a constant to be determined. Use Gauss’s law to find an expression for the electric field strength E inside the cylinder, r ≤ R, in terms of λ and R.

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.

An infinite cylinder of radius R has a linear charge density λ . The volume charge density (C/m³) within the cylinder (r ≤ R ) is p (r) = rp₀ / R, where p₀ is a constant to be determined. The charge within a small volume dV is dq = pdV. The integral of pdV over a cylinder of length L is the total charge Q = λL within the cylinder. Use this fact to show that p₀ = 3λ / 2πR² Hint: Let dV be a cylindrical shell of length L, radius r, and thickness dr. What is the volume of such a shell?

A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball (r ≤ R ) is $E\left(r\right)=r^4{E}_{max}/R^4$. Find an expression for the volume charge density ρ(r) inside the ball as a function of r.

A sphere of radius R has total charge Q. The volume charge density (C/m³) within the sphere is p(r) = C/r², where C is a constant to be determined. Use Gauss’s law to find an expression for the electric field strength E inside the sphere, r ≤ R, in terms of Q and R.

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 ${\Phi }_e=Q_{in}/{ϵ}_0$.