Charge density in a rotating cylinder.

First, we have that the magnetic force and the electric force have the same magnitude at the equilibrium, so that the charges are static. Then: Eq = qvB E = vB = wRB Now we can use the electric flux formula: Electric field X Area = internal charge of that area / electric permissivity,

$E(2\pi Rz) = \rho(\pi(R^2)z)/e \rightarrow \rho = 2wBe = 7.97E-12$

Forces acting on electrons are: Lorentz force and centrifugal force. But, since we neglect centrifugal force, we are left with Lorentz force only, which we can express by: F1=B∙e∙v. And that creates inducted electrial field, which acts with force F2=e∙E. We can equal these two forces: F1=F2. By doing this, we obtain: B∙e∙v =e∙E, and by reducing e, we get B∙v =E. v is equal to ω∙r, so, we can write: E= B∙ω∙r. Form Gauss's Law, we have: E∙S=Q/ε, where S is the area of cylinder, Q total charge and ε electrical permitivity ( I will use the value of 8.85 ∙10^(-12) C^2/(N∙m^2). And Q=ρ∙V, where ρ represents charge density. So we have: E∙2∙r∙π∙l=(ρ∙r^2∙π∙l)/ε , and by dividing the equation with 2∙r∙π∙l, we get: E=(ρ∙r)/(2∙ε) . And finally, by equalizing E= B∙ω∙r and E=(ρ∙r)/(2∙ε) , we get: B∙ω∙r= E=(ρ∙r)/(2∙ε), and from this we get: ρ=2∙ε∙B∙ω=7.965∙10^(-12) C/m^3 ≈8∙10^(-12) C/m^3.

The magnetic force acting on a free electron is given by $\vec{F}_{m}= q \vec{v}\times \vec{B}.$ Hence, the magnetic force is oriented radially and its magnitude is $F_{m}=q \omega r B$ . Since, we neglect the centrifugal force $F_{c}=m_{e} \omega^{2} r$ (we can do so because the mass of the electron is small), the magnetic force must be equal to the electric force $F_{e}= q E(r)$ which implies $E(r)= \omega B r.$ Now, we must determine the charge distribution that generates this electric field. In order to do so, we can apply Gauss's low to a cylindrical shell of inner radius r, outer radius $r+\Delta r$ and height h. Then $2\pi (r+\Delta r)h (E+\Delta E) - 2\pi r h E = \frac{ \rho(r) 2\pi r \Delta r h}{\epsilon_{0}}$ where $\Delta E= \omega B \Delta r$ . In the limit $\Delta r \rightarrow 0$ we obtain $\rho(r)=2 \epsilon_{0} \omega B= 7.965 \times 10^{-12} C/m^{3}$ Note that the charge density is constant and it dependes on the direction of the cylinder's rotation (clockwise or counterclockwise).